22.19 - Miscellaneous Examples on Applications of Derivatives - Part 1

In the previous section, we completed a discussion on the various applications of derivatives. In this section, we will see some miscellaneous examples.

Solved example 22.65
A car starts from a point P at time t = 0 seconds and stops at point Q. The distance x, in meters, covered by it, in t seconds is given by $\rm{x \,=\,t^2 \left(2 - \frac{t}{3} \right) }$.
Find the time taken by it to reach Q and also find the distance between P and Q.
Solution:
1. Consider the given equation: $\rm{x \,=\,t^2 \left(2 - \frac{t}{3} \right) \,=\,2t^2 - \frac{t^3}{3}}$
• This equation gives the total distance traveled when the reading in the stop-watch is 't'.
2. We know that for uniform motion, velocity = distance/time 
• So for non-uniform motion, the instantaneous velocity at any instant 't' is given by dx/dt. Thus we get:
$\rm{v\,=\,\frac{dx}{dt}\,=\,4t - t^2}$
3. When the car reaches Q, it stops. So it's velocity at Q is zero.
• Equating the above equation to zero, we get:
4t - t² = 0
⇒ t(4 - t) = 0
⇒ t = 0 and t = 4
• t = 0 is the instant at which the car is at P.
• t = 4 is the instant at which the car is at Q.
• So we can write:
The car takes 4 seconds to reach Q.
4. Using the given equation, we can write:
Distance traveled in the 4 seconds from P to Q =
$\rm{2(4)^2 - \frac{(4)^3}{3}\,=\,\frac{96 - 64}{3}\,=\,\frac{32}{3}~\text{m}}$

Solved example 22.66
A water tank has the shape of an inverted right circular cone with it's axis vertical and vertex lowermost. It's semi-vertical angle is tan⁻¹ (0.5). Water is poured into it at a constant rate of 5 cubic meter per hour. Find the rate at which the level of water is rising at the instant when the depth of water in the tank is 4 m.
Solution:
1. In the fig.22.68 below,
• The water tank is drawn in red color. It is an inverted cone. The center of the circular base is O.
• The green dotted circle indicates the top surface of water.
    ♦ The center of this top surface of water, is C.
    ♦ Radius of this top surface of water, is r.
• The volume of water in the tank forms an inner inverted cone. Both cones have the same axis and the same vertex A.

Fig.22.68

• When the falling water forms the inner cone, three items will change continuously. They are:
    ♦ Volume of water (volume of inner cone) V
    ♦ Height of water (height of inner cone) h
    ♦ Radius of top surface of water (radius of the base of inner cone) r.
2. We have the formula for volume of a cone:
$\rm{V\,=\,\frac{1}{3} \pi r^2 h}$
• To eliminate r, we will use the semi-vertical angle.
Semi-vertical angle = ∠OAP = ∠CAB = tan⁻¹ (0.5)
• So we can write:
$\rm{\tan (\angle CAB) \,=\,\frac{BC}{AC}\,=\,\frac{r}{h}}$
⇒ $\rm{\tan (\tan^{-1} (0.5)) \,=\,\frac{r}{h}}$
⇒ $\rm{0.5 \,=\,\frac{r}{h}}$
⇒ $\rm{r \,=\,\frac{h}{2}}$
• Now the expression for volume becomes:
$\rm{V\,=\,\frac{1}{3} \pi \left(\frac{h}{2} \right)^2 h}$
⇒ $\rm{V\,=\,\frac{1}{3} \pi \left(\frac{h^3}{4} \right)}$
(Note that, using the semi-vertical angle, we can eliminate either h or r. We must not eliminate h because, we want to find the rate of change of h)
3. Given that, water is being poured at a constant rate of 5 cubic meter per hour. So volume at any time 't' will be 5t, where 't' is in hours.
4. Equating the results in (2) and (3), we get:
$\rm{5t\,=\,\frac{1}{3} \pi \left(\frac{h^3}{4} \right)}$
⇒ $\rm{60t\,=\, \pi h^3}$
⇒ $\rm{h\,=\,\left(\frac{60t}{\pi}\right)^{1/3}}$
5. From this we get:
$\rm{\frac{dh}{dt}\,=\,\left(\frac{60}{\pi}\right)^{1/3} \left(\frac{1}{3} \right)t^{-2/3}}$
• This result will give the rate of change of h w.r.t time at any time 't'.
6. We want the rate at the time when the depth of water (height of inner cone) is 4 m. So we need to find that time.
• Using the result in (4), we can write:
$\rm{h\,=\,\left(\frac{60t}{\pi}\right)^{1/3}}$
⇒ $\rm{4\,=\,\left(\frac{60t}{\pi}\right)^{1/3}}$
⇒ $\rm{4^3 \,=\,\frac{60t}{\pi}}$
⇒ $\rm{64 \pi \,=\,60t}$
⇒ $\rm{t\,=\,\frac{64 \pi}{60}\,=\,\frac{16 \pi}{15}}$
7. Now we substitute this result in (5). We get:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dh}{dt}}    & {~=~}    &{\left(\frac{60}{\pi}\right)^{1/3} \left(\frac{1}{3} \right)t^{-2/3}}    \\
{~\color{magenta}    2    }    &{{\implies}}    &{{\left.\frac{dh}{dt} \right|_{h=4}}}    & {~=~}    &{\left(\frac{60}{\pi}\right)^{1/3} \left(\frac{1}{3} \right)\left(\frac{16 \pi}{15} \right)^{-2/3}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\left(\frac{(15)(4)}{\pi}\right)^{1/3} \left(\frac{1}{3} \right)\left(\frac{15}{16 \pi} \right)^{2/3}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\left(\frac{(15)(4)}{\pi}\right)^{1/3} \left(\frac{1}{3} \right)\left(\frac{(15)(15)}{(16 \pi)(16 \pi)} \right)^{1/3}}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\left(\frac{(15)(4)}{\pi}\right)^{1/3} \left(\frac{1}{3} \right)\left(\frac{(15)(15)}{4^3 (4) (\pi)(\pi)} \right)^{1/3}}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\left(\frac{(15)}{\pi}\right)^{1/3} \left(\frac{1}{3} \right)\left(\frac{(15)(15)}{4^3 (\pi)(\pi)} \right)^{1/3}}    \\
{~\color{magenta}    7    }    &{{}}    &{{}}    & {~=~}    &{\frac{15}{4} \left(\frac{1}{\pi}\right)^{1/3} \left(\frac{1}{3} \right)\left(\frac{1}{(\pi)(\pi)} \right)^{1/3}}    \\
{~\color{magenta}    8    }    &{{}}    &{{}}    & {~=~}    &{\frac{5}{4} \left(\frac{1}{\pi}\right)^{1/3} \left(\frac{1}{(\pi)(\pi)} \right)^{1/3}}    \\
{~\color{magenta}    9    }    &{{}}    &{{}}    & {~=~}    &{\frac{5}{4 \pi}~=~\frac{5}{4(22/7)}~=~\frac{35}{88}~\text{m/h}}    \\
\end{array}$

Solved example 22.67
A man of height 2 meters walks at a uniform speed of 5 km/h away from a lamp post which is 6 meters high. Find the rate at which the length of his shadow increases.
Solution:
1. In the fig.22.69(a) below,
• AB is the lamp post. A is the position of the lamp. A ray of light from the lamp is shown as a yellow dotted line.

Fig.22.69

• CD is the initial position of the man just before he starts to walk. So BD is the initial distance of the man from the lamp post. It is denoted as d₀.
• The ray of light passes just above C and creates the shadow of C at E. So DE is the initial length of the shadow. It is denoted as l₀.
2. Let us assume that, in t hours, the man reaches the position C'D'. This is shown in fig.b.
• Then the distance DD' = 5t
• Also, length of shadow at that time will be D'E'. This length is denoted as l.
3. Consider the two triangles ABE' and C'D'E' in fig.b. These two triangles are similar. So we get:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{AB}{BE'}}    & {~=~}    &{\frac{C'D'}{D'E'}}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{6}{BD + DD' + D'E'}}    & {~=~}    &{\frac{2}{l}}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{6}{d_0 + 5t + l}}    & {~=~}    &{\frac{2}{l}}    \\
{~\color{magenta}    4    }    &{\implies}    &{\frac{d_0 + 5t + l}{6}}    & {~=~}    &{\frac{l}{2}}    \\
{~\color{magenta}    5    }    &{\implies}    &{\frac{d_0 + 5t}{6}~+~\frac{l}{6}}    & {~=~}    &{\frac{l}{2}}    \\
{~\color{magenta}    6    }    &{\implies}    &{\frac{d_0 + 5t}{6}}    & {~=~}    &{\frac{l}{3}}    \\
{~\color{magenta}    7    }    &{\implies}    &{\frac{d_0 + 5t}{2}}    & {~=~}    &{l}    \\
{~\color{magenta}    8    }    &{\implies}    &{l}    & {~=~}    &{\frac{d_0}{2}~+~\frac{5t}{2}}    \\
{~\color{magenta}    9    }    &{\implies}    &{\frac{dl}{dt}}    & {~=~}    &{\frac{5}{2}~\text{km/h}}    \\
\end{array}$

Solved example 22.68
The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?
Solution:
1. In the fig.22.70 (a) below,
• ABC is an isosceles triangle. The base AB is fixed. Sides AC and BC have an initial length of l₀ cm.

