22.21 - Miscellaneous Examples on Applications of Derivatives - Part 3

In the previous section, we saw some miscellaneous examples. In this section, we will see a few more miscellaneous examples.

Solved example 22.77
Find the intervals in which the function given by
$\rm{f(x)\,=\,\frac{3}{10} x^4 - \frac{4}{5} x^3 - 3 x^2 + \frac{36}{5} x + 11}$
is (a) strictly increasing (b) strictly decreasing.
Solution:
1. First we write the derivative:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(x)}    & {~=~}    &{\frac{3}{10} x^4 - \frac{4}{5} x^3 - 3 x^2 + \frac{36}{5} x + 11}    \\
{~\color{magenta}    2    }    &{\implies}    &{f'(x)}    & {~=~}    &{\frac{3(4)}{10} x^3 - \frac{4(3)}{5} x^2 - 3(2) x + \frac{36}{5} + 0}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{6}{5} x^3 - \frac{12}{5} x^2 - 6 x + \frac{36}{5}}    \\
\end{array}$

2. Equating the above result to zero, we get:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f'(x)}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{6}{5} x^3 - \frac{12}{5} x^2 - 6 x + \frac{36}{5}}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{\implies}    &{6 x^3 - 12 x^2 - 30 x + 36}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{\implies}    &{x^3 - 2 x^2 - 5 x + 6}    & {~=~}    &{0}    \\
\end{array}$

3. Solving the above third degree equation, we get:
x = −2, x = 1 and x = 3

• So the domain R can be divided into four intervals:
(−∞,−2), (−2,1), (1,3) and (3,∞)

4. Consider the first interval (−∞,−2)
−3 is a convenient point in this interval.

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f'(-3)}    & {~=~}    &{\frac{6}{5} (-3)^3 - \frac{12}{5} (-3)^2 - 6 (-3) + \frac{36}{5}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{-28.8}    \\
\end{array}$                           

• So f '(x) is −ve in this interval. That means, f(x) is decreasing in this interval.
• Note that, f '(−2) will be zero. But −2 is not an input value. So we can write:
The given f is strictly decreasing in (−∞,−2)

5. Consider the second interval (−2,1)
0 is a convenient point in this interval.

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f'(0)}    & {~=~}    &{\frac{6}{5} (0)^3 - \frac{12}{5} (0)^2 - 6 (0) + \frac{36}{5}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{7.2}    \\
\end{array}$                           

• So f '(x) is +ve in this interval. That means, f(x) is increasing in this interval.
• Note that, f '(−2) and f '(0) will be zero. But −2 and 1 are not an input values. So we can write:
The given f is strictly increasing in (−2,1)

6. Consider the third interval (1,3)
2 is a convenient point in this interval.

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f'(2)}    & {~=~}    &{\frac{6}{5} (2)^3 - \frac{12}{5} (2)^2 - 6 (2) + \frac{36}{5}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{-4.8}    \\
\end{array}$                           

• So f '(x) is −ve in this interval. That means, f(x) is decreasing in this interval.
• Note that, f '(1) and f '(3) will be zero. But 1 and 3 are not an input values. So we can write:
The given f is strictly decreasing in (1,3)

7. Consider the fourth interval (3,∞)
4 is a convenient point in this interval.

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f'(4)}    & {~=~}    &{\frac{6}{5} (4)^3 - \frac{12}{5} (4)^2 - 6 (4) + \frac{36}{5}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{21.6}    \\
\end{array}$                           

• So f '(x) is +ve in this interval. That means, f(x) is increasing in this interval.
• Note that, f '(3) will be zero. But 3 is not an input value. So we can write:
The given f is strictly increasing in (3,∞)

Solved example 22.78
Show that the function f given by
f(x) = tan⁻¹(sin x + cos x), x > 0 is always a strictly increasing function in (0, π/4).
Solution:
1. First we write the derivative:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(x)}    & {~=~}    &{\tan^{-1}(\sin x + \cos x)}    \\
{~\color{magenta}    2    }    &{\implies}    &{f'(x)}    & {~=~}    &{\left(\frac{1}{1 + (\sin x + \cos x)^2} \right) (\cos x - \sin x)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\left(\frac{1}{1 + \sin^2 x + 2 \sin x \cos x + \cos^2 x} \right) (\cos x - \sin x)}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\left(\frac{1}{2 + \sin(2 x)} \right) (\cos x - \sin x)}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{\cos x - \sin x}{2 + \sin(2 x)}}    \\
\end{array}$

