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 '(x), we know that,
calculation of f ''(x) 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.
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 maximum 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 k:
$\rm{2(h-3)+ 4h^3 = 0}$
• 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'.
The link below gives a few more solved examples:
In the next section, we will see some miscellaneous examples.
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