Fig.22.70

2. The sides are decreasing at the rate of 3 cm/s. So at the instant when the stop-watch reading is t, the length of both sides will be (l₀ − 3t). This is shown in fig.b
3. In fig.b, the altitude (dashed yellow line) will have a length equal to $\rm{\sqrt{(l_0 - 3t)^2 ~-~(b/2)^2}}$
• This is the altitude at any instant t.
• So the area of the triangle at any instant t will be given by: $\rm{A = (1/2) (b) \sqrt{(l_0 - 3t)^2 ~-~(b/2)^2}}$
4. Now the rate of change of area can be obtained as:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{A}    & {~=~}    &{(1/2) (b) \sqrt{(l_0 - 3t)^2 ~-~(b/2)^2}}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{dA}{dt}}    & {~=~}    &{(1/2) (b) (1/2)((l_0 - 3t)^2 ~-~(b/2)^2)^{-1/2} 2 (l_0 – 3t)(-3)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{(-3/2) (b)((l_0 - 3t)^2 ~-~(b/2)^2)^{-1/2} (l_0 – 3t)}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{-3 b (l_0 – 3t)}{2 \sqrt{(l_0 - 3t)^2 ~-~(b/2)^2}}}    \\
\end{array}$

• This result gives the rate of change at any instant t.

5. So we need the instant at which the sides become equal to base. For that, we use the result in (2):
Length of side at any instant = l₀ − 3t
• So at the required instant, l₀ − 3t = b
6. The above result can be used directly because (l₀ − 3t) is available in the rate. We get:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dA}{dt}}    & {~=~}    &{\frac{-3 b (l_0 – 3t)}{2 \sqrt{(l_0 - 3t)^2 ~-~(b/2)^2}}}    \\
{~\color{magenta}    2    }    &{\implies}    &{\left. \frac{dA}{dt} \right|_b}    & {~=~}    &{\frac{-3 b (b)}{2 \sqrt{(b)^2 ~-~(b/2)^2}}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{-3 b^2}{2 \sqrt{(3b^2)/4}}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{-3 b^2}{\sqrt{3b^2}}}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{-b \sqrt{3}~~\text{square  cm per sec}}    \\
\end{array}$                            

Solved example 22.69
A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic meter per hour. Then the depth of wheat is increasing at the rate of
(A) 1 m³/h    (B) 0.1 m³/h    (C) 1.1 m³/h    (D) 0.5 m³/h
Solution:
1. Volume V of wheat in the cylinder, at t hours will be given by:
V = πr²h where r is the radius of the cylinder and h is the depth of wheat at t hours.
2. The above volume is equal to 314 t.
3. Equating the two results, we get: πr²h = 314 t
⇒ 3.14 r²h = 314 t
⇒ r²h = 100 t
⇒ 10²h = 100 t
⇒ h = t
⇒ dh/dt = 1

• So the correct option is (A).

---

In the next section, we will see a few more miscellaneous examples.

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22.18 - Solved Examples on Second Derivative Test

In the previous section, we completed a discussion on second derivative test. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved example 22.60
Find the shortest distance of the point (0,c) from the parabola y = x², where 0 ≤ c ≤ 5.
Solution:
Step I: Writing the problem as a function
1. Let (h,k) be any point on the parabola.
• Let D be the distance between (0,c) and (h,k).
• Then we can write:
$\rm{D = \sqrt{(h-0)^2+(k-c)^2} = \sqrt{h^2+(k-c)^2}}$
2. (h,k) is a point on the parabola y = x². So we can write: k = h²
• Substituting this in (1), we get:
$\rm{D = \sqrt{k+(k-c)^2}}$
3. (0,c) is a constant point on the y-axis. It's distance D from (h,k) will depend on the position of (h,k).
• We want such a position for (h,k) that, D is minimum.
4. From (2), it is clear that, D depends on k. So D is a function of k.
We can write: $\rm{D = f(k) = \sqrt{k+(k-c)^2}}$
5. We want D to have the minimum possible value. That means, we want the least of all the local minima of f. That means, we need to find all local minima and compare them.

Step II: Finding the critical points
1. Write the first two derivatives:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(k)}    & {~=~}    &{\sqrt{k+(k-c)^2}}    \\
{~\color{magenta}    2    }    &{\implies}    &{f'(k)}    & {~=~}    &{(1/2)[k+(k-c)^2]^{(-1/2)} [1+2(k-c)]}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{1+2(k-c)}{2 \sqrt{k+(k-c)^2}}}    \\
\end{array}$                           

• After examining f '(k), we know that, calculation of f ''(k) will be a lengthy process. So we will use the first derivative test for this problem.
2. Equate the first derivative to zero and solve for k:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{1+2(k-c)}{2 \sqrt{k+(k-c)^2}}}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{\implies}    &{1 + 2(k-c)}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{\implies}    &{k-c}    & {~=~}    &{-(1/2)}    \\
{~\color{magenta}    4    }    &{\implies}    &{k}    & {~=~}    &{c-(1/2)}    \\
{~\color{magenta}    5    }    &{\implies}    &{k}    & {~=~}    &{\frac{2c - 1}{2}}    \\
\end{array}$                           
3. So the only one point in category I is: k = c − (1/2)
4. We obtained $\rm{f'(k) = \frac{1+2(k-c)}{2 \sqrt{k+(k-c)^2}}}$
• k will be always +ve because, it is the square of h. So f '(k) is defined for all real numbers. That means, there is no input k at which f '(k) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: k = c − (1/2)

Step III: Dividing the number line
• The critical point is x = 1. Therefore, the number line can be divided into two intervals:
(−∞, c − (1/2)) and (c − (1/2),∞)

Step IV: Finding the sign of f '(x) in each interval
1. First interval:
When k is from the first interval (−∞, c − (1/2)), we have:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{k}    & {~<~}    &{c-(1/2)}    \\
{~\color{magenta}    2    }    &{\implies}    &{k}    & {~<~}    &{\frac{2c - 1}{2}}    \\
{~\color{magenta}    3    }    &{\implies}    &{2k}    & {~<~}    &{2c - 1}    \\
{~\color{magenta}    4    }    &{\implies}    &{2k - 2c +1}    & {~<~}    &{0}    \\
{~\color{magenta}    5    }    &{\implies}    &{2(k - c) +1}    & {~<~}    &{0}    \\
\end{array}$
• We obtained: 2(k−c) + 1 < 0
That means, the numerator of f '(k) is −ve
• The denominator of f '(k) is +ve because, it is twice the distance D. Distance is always +ve.
• So we get:
In the first interval, f '(k) is −ve

2. Second interval:
When k is from the second interval (c − (1/2),∞), we have:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{k}    &{~>~}    &{c-(1/2)}    \\
{~\color{magenta}    2    }    &{\implies}    &{k}    & {~>~}    &{\frac{2c - 1}{2}}    \\
{~\color{magenta}    3    }    &{\implies}    &{2k}    & {~>~}    &{2c - 1}    \\
{~\color{magenta}    4    }    &{\implies}    &{2k - 2c +1}    & {~>~}    &{0}    \\
{~\color{magenta}    5    }    &{\implies}    &{2(k - c) +1}    & {~>~}    &{0}    \\
\end{array}$                           
• We obtained: 2(k−c) + 1 > 0
That means, the numerator of f '(k) is +ve
• The denominator of f '(k) is +ve because, it is twice the distance D. Distance is always +ve.
• So we get:
In the second interval, f '(k) is +ve

Step V: Analyzing the change of signs at critical point
• At the critical point, the sign of f '(k) changes from −ve to +ve. So this critical point is a point of local minimum.
• That means, if we put k = c −(1/2), f(k) will be minimum.
• That means, if we put k = c −(1/2), D will be minimum.

Step VI: Final result
• We have: $\rm{D = f(k) = \sqrt{k+(k-c)^2}}$
• So minimum distance can be obtained as:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(c-(1/2))}    & {~=~}    &{\sqrt{c-(1/2)+[c-(1/2)-c]^2}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\sqrt{c-(1/2)+[-(1/2)]^2}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\sqrt{c-\frac{1}{2}+\frac{1}{4}}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\sqrt{c-\frac{1}{4}}}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\sqrt{\frac{4c - 1}{4}}}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\frac{\sqrt{4c - 1}}{2}}    \\
\end{array}$

Check: This can be written in 4 steps.
1. Assume that, c = 4.5. Then the constant point on the y-axis is (0,4.5). It is marked as A in the fig.22.60 below:

Fig.22.60

2. Based on the above solution, we can write:
k = c − (1/2) = 4.5 − (1/2) = 4
• So h = √4 = 2
• So (h,k) is (2,4). This is marked as B in the fig.22.60 above.

3. We must prove that, the distance AB is the minimum possible distance.
• To prove that, we draw a tangent at B (green line). Then we draw AB (yellow dashed line).
• We see that, AB is perpendicular to the tangent. So AB is the minimum possible distance from A to any point on the parabola.

4. We can draw a line from A to any other point B' of the positive side of the parabola. Then AB' will not be perpendicular to the tangent at B'

Solved example 22.61
Let AP and BQ be two vertical poles at points A and B respectively. If AP = 16 m, BQ = 22 m and AB = 20 m, find the distance of a point R on AB from the point A such that RP² +RQ² is minimum.
Solution:
Step I: Writing the problem as a function
1. We want the distance of R from A. Let this distance be x. It is marked in fig.22.61 below:

Fig.22.61

2. Applying Pythagoras theorem,we get:
• RP² = AR² + AP²
• RQ² = RB² + BQ²
⇒ RP² + RQ² = AR² + AP² + RB² + BQ²
= x² + 16² + (20-x)² + 22²
= x² + 16² + 20² − 40x + x² + 22²
= 2x²  − 40x + 16² + 20²+ 22²
3. We want the above sum to be a minimum.
4. From (2), it is clear that, the sum S depends on x. So S is a function of x.
We can write: S = f(x) = 2x²  − 40x + 16² + 20²+ 22²
5. We want S to have the minimum possible value. That means, we want the least of all the local minima of f. That means, we need to find all local minima and compare them.

Step II: Finding the critical points
1. Write the first two derivatives:
• f '(x) = 4x − 40
• f ''(x) = 4
2. Equate the first derivative to zero and solve for x:
4x −40 = 0
⇒ 4x = 40
⇒ x = 10
3. So the only one point in category I is: x = 10
4. We obtained f '(x) = 4x −40
• This function is defined for all real numbers. That means, there is no input x at which f '(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: x = 10

Step III: Find f '' at the critical point
• f ''(10) =  4. This is because, f ''(x) is a constant 

Step IV: Applying the second derivative test
• Since f ''(10) is +ve, there is a local minimum at x = 10
• Since there is only one critical point, there are no other points to compare. We can write:
The minimum point of the function occurs at x = 10

Step V: Finding the actual minimum value:
• The local minimum value at x = 10 is given by:
f(10) = 2(10)²  − 40(10) + 16² + 20²+ 22² = 940

Step VI: Final result:
• The distance AR should be 10 m. Then we will get the minimum sum of squares.
• The minimum sum of squares = 940

Check:
• Let us put x = 8.
• Then S = f(8) = 2(8)²  − 40(8) + 16² + 20²+ 22² = 948
• So, even if we decrease the distance AR, the sum of squares will increase.

Step VI (optional): Drawing the graph

• The graph is shown in fig.22.62 below:

Fig.22.62

   ♦ f is drawn in red color.
   ♦ (10,940) is a local minimum.
   ♦ So when x = 10, the function will have the minimum value, which is 940

Solved example 22.62
If length of three sides of a trapezium other than the base are equal to 10 cm, then find the area of the trapezium when it is maximum.
Solution:
Step I: Writing the problem as a function
1. Fig.22.63 below is drawn based on the given data.

Fig.22.63

• The trapezium will have two identical triangles on the sides. The base of the triangles is the unknown quantity. It is marked as x.
• Note that, as x increases, the area of the trapezium will increase. But x cannot increase indefinitely because, the side 10 cm is a constant. If x go on increasing, the height of the trapezium will have to be reduced. This will decrease the area. So we must find the optimum value of x.
2. Area A of the trapezium is given by:
$\rm{A = \text{Average length of parallel sides × height} = \frac{10+(10+2x)}{2}  × PD}$
3. Applying Pythagoras theorem, we have:
PD = QC = √(10² − x2)
4. So from (2), we get:
$\rm{A = \frac{10+(10+2x)}{2}  × \sqrt{10^2 - x^2} = (10+x)  × \sqrt{10^2 - x^2}}$
5. We want the above area to be maximum.
6. From (4), it is clear that, A depends on x. So A is a function of x.
• We can write: $\rm{A = f(x) = (10+x)  × \sqrt{10^2 - x^2}}$
7. We want A to have the maximum possible value. That means, we want the largest of all the local maxima of f. That means, we need to find all local maxima and compare them.

Step II: Finding the critical points
1. Write the first two derivatives:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(x)}    & {~=~}    &{ (10+x)  × \sqrt{10^2 - x^2}}    \\
{~\color{magenta}    2    }    &{\implies}    &{f'(x)}    & {~=~}    &{(10+x) × \frac{1}{2 \sqrt{10^2 - x^2}} (-2x)~+~\sqrt{10^2 - x^2}(1)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{-2x(10+x)}{2 \sqrt{10^2 - x^2}}~+~\sqrt{10^2 - x^2}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{-x(10+x)}{\sqrt{10^2 - x^2}}~+~\sqrt{10^2 - x^2}}    \\
\end{array}$                           

• After examining f '(x), we know that, calculation of f ''(x) will be a lengthy process. So we will use the first derivative test.

2. Equate the first derivative to zero and solve for x:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{-x(10+x)}{\sqrt{10^2 - x^2}}~+~\sqrt{10^2 - x^2}}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{\implies}    &{-x(10+x)~+~10^2 - x^2}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{\implies}    &{-10 x - x^2~+~10^2 - x^2}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{\implies}    &{-10 x -2 x^2 +100}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{\implies}    &{2 x^2 + 10 x - 100}    & {~=~}    &{0}    \\
{~\color{magenta}    6    }    &{\implies}    &{x^2 + 5 x - 50}    & {~=~}    &{0}    \\
\end{array}$                           

• Solving the quadratic equation, we get:
x = 5 and x = −10
3. So the only two points in category I are: x = 5 and x = −10
4. We obtained $\rm{f'(x) = \frac{-x(10+x)}{\sqrt{10^2 - x^2}}~+~\sqrt{10^2 - x^2}}$
• This function is not defined at x = 10.
• Therefore, there is one point in category II, which is x = 10
5. So the three critical points are:
x = 5, x = −10 and x = 10

Step III: Finding the f values at the critical points
Normally, in the first derivative test, we divide the number line into various intervals and then find the sign of f ' in each interval. But in this problem, since there are only three critical points, it is easier to find the actual f values and then to compare them.
1. $\rm{f(5) = (10+5)  × \sqrt{10^2 - 5^2} = 15 \sqrt{75}}$
2. $\rm{f(-10) = (10+(-10))  × \sqrt{10^2 - (-10)^2} = 0}$
3. $\rm{f(10) = (10+10)  × \sqrt{10^2 - 10^2} = 0}$

Step IV: Comparing the f values:
We see that, f is maximum at x = 5 and the maximum area is 15√75 cm².

Step VI (optional): Drawing the graph

• The graph is shown in fig.22.64 below:

Fig.22.64

   ♦ f is drawn in red color.
   ♦ (5,129.9) is a local maximum.
   ♦ So when x = 5, the function will have the minimum value, which is 15√75 = 129.9 cm²
   ♦ f is not defined in the intervals (−∞, −10) and (10,∞).
        ✰ This is because, the square root in f will then give complex numbers.

Solved example 22.63
Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.
Solution:
Step I: Writing the problem as a function
1. Fig.22.65 below is drawn based on the given data.
 

Fig.22.65

    ♦ OE = x = radius of the cylinder
    ♦ OC = r = radius of the cone
    ♦ OA = h = height of the cone
• The triangles AOC and QEC are similar triangles.
So we can write:
$\rm{\frac{QE}{OA} = \frac{EC}{OC}}$
⇒ $\rm{\frac{QE}{h} = \frac{r-x}{r}}$
⇒ $\rm{QE = \frac{h(r-x)}{r}}$
• Curved surface area of the cylinder, S = Perimeter × height = 2π × OE × QE
⇒ $\rm{S = 2 \pi \times x \times \frac{h(r-x)}{r}}$
⇒ $\rm{S = \frac{2 \pi x h(r-x)}{r}}$
2. We want the above area to be maximum.
3. From (1), it is clear that, S depends on x. So S is a function of x.
• We can write: $\rm{S = f(x) = \frac{2 \pi x h(r-x)}{r}}$
7. We want S to have the maximum possible value. That means, we want the largest of all the local maxima of f. That means, we need to find all local maxima and compare them.

Step II: Finding the critical points
1. Write the first two derivatives:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(x)}    & {~=~}    &{\frac{2 \pi x h(r-x)}{r}}    \\
{~\color{magenta}    2    }    &{\implies}    &{f'(x)}    & {~=~}    &{\frac{2 \pi h}{r} . \frac{d}{dx} [x(r-x)]}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{2 \pi h}{r} .[x(-1) ~+~(r-x)]}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{2 \pi h}{r} .[r – 2x]}    \\
{~\color{magenta}    5    }    &{\implies}    &{f''(x)}    & {~=~}    &{\frac{2 \pi h}{r} .[ – 2]}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\frac{-4 \pi h}{r}}    \\
\end{array}$                           

2. Equate the first derivative to zero and solve for x:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{2 \pi h}{r} .[r – 2x]}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{\implies}    &{r-2x}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{{}}    &{2x}    & {~=~}    &{r}    \\
{~\color{magenta}    4    }    &{{}}    &{x}    & {~=~}    &{\frac{r}{2}}    \\
\end{array}$                           

3. So the only one point in category I is: x = r/2
4. We obtained $\rm{f'(x) = \frac{2 \pi h}{r} .[r – 2x]}$
• This function is defined at all x.
• Therefore, there is no point in category II.
5. So the only one critical point is:
x = r/2

Step III: Applying the second derivative test
1. We obtained f ''(x) = $\rm{\frac{-4 \pi h}{r}}$
h and r are +ve constants. So f '' is always −ve
2. Therefore, f '' is −ve at the critical point.
⇒ f has a local maximum at the critical point.

Step IV: Final result:
The curved surface area will be maximum when x = r/2.
That is when radius of the cylinder is half of that of the cone.

Step V (optional): Drawing the graph

• The graph is shown in fig.22.66 below:
    ♦ h is assumed to be 10 cm.
    ♦ r is assumed to be 4 cm.

Fig.22.66

 

   ♦ f is drawn in red color.
   ♦ (2,64.8) is a local maximum.
   ♦ So when x = 2, the function will have the minimum value, which is 62.8 cm².

Solved example 22.64
An Apache helicopter of enemy is flying along the curve given by y = x² + 7. A soldier placed at (3,7), wants to shoot down the helicopter when it is nearest to him. Find the nearest distance.
Solution:
Step I: Writing the problem as a function
1. Let (h,k) be any point on the curve.
• Let D be the distance between (3,7) and (h,k).
• Then we can write:
$\rm{D = \sqrt{(h-3)^2+(k-7)^2}}$
2. (h,k) is a point on the curve y = x² + 7. So we can write: k = h² + 7
• Substituting this in (1), we get:
$\rm{D = \sqrt{(h-3)^2+(h^2 + 7-7)^2}= \sqrt{(h-3)^2+ h^4}}$
3. (3,7) is a constant point. It's distance D from (h,k) will depend on the position of (h,k).
• We want such a position for (h,k) that, D is minimum.
4. From (2), it is clear that, D depends on h.
• D is minimum when $\rm{\sqrt{(h-3)^2+ h^4}}$ is minimum.
⇒ D is minimum when $\rm{(h-3)^2+ h^4}$ is minimum.
• $\rm{(h-3)^2+ h^4}$ can be considered as a function of h. We can write:
f(h) = (h−3)² + h⁴.
5. We want D to have the minimum possible value. That means, we want the least of all the local minima of f. That means, we need to find all local minima and compare them.

Step II: Finding the critical points
1. Write the first two derivatives:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(k)}    & {~=~}    &{(h-3)^2+ h^4}    \\
{~\color{magenta}    2    }    &{\implies}    &{f'(k)}    & {~=~}    &{2(h-3)+ 4h^3}    \\
{~\color{magenta}    3    }    &{\implies}    &{f''(k)}    & {~=~}    &{2+ 12h^2}    \\
\end{array}$                           

2. Equate the first derivative to zero and solve for h:
$\rm{2(h-3)+ 4h^3 = 0}$
• This is a cubic equation. The only real solution is h = 1                  
3. So the only one point in category I is: k = 1
4. We obtained $\rm{f'(h) = 2(h-3)+ 4h^3}$
• So f '(k) is defined for all real numbers. That means, there is no input h at which f '(h) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: h = 1

Step III: Applying the second derivative test
• f ''(1) = 2 + 12(1)² = 2 + 12 = 14
• f ''(1) is +ve. That means, the critical point h = 1 is a local minimum.

Step IV: Final result:
• f is minimum when h = 1
⇒ D is minimum when h = 1
• When h = 1, k = 1² + 7 = 8
⇒ D is minimum when the helicopter is at (1,8)
⇒ The minimum value of D = $\rm{\sqrt{(1-3)^2+(8-7)^2} = \sqrt{4 + 1} = \sqrt{5}}$

Check: This can be written in 4 steps.
1. The constant point is (3,7). It is marked as A in the fig.22.67 below:

Fig.22.67

2. We obtained the point of minimum distance (h,k) as: (1,8). This is marked as B in the fig.22.67 above.

3. We must prove that, the distance AB is the minimum possible distance.
• To prove that, we draw a tangent at B (green line). Then we draw AB (yellow dashed line).
• We see that, AB is perpendicular to the tangent. So AB is the minimum possible distance from A to any point on the curve.

4. We can draw a line from A to any other point B' of the positive side of the parabola. Then AB' will not be perpendicular to the tangent at B'.

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The link below gives a few more solved examples:

Exercise 22.5

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In the next section, we will see some miscellaneous examples.

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22.17 - The Second Derivative Test

In the previous section, we saw the concavity test. In this section, we will see the second derivative test for finding local extrema.

1. Consider the graph in fig.22.55 of the previous section. With some modifications, it is shown again below as fig.22.57.
• Two modifications are made to the earlier fig.22.55
(i) The point (0,0) is specially marked because, it is a local maximum.
(ii) The vertical white dashed line is drawn through the local minimum. This is for getting a better understanding about the change of sign of f ' on either sides of the local minimum. Such a line should have to be drawn through the local maximum also. But there, we already have the y-axis.

Fig.22.57

• The vertical magenta dashed line divides the graph into two portions: concave down portion and concave up portion.

2. Consider the concave down portion. Any concave down portion will contain a peak point. This peak point will be a local maximum. We want a relation between this local maximum and the second derivative f ''. The relation can be established in 4 steps:
(i) We see that, in the concave down portion:
   ♦ The green line is below the x-axis.
   ♦ This happens because, in this portion, all f '' values are −ve. We have seen the reason in the previous section 22.16.
(ii) We see that, in the concave down portion:
• The yellow curve is above x-axis up to the local maximum.
   ♦ This happens because, in this portion, all f ' values are +ve upto the local maximum. We have seen the reason in a previous section. See fig.22.46(b) of section 22.15.
• The yellow curve is below x-axis after the local maximum.
   ♦ This happens because, in this portion, all f ' values are −ve after the local maximum. We have seen the reason in a previous section. See fig.22.46(b) of section 22.15.
(iii) We see that, in the concave down portion:
• The yellow curve changes sign at the local maximum. That means, f ' changes sign at the local maximum.
(We know that, at the point where the "change in sign" occurs, f ' will be zero)
(iv) Based on the above three steps, we can establish the relation:
• At the local maximum:
   ♦ f ' will be zero.
   ♦ f '' will be −ve.

3. Consider the concave up region. Any concave up region will contain a valley point. This valley point will be a local minimum . We want a relation between this local minimum and the second derivative f ''. The relation can be established in 4 steps:
(i) We see that, in the concave up portion:
   ♦ The green line is above the x-axis.
   ♦ This is because, in this portion, all f '' values are +ve. We have seen the reason in the previous section 22.16.
(ii) We see that, in the concave up portion:
• The yellow curve is below x-axis up to the local minimum.
   ♦ This happens because, in this portion, all f ' values are −ve up to the local minimum. We have seen the reason in a previous section. See fig.22.46(a) of section 22.15.
• The yellow curve is above x-axis after the local maximum.
   ♦ This happens because, in this portion, all f ' values are +ve after the local maximum. We have seen the reason in a previous section. See fig.22.46(a) of section 22.15.
(iii) We see that, in the concave up portion:
• The yellow curve changes sign at the local minimum. That means, f ' changes sign at the local minimum.
(We know that, at the point where the "change in sign" occurs, f ' will be zero)
(iv) Based on the above three steps, we can establish the relation:
• At the local minimum:
   ♦ f ' will be zero.
   ♦ f '' will be +ve.
4. So now we have an effective test to determine whether a critical point is a local maximum or a local minimum. The test can be done in 2 steps:
Step I: Determine all critical points
Step II: Determine the sign of f '' at each of those points.
   ♦ If f '' is −ve at a critical point, then that point is a local maximum.
   ♦ If f '' is +ve at a critical point, then that point is a local minimum.
◼ This test is known as the second derivative test.

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• Note that, the second derivative test does not tell us anything about the case when f '' = 0
• That means, the test is not applicable at points where f '' = 0. At such a point, we must use the first derivative test to find whether that point is a local maximum or local minimum. It is also possible that, such a point is neither a local maximum nor a local minimum.

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Solved example 22.56
Find local extrema of the function f given by f(x) = 3x⁴ + 4x³ − 12x² + 12
Solution:
Step I: Finding the critical points
1. Write the first two derivatives:
f '(x) = 12x³ + 12x² − 24x
f ''(x) = 36x² + 24x − 24
2. Equate the first derivative to zero and solve for x:
12x³ + 12x² − 24x = 0
⇒ 12x(x² + x − 2) = 0
⇒ x(x² + 2x − x − 2) = 0
⇒ x[x(x + 2) − (x + 2)] = 0
⇒ x[(x + 2) (x − 1)] = 0
⇒ x = 0, x = −2 and x = 1
3. So the points in category I are: x = 0, x = −2 and x = 1
4. We obtained f '(x) = 12x³ + 12x² − 24x
• This function is defined for all real numbers. That means, there is no input x at which f '(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only three critical points are: x = 0, x = −2 and x = 1 

Step II: Find f '' at the critical points
1. f ''(0) =  36(0)² + 24(0) − 24 = −24
2. f ''(−2) =  36(−2)² + 24(−2) − 24 = 36(4) − 48 −24 = 144 − 72 = 72
3. f ''(1) =  36(1)² + 24(1) − 24 = 36(1) + 24 −24 = 36

Step III: Applying the second derivative test
1. Since f ''(0) is −ve, there is a local maximum at x = 0.
2. Since f ''(−2) is +ve, there is a local minimum at x = −2.
3. Since f ''(1) is +ve, there is a local minimum at x = 1.

Step IV: Finding the actual extrema values:
1. The local maximum value at x = 0 is given by:
f(0) = 3(0)⁴ + 4(0)³ − 12(0)² + 12 = 12
2. The local minimum value at x = −2 is given by:
f(−2) = 3(−2)⁴ + 4(−2)³ − 12(−2)² + 12 = −20
3. The local minimum value at x = 1 is given by:
f(1) = 3(1)⁴ + 4(1)³ − 12(1)² + 12 = 7

Step V (optional): Drawing the graph

• The graph is shown in fig.22.58 below:

Fig.22.56

   ♦ f is drawn in red color.
   ♦ (−1,7) is a local minimum.
   ♦ (0,12) is a local maximum.
   ♦ (2,20) is a local minimum.

Solved example 22.57
Find all points of local extrema of the function f given by f(x) = 2x³ − 6x² + 6x + 5
Solution:
Step I: Finding the critical points
1. Write the first two derivatives:
f '(x) = 6x² − 12x + 6
f ''(x) = 12x − 12
2. Equate the first derivative to zero and solve for x:
6x² − 12x + 6 = 0
⇒ 6(x² −2x + 1) = 0
⇒ x² −2x + 1 = 0
⇒ (x − 1)² = 0
⇒ x = 1
3. So the only one point in category I is: x = 1
4. We obtained f '(x) = 6x² − 12x + 6
• This function is defined for all real numbers. That means, there is no input x at which f '(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: x = 1 

Step II: Find f '' at the critical points
• f ''(1) = 12(1) − 12 = 0

Step III: Applying the second derivative test
1. The second derivative test is not applicable at the point where f '' = 0.
2. So we have to apply the first derivative test for this problem
(We have already solved this problem by applying the first derivative test. See solved example 22.52 in section 22.15. So we can use those steps to complete this problem)

Step IV: Dividing the number line
• The critical point is x = 1. Therefore, the number line can be divided into two intervals:
(−∞, 1) and (1,∞)

Step V: Finding the sign of f '(x) in each interval
First interval:
• A convenient number in the first interval is −1.
• f '(−1) = 6(−1)2 − 12(−1) + 6 = 6 + 12 + 6 = 24
• Therefore, the sign of f '(x) in the first interval is +ve.

Second interval:
• A convenient number in the second interval is 2.
• f '(2) = 6(2)2 − 12(2) + 6 = 24 − 24 + 6 = 6
• Therefore, the sign of f '(x) in the second interval is +ve.

Step VI: Analyzing the change of signs at critical point
• At the critical point 1, there is no change of sign for f '(x). So this critical point is neither a point of local maximum nor a point of local minimum.

Step VII (optional): Drawing the graph

• The graph is drawn for the solved example 22.52 in section 22.15.

Solved example 22.58
Find all points of local extrema of the function f given by f(x) =x⁵− 5x³
Solution:
Step I: Finding the critical points
1. Write the first two derivatives:
f '(x) = 5x⁴ − 15x²
f ''(x) = 20x³ − 30x
2. Equate the first derivative to zero and solve for x:
5x⁴ − 15x² = 0
⇒ 5x²(x² − 3) = 0
⇒ 5x² = 0 and (x² − 3) = 0
⇒ x = 0, x = √3 and x = −√3
3. So the three points in category I are:
x = 0, x = √3 and x = −√3
4. We obtained f '(x) = 5x⁴ − 15x²
• This function is defined for all real numbers. That means, there is no input x at which f '(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only three critical points are:
x = 0, x = √3 and x = −√3

Step II: Find f '' at the critical points
• f ''(0) = 20(0)³ − 30(0) = 0
• f ''(√3) = 20(√3)³ − 30(√3) = √3[20(√3)² − 30]
= √3[20(3) − 30] = √3[60 − 30] = 30√3
• f ''(−√3) = 20(−√3)³ − 30(−√3) = −√3[20(−√3)² − 30]
= −√3[20(3) − 30] = −√3[60 − 30] = −30√3

Step III: Applying the second derivative test
1. Since f ''(0) is zero, the second derivative test is not applicable at x = 0.
2. Since f ''(√3) is +ve, there is a local minimum at x = √3.
3. Since f ''(−√3) is −ve, there is a local maximum at x = −√3.

Step IV: Since the second derivative is zero at x = 0, we have to apply the first derivative test at that point. For that, we divide the number line
• The critical points are x = 0, x = √3 and x = −√3. Therefore, the number line can be divided into four intervals:
(−∞, −√3), (−√3,0), (0,√3) and (√3,∞)
• Since we are checking x = 0, we need to consider only two intervals: (−√3,0) and (0,√3)

Step V: Finding the sign of f '(x) in each interval
First interval: (−√3,0)
• A convenient number in the first interval is −1.
• f '(−1) = 5(−1)⁴ − 15(−1)² = 5 − 15 = −10
• Therefore, the sign of f '(x) in the first interval is −ve.

Second interval: (0,√3)
• A convenient number in the second interval is 1.
• f '(1) = 5(1)⁴ − 15(1)² = 5 − 15 = −10
• Therefore, the sign of f '(x) in the second interval is −ve.

Step VI: Analyzing the change of signs at critical point
• At the critical point 0, there is no change of sign for f '(x). So this critical point is neither a point of local maximum nor a point of local minimum.

Step VII (optional): Drawing the graph

Fig.22.58

   ♦ f is drawn in red color.
   ♦ (−1.73,10.39) is a local maximum.
   ♦ (0,0) is neither a local maximum nor a local minimum.
         ✰ This is a point of inflection.
   ♦ (1.73,−10.39) is a local minimum.

Solved example 22.59
Find two positive numbers whose sum is 15 and the sum of whose squares is minimum.
Solution:
Step I: Writing the problem as a function
1. Sum of two positive numbers is to be 15
So if one number is x, then the other number is (15−x)
2. If S is the sum of the squares, then we can write:
S = x² + (15 − x)2
3. The value of S varies depending on the value of x. So S is a function of x. We can write:
S = f(x) = x² + (15 − x)²
4. We want S to have the minimum possible value. That means, we want the least of all the local extrema of f. That means, we need to find all local extrema and compare them.

Step II: Finding the critical points
1. Write the first two derivatives:
• f '(x) = 2x + 2(15 − x)(−1)
= 2x − 2(15 − x)
= 2x −30 + 2x
= 4x −30
• f ''(x) = 4
2. Equate the first derivative to zero and solve for x:
4x −30 = 0
⇒ 4x = 30
⇒ x = 7.5
3. So the only one point in category I is: x = 7.5
4. We obtained f '(x) = 4x −30
• This function is defined for all real numbers. That means, there is no input x at which f '(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: x = 7.5

Step III: Find f '' at the critical point
• f ''(7.5) =  4. This is because, f ''(x) is a constant 

Step IV: Applying the second derivative test
• Since f ''(7.5) is +ve, there is a local minimum at x = 7.5
• Since there is only one critical point, there are no other points to compare. We can write:
The minimum value of the function occurs at x = 7.5

Step V: Finding the actual minimum value:
• The local maximum value at x = 7.5 is given by:
f(7.5) = (7.5)² + (15 − 7.5)² = 2(7.5)² = 112.5

Step VI: Final result:
• Out of the two numbers, the first number x = 7.5
• So second number (15 − x) = 7.5
• The minimum sum = (7.5)² + (7.5)² = 2(7.5)² = 112.5 

Check:
• Let us put x = 8. Then (x − 8) = 7
• Then S = 8² + 7² = 113. This is greater than 112.5

Step VI (optional): Drawing the graph

• The graph is shown in fig.22.59 below:

Fig.22.59

   ♦ f is drawn in red color.
   ♦ (7.5,112.5) is the local minimum.

◼ In general, if sum of two positive numbers is k and sum of their squares is to be minimum, then there is only one option:
Both numbers must be equal to k/2.

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Based on the above solved examples, we can write a comparison between first derivative test and second derivative test. It can be written in 2 steps:
1. In the first derivative test, we check the sign of f ' on either sides of each critical point. Based on the signs, we decide whether that critical point is a local maximum or a local minimum.
2. In the second derivative test, we check the sign of f '' at each critical point. Based on the sign, we decide whether that critical point is a local maximum or a local minimum.
3. So in many cases, the second derivative test will be easier to apply.

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In the next section, we will see a few more solved examples.

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22.16 - Concavity Test For Finding The Shape of Graph

In the previous section, we we saw the first derivative test for finding local extrema. In this section, we will see the concavity test for finding shape of graph.

First we will see upward concavity. It can be written in 5 steps:
1. The red curve in fig.22.52 below shows the graph of a function f.

Fig.22.52

• We see that, the graph has an upward concavity. In such a situation, we say: f is concave up.

2. Let us analyze three random tangents of f.
Three random points are marked on f. They are: green, magenta and yellow.
• Consider the tangent at the green point.
   ♦ This tangent is drawn in green color.
   ♦ This tangent makes an angle of 110.6^o with the +ve side of the x-axis.
• Consider the tangent at the magenta point.
   ♦ This tangent is drawn in magenta color.
   ♦ This tangent makes an angle of 137.6^o with the +ve side of the x-axis.
• Consider the tangent at the yellow point.
   ♦ This tangent is drawn in yellow color.
   ♦ This tangent makes an angle of 65.2^o with the +ve side of the x-axis.

3. We know that, slope of a line is equal to the tangent of the angle which the line makes with the +ve side of the x-axis.
• So let us write the slopes:
   ♦ Slope of green tangent = tan(110.6) = −2.66
   ♦ Slope of magenta tangent = tan(137.6) = −0.913
   ♦ Slope of yellow tangent = tan(65.2) = 2.16
• We see that:
   ♦ Magenta has a larger slope than green.
   ♦ Yellow has a larger slope than magenta.
• So the slope increases in the order:
Green < Magenta < Yellow

4. Now consider the green, magenta and yellow points.
   ♦ yellow is at the right side of magenta.
   ♦ magenta is at the right side of green.
• So we can write:
For a function which is concave up, the slope of tangent increases as we move from left to right.

5. Recall that, slope of the tangent is given by f '. This f ' is also a function.
• So, if f ' is increasing, the second derivative f '' must be +ve.
• Therefore we can write:
A function f is concave up in an interval (a,b), if f '' is +ve at all points in that interval.

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Now we will see downward concavity. It can be written in 5 steps:
1.The red curve in fig.22.53 below shows the graph of a function f.

Fig.22.53

• We see that, the graph has a downward concavity. In such a situation, we say: f is concave down.

2. Let us analyze three random tangents of f.
Three random points are marked on f. They are: yellow, magenta and green.
• Consider the tangent at the yellow point.
   ♦ This tangent is drawn in yellow color.
   ♦ This tangent makes an angle of 65.2^o with the +ve side of the x-axis.
• Consider the tangent at the magenta point.
   ♦ This tangent is drawn in magenta color.
   ♦ This tangent makes an angle of 137.6^o with the +ve side of the x-axis.
• Consider the tangent at the green point.
   ♦ This tangent is drawn in green color.
   ♦ This tangent makes an angle of 110.6^o with the +ve side of the x-axis.

3. We know that, slope of a line is equal to the tangent of the angle which the line makes with the +ve side of the x-axis.
• So let us write the slopes:
   ♦ Slope of yellow tangent = tan(65.2) = 2.16
   ♦ Slope of magenta tangent = tan(137.6) = −0.913
   ♦ Slope of green tangent = tan(110.6) = −2.66

• We see that:
   ♦ Magenta has a smaller slope than yellow.
   ♦ Green has a smaller slope than magenta.
• So the slope decreases in the order:
Yellow < Magenta < Green

4. Now consider the green, magenta and yellow points.
   ♦ yellow is at the left side of magenta.
   ♦ magenta is at the left side of green.
• So we can write:
For a function which is concave down, the slope of tangent decreases as we move from left to right.

5. Recall that, slope of the tangent is given by f '. This f ' is also a function.
• So, if f ' is decreasing, the second derivative f '' must be −ve.
• Therefore we can write:
A function f is concave down in an interval (a,b), if f '' is −ve at all points in that interval.

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Based on the above two examples, we can write the test for concavity:
Let f be a twice differentiable function in the interval (a,b).
(i) If f '' > 0 for all points in (a,b), then f is concave up.
(ii) If f '' < 0 for all points in (a,b), then f is concave down.

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Now we can learn about inflection point. It can be written in 3 steps:
1. Fig.22.54 below, shows the graph of a function f.

Fig.22.54

• We see that:
   ♦ f is concave up in the interval (a,c)
   ♦ f is concave down in the interval (c,a)
2. Let us analyze the change in sign of f '':
(i) We know that, if f is concave up, all f '' will be +ve.
• So all f '' in (a,c), will be +ve.
(ii) We also know that, if f is concave down, all f '' will be −ve.
• So all f '' in (c,a), will be −ve.
(iii) Such a change in sign for f '' is possible only if:
f '' becomes zero at some point.
(iv) Obviously, that zero point will be at the junction between "concave up portion" and "concave down portion".
• That means, f ''(c) = 0
3. So we can write the definition of inflection point:
Inflection point of a function f is the point at which f changes concavity.
4. Based on the above definition, we can write:
The inflection point in our present case is (c,f(c)).

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Now we will see a solved example
Solved example 22.54
For the function f(x) = x³ − 6x², determine all intervals where f is concave up and all intervals where f is concave down.
Solution:
Step I: Finding the inflection points and dividing the number line
1. First we will write f '(x):
f '(x) = 3x² − 12x

2. Next we will write f ''(x):
f '' = 6x − 12

3. Equating f''(x) to zero, we get:
6x − 12 = 0
⇒ 6x = 12
⇒ x = 2

4. So the only one inflection point is at x = 2

5. Now the number line can be divided into two intervals:
(−∞,2) and (2,∞)

Step II: Finding the sign of f '' in each interval and assessing the concavity
First interval: (−∞,2)
• A convenient number in the first interval is 0.
• f ''(0) = 6(0) − 12 =  −12
• Therefore, the sign of f ''(x) in the first interval is −ve.
• Then f is concave down in the first interval.

Second interval: (2,∞)
• A convenient number in the second interval is 3.
• f ''(3) = 6(3) − 12 = 18 −12 = 6
• Therefore, the sign of f ''(x) in the second interval is +ve.
• Then f is concave up in the second interval.

Step III (optional): Drawing the graph

• The graph is shown in fig.22.55 below:

Fig.22.55

   ♦ f is drawn in red color.
   ♦ f ' is drawn in yellow color
   ♦ f '' is drawn in green color
• The vertical magenta dashed line divides the number line into the two intervals (−∞,2) and (2,∞). The following five facts are in agreement with our discussion on concavity test:
(i) For all inputs from the left of this vertical line, f is concave down.
(ii) For all inputs from the right of this vertical line, f is concave up.
(iii) For all inputs from the left of this vertical line, f ' is decreasing.
(iv) For all inputs from the right of this vertical line, f ' is increasing.
(v) f '' and this vertical line intersects at the x-axis. That means, value of f '' at x = 2 (the point of inflection), is zero.

Solved example 22.55
For the function f(x) = x + sin(2x), x ∈ [−π/2,π/2], determine all intervals where f is concave up and all intervals where f is concave down.
Solution:
Step I: Finding the inflection points and dividing the number line
1. First we will write f '(x):
f '(x) = 1 + 2 cos(2x)

2. Next we will write f ''(x):
f '' = −4 sin(2x)

3. Equating f''(x) to zero, we get:
−4 sin(2x) = 0
⇒ −8 sin x cos x = 0
⇒ sin x = 0 or cos x = 0
• For the equation sin x = 0, the only one solution in [−π/2,π/2], is x = 0
• For the equation cos x = 0, the only two solutions in [−π/2,π/2], are x = −π/2 and x = π/2


4. So the three inflection points are:
x = −π/2, x = 0 and x = π/2
5. Now the given domain [−π/2,π/2] can be divided into two intervals:
[−π/2,0] and [0,π/2]

Step II: Finding the sign of f '' in each interval and assessing the concavity
First interval: [−π/2,0]
• A convenient number in the first interval is −π/8.
• f ''(−π/8) = −4 sin(2(−π/8))
= −4 sin (−π/4) = 4 sin (π/4) = 4/(√2) 
• Therefore, the sign of f ''(x) in the first interval is +ve.
• Then f is concave up in the first interval.

Second interval: [0,π/2]
• A convenient number in the second interval is π/8.
• f ''(π/8) = −4 sin(2(π/8))
= −4 sin (π/4) = −4/(√2) 
• Therefore, the sign of f ''(x) in the second interval is −ve.
• Then f is concave down in the second interval.

Step III (optional): Drawing the graph

• The graph is shown in fig.22.56 below:

Fig.22.56

   ♦ f is drawn in red color.
   ♦ f ' is drawn in yellow color
   ♦ f '' is drawn in green color
• The two vertical magenta dashed lines are the boundaries of the domain [−π/2,π/2]. Those two vertical lines indicate the position of two inflection points x = −π/2 and x = π/2 also. Position of the third inflection point x = 0, is indicated by the y-axis. The following five facts are in agreement with our discussion on concavity test:
(i) For all inputs from between the left vertical line and y-axis, f is concave up.
(ii) For all inputs from between y-axis and the right vertical line, f is concave down.
(iii) For all inputs from between the left vertical line and y-axis, f ' is increasing.
(iv) For all inputs from between y-axis and the right vertical line, f ' is decreasing.
(v) Consider the point of intersection of  f '' with the two vertical lines and the y-axis. All those three points of intersection are at the x-axis. That means, value of f '' at those points (the points of inflection), is zero.

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In the next section, we will see the second derivative test.

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22.15 - First Derivative Test For Local Extrema

In the previous section, we we saw the analytical method for finding absolute extrema. In this section, we will learn how to test whether, a critical point is a local extremum.

• We have seen the method for finding the critical points. But all those critical points, need not be local extrema points.
• We want an analytical method to test whether a critical point is a local extremum. The method can be explained in 12 steps:
1. In fig.22.44(a) below, the red curve is the graph of a function f in the interval (a,b).

Fig.22.44

• From the graph, we see that, f is decreasing in (a,b).
• Take any point from (a,b). Draw a tangent at that point.
    ♦ That tangent will be sloping downwards as we move from left to right.
    ♦ That means, that tangent has a negative slope.
(Two sample tangents are drawn in the fig.a)
    ♦ Negative slope indicates that, derivative at that point is negative.

2. In fig.22.44(b) above, the red curve is the graph of a function f in the interval (a,b).
• From the graph, we see that, f is decreasing in (a,b).
• Take any point from (a,b). Draw a tangent at that point.
    ♦ That tangent will be sloping downwards as we move from left to right.
    ♦ That means, that tangent has a negative slope.
(Two sample tangents are drawn in the fig.a)
    ♦ Negative slope indicates that, derivative at that point is negative.

3. In fig.22.44, we saw two functions. They have different shapes. But both are decreasing functions. They give the same result.
• The result can be summarized in three steps:
(i) Take any function.
(ii) Suppose that, the function is decreasing in (a,b).
(iii) Then all derivatives in (a,b) will be negative.

4. In fig.22.45(a) below, the red curve is the graph of a function f in the interval (a,b).

Fig.22.45

• From the graph, we see that, f is increasing in (a,b).
• Take any point from (a,b). Draw a tangent at that point.
    ♦ That tangent will be sloping upwards as we move from left to right.
    ♦ That means, that tangent has a positive slope.
(Two sample tangents are drawn in the fig.a)
    ♦ Positive slope indicates that, derivative at that point is positive.

5. In fig.22.45(b) above, the red curve is the graph of a function f in the interval (a,b).
• From the graph, we see that, f is increasing in (a,b).
• Take any point from (a,b). Draw a tangent at that point.
    ♦ That tangent will be sloping upwards as we move from left to right.
    ♦ That means, that tangent has a positive slope.
(Two sample tangents are drawn in the fig.b)
    ♦ Positive slope indicates that, derivative at that point is positive.

6. In fig.22.45, we saw two functions. They have different shapes. But both are increasing functions. They give the same result.
• The result can be summarized in three steps:
(i) Take any function.
(ii) Suppose that, the function is increasing in (a,b).
(iii) Then all derivatives in (a,b) will be positive.

7. In fig.22.46(a) below, the red curve is the graph of a function f in the interval (a,b).

Fig.22.46

• The point c in the interval (a,b), is a critical point.
• We can note five important facts:
(i) Point c is a local minimum.
(ii) To the left of c, all tangents will be having a negative slope.
(One sample tangent is drawn on the left side)
(iii) So all derivatives on the left are negative. That means, the portion between points a and c, is of decreasing nature.
(iv) To the right of c, all tangents will be having a positive slope.
(One sample tangent is drawn on the right side)
(v) So all derivatives on the right are positive. That means, the portion between points c and b, is of increasing nature.
• Based on the above five facts, we can write:
If c is a local minimum in (a,b), then:
    ♦ Sign of the derivative changes at c
    ♦ This change is from negative to positive.
    ♦ Portion between a and c is decreasing in nature.
    ♦ Portion between c and b is increasing in nature.

8. In fig.22.46(b) above, the red curve is the graph of a function f in the interval (a,b).
• The point c in the interval (a,b), is a critical point.
• We can note five important facts:
(i) Point c is a local maximum.
(ii) To the left of c, all tangents will be having a positive slope.
(One sample tangent is drawn on the left side)
(iii) So all derivatives on the left are positive. That means, the portion between points a and c, is of increasing nature.
(iv) To the right of c, all tangents will be having a negative slope.
(One sample tangent is drawn on the right side)
(v) So all derivatives on the right are negative. That means, the portion between points c and b, is of decreasing nature.
• Based on the above five facts, we can write:
If c is a local maximum in (a,b), then:
    ♦ Sign of the derivative changes at c
    ♦ This change is from positive to negative.
    ♦ Portion between a and c is increasing in nature.
    ♦ Portion between c and b is decreasing in nature.

9. In both figs.22.46(a) and 22.46(b), the derivative exists at the point c. Let us see the cases where the derivative does not exist.
• In fig.22.47(a) below, c is a cusp point. Every thing that we wrote in (7) is applicable for this fig.
• In fig.22.47(b) below also, c is a cusp point. Every thing that we wrote in (8) is applicable for this fig.

Fig.22.47

• So we can write:
It does not matter whether the derivative exists at c or not. We can find whether c is a local maximum or local minimum, just by finding the signs of the derivative on either sides of c.

10. Now let us see the case where c is neither a local maximum nor a local minimum. In fig.22.48(a) below, the red curve is the graph of a function f in the interval (a,b).

Fig.22.48

• The point c in the interval (a,b), is a critical point.
• We can note six important facts:
(i) Point c is a not a local minimum.
(ii) Point c is a not a local maximum.
(iii) To the left of c, all tangents will be having a negative slope.
(One sample tangent is drawn on the left side)
(iv) So all derivatives on the left are negative. That means, the portion between points a and c, is of decreasing nature.
(v) To the right of c also, all tangents will be having a negative slope.
(One sample tangent is drawn on the right side)
(vi) So all derivatives on the right are negative. That means, the portion between points c and b, is also of decreasing nature.
• Based on the above six facts, we can write:
If c is neither a local minimum nor a local maximum in (a,b), then:
    ♦ Sign of the derivative will not change at c
    ♦ If it is −ve on the left, it will be −ve on right also.
    ♦ If the portion between a and c is decreasing in nature, the portion between c and b will also be decreasing in nature.

11. In fig.22.48(b) above, the red curve is the graph of a function f in the interval (a,b).
• The point c in the interval (a,b), is a critical point.
• We can note six important facts:
(i) Point c is a not a local minimum.
(ii) Point c is a not a local maximum.
(iii) To the left of c, all tangents will be having a +ve slope.
(One sample tangent is drawn on the left side)
(iv) So all derivatives on the left are +ve. That means, the portion between points a and c, is of increasing nature.
(v) To the right of c also, all tangents will be having a +ve slope.
(One sample tangent is drawn on the right side)
(vi) So all derivatives on the right are +ve. That means, the portion between points c and b, is also of increasing nature.
• Based on the above six facts, we can write:
If c is neither a local minimum nor a local maximum in (a,b), then:
    ♦ Sign of the derivative will not change at c
    ♦ If it is +ve on the left, it will be +ve on right also.
    ♦ If the portion between a and c is increasing in nature, the portion between c and b will also be increasing in nature.

12. Based on the above 11 steps, we can write the test. It can be written in four steps:
(i) f is a continuous function in the interval (a,b). Point c falls with in (a,b).
(ii) f(c) is a local minimum if:
    ♦ The derivative is −ve between a and c.
    ♦ The derivative is +ve between c and b.
(iii) f(c) is a local maximum if:
    ♦ The derivative is +ve between a and c.
    ♦ The derivative is −ve between c and b.
(iv) f(c) is neither a local maximum nor a local minimum if:
    ♦ The derivative has the same sign between [a and c] and [c and b].

• The above four steps taken together, is known as the first derivative test for finding local extrema.

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Now we will see some solved examples.

Solved example 22.51
Find all points of local maxima and local minima of the function f given by f(x) = x³ − 3x + 3
Solution:
Step I: Finding the critical points and dividing the number line
1. First we write the derivative:
f '(x) = 3x² − 3

2. This derivative must be equated to zero.
3x² − 3 = 0
⇒ 3(x² − 1) = 0
⇒ x² = 1
⇒ x = +1 and x = −1

3. So the points in category I are: x = 1 and x = −1
4. We obtained f '(x) = 3x² − 3
• This function is defined for all real numbers. That means, there is no input x at which f '(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only two critical points are: x = 1 and x = −1
6. Therefore, the number line can be divided into three intervals:
(−∞,−1), (−1,1) and (1,∞)

Step II: Finding the sign of f '(x) in each interval
First interval:
• A convenient number in the first interval is −2.
• f '(−2) = 3(−2)² − 3 = 3(4) − 3 = 12 − 3 = 9
• Therefore, the sign of f '(x) in the first interval is +ve. 

Second interval:
• A convenient number in the second interval is 0.2.
• f '(0.2) = 3(0.2)² − 3 = 3(0.04) − 3 = 0.12 − 3 = −2.88
• Therefore, the sign of f '(x) in the second interval is −ve.

Third interval:
• A convenient number in the third interval is 2.
• f '(2) = 3(2)² − 3 = 3(4) − 3 = 12 − 3 = 9
• Therefore, the sign of f '(x) in the third interval is +ve.

Step III: Analyzing the change of signs at critical points
• At the first critical point −1, the sign of f '(x) changes from +ve to −ve. So this critical point is a point of local maximum.
• At the second critical point +1, the sign of f '(x) changes from −ve to +ve. So this critical point is a point of local minimum.

• Fig.22.49 below shows the graph

Fig.22.49

Solved example 22.52
Find all points of local maxima and local minima of the function f given by f(x) = 2x³ − 6x² + 6x + 5
Solution:
Step I: Finding the critical points and dividing the number line
1. First we write the derivative:
f '(x) = 6x² − 12x + 6

2. This derivative must be equated to zero.
6x² − 12x + 6 = 0
⇒ x² − 2x + 1 = 0
⇒ (x − 1)² = 0
⇒ (x − 1) = 0
⇒ x = +1

3. So the only one point in category I is: x = 1
4. We obtained f '(x) = 6x² − 12x + 6
• This function is defined for all real numbers. That means, there is no input x at which f '(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: x = 1
6. Therefore, the number line can be divided into two intervals:
(−∞, 1) and (1,∞)

Step II: Finding the sign of f '(x) in each interval
First interval:
• A convenient number in the first interval is −1.
• f '(−1) = 6(−1)² − 12(−1) + 6 = 6 + 12 + 6 = 24
• Therefore, the sign of f '(x) in the first interval is +ve. 

Second interval:
• A convenient number in the second interval is 2.
• f '(2) = 6(2)² − 12(2) + 6 = 24 − 24 + 6 = 6
• Therefore, the sign of f '(x) in the second interval is +ve.

Step III: Analyzing the change of signs at critical point
• At the critical point 1, there is no change of sign for f '(x). So this critical point is neither a point of local maximum nor a point of local minimum.

• Fig.22.50 below shows the graph

Fig.22.50

• We see that:
There is only one critical point, and it is neither a point of local maximum nor a point of local minimum.

Solved example 22.53
Find all points of local maxima and local minima of the function f given by f(x) = x¹¹ − 6x¹⁰.
Solution:
Step I: Finding the critical points and dividing the number line
1. First we write the derivative:
f '(x) = 11x¹⁰ −60x⁹.

2. This derivative must be equated to zero.
11x¹⁰ −60x⁹ = 0
⇒ x⁹(11x − 60) = 0
⇒ x⁹ = 0 and 11x − 60 = 0
⇒ x = 0 and 11x  = 60
⇒ x = 0 and x = 60/11

3. So the points in category I are: x = 0 and x = 60/11
4. We obtained f '(x) = 11x¹⁰ −60x⁹.
• This function is defined for all real numbers. That means, there is no input x at which f '(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only two critical points are: x = 0 and x = 60/11
6. Therefore, the number line can be divided into three intervals:
(−∞,0), (0,60/11) and (60/11,∞)

Step II: Finding the sign of f '(x) in each interval
First interval:
• A convenient number in the first interval is −1.
• f '(−1) = 11(−1)¹⁰ −60(−1)⁹ = 11+60 = 71
• Therefore, the sign of f '(x) in the first interval is +ve. 

Second interval:
• A convenient number in the second interval is 1.
• f '(1) = 11(1)¹⁰ −60(1)⁹ = 11−60 = −49
• Therefore, the sign of f '(x) in the second interval is −ve.

Third interval:
• A convenient number in the third interval is 10.
• f '(10) = 11(10)¹⁰ −60(10)⁹ = (10)⁹ [11(10) −60]
= (10)⁹ [110 −60] = (10)⁹ [50]
• Therefore, the sign of f '(x) in the third interval is +ve.

Step III: Analyzing the change of signs at critical points
• At the first critical point 0, the sign of f '(x) changes from +ve to −ve. So this critical point is a point of local maximum.
• At the second critical point 60/11 (=5.455), the sign of f '(x) changes from −ve to +ve. So this critical point is a point of local minimum.

• Fig.22.51 below shows the graph

Fig.22.51

• We see that:
Some portion near the critical point x = 0, is horizontal. In fact it is not horizontal. It appears to be horizontal because of the large scale of the graph. This can be proved in two steps:
(i) f(−0.1) = (−0.1)¹¹ − 6(−0.1)¹⁰
= (−1)(0.1)¹¹ − 6(0.1)¹⁰
= (0.1)¹⁰ [(−1)(0.1) − 6]
= (0.1)¹⁰ [−0.1 − 6]
= (0.1)¹⁰ [−6.1]
= (−1)(0.1)¹⁰ [6.1]
• This is a −ve quantity. That means, just to the left of the critical point x=0, the point on the graph is below the x-axis.
(ii) f(0.1) = (0.1)¹¹ − 6(0.1)¹⁰
= (0.1)¹⁰ [0.1 − 6]
= (0.1)¹⁰ [−5.9]
= (−1)(0.1)¹⁰ [5.9]
• This is a −ve quantity. That means, just to the right of the critical point x=0, the point on the graph is below the x-axis.

• We also see that, point x = 5.455 appears to be a cusp point. But in reality, there is a smooth transition at that point. If we enlarge the surrounding region of that point, we will be able to see the smooth transition. 

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In the next section, we will see the concavity test.

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