2. In this problem, there is no need to equate f'(x) to zero and find the intervals. This is because, we are already given an interval. We are asked to show that, the function is always strictly increasing in that interval.
• The given interval is (0, π/4). The boundaries of this interval may not be consecutive critical points. So the usual method of checking using a 'convenient number' cannot be used. We must show the general case.

3. We obtained the derivative as: $\frac{\cos x - \sin x}{2 + \sin(2 x)}$
• The largest input possible from the given interval is a value just below π/4.
• Even if the input is π/4, sin(2x) will be in the first quadrant. Sine is always +ve in the first quadrant. So (2 + sin(2x)) is +ve in the given interval.
• That means, denominator of the derivative is +ve in the given interval.

4. Now we can write, the derivative will be +ve if the numerator is also +ve.
That means: Derivative will be +ve if cos x − sin x > 0.
⇒ cos x / sin x  − sin x / sin x > 0
⇒ cot x  − 1 > 0
⇒ cot x  > 1

• In the given interval (0, π/4), tan x is always less than 1. So cot x is always greater than 1.
• So we can write: the numerator is also +ve.

5. We see that, f '(x) is +ve in the interval (0, π/4).
Therefore, f(x) is increasing in this interval.

6. Note that,
   ♦ f '(0) is 1/2. It is +ve.
   ♦ f '(π/4) is zero.
• But π/4 is not an input value. So we can write:
The given f is strictly increasing in (0, π/4)

7. Fig.22.75 below shows the graph.

Fig.22.75

Solved example 22.79
Find the intervals in which the function f given by
$f(x)\,=\,\frac{4 \sin x - 2x - x \cos x}{2 + \cos x}$
is (i) increasing (ii) decreasing.
Solution:
1. First we write the derivative:
Using quotient rule, we get:
• $f'(x)\,=\,-{\frac{\cos x(\cos x - 4)}{(2 + \cos x)^2}}$

2. Next, we equate the above result to zero:
• The denominator cannot be zero. So we get:
cos x (cos x − 4) = 0
• cos x cannot be 4. So we get: cos x = 0

3. Solving the equation cos x = 0, we get:
x = π/2 and x = 3π/2 as the principal solutions

• So the angle of the unit circle can be divided into three intervals:
(0, π/2), (π/2, 3π/2) and (3π/2, 2π)

4. Consider the first interval (0, π/2)
• π/4 is a convenient point in this interval.
$f'(\pi/4)\,=\,-{\frac{\cos (\pi/4)(\cos (\pi/4) - 4)}{(2 + \cos (\pi/4))^2}}$
• Here, the only −ve term is (cos(π/4) − 4).
• So f '(x) is +ve in this interval. That means, f(x) is increasing in this interval.

5. Consider the second interval (π/2, 3π/2)
• π is a convenient point in this interval.
$f'(\pi)\,=\,\frac{\cos (\pi)(\cos (\pi) - 4)}{(2 + \cos (\pi))^2}$
• Here, both terms in the numerator are −ve. The denominator is +ve.
• So f '(x) is −ve in this interval. That means, f(x) is decreasing in this interval.

6. Consider the third interval (3π/2, 2π)
• 7π/4 is a convenient point in this interval.
$f'(7 \pi/4)\,=\,\frac{\cos (7\pi/4)(\cos (7\pi/4) - 4)}{(2 + \cos (7\pi/4))^2}$
• cos(7π/4) = cos(π + 3π/4) = −cos(3π/4)
= −[cos(π/2 + π/4)] = −[−sin(π/4)] = sin(π/4)

• So here, the only −ve term is (cos(7π/4) − 4).
• Thus f '(x) is +ve in this interval. That means, f(x) is increasing in this interval.

7. Now we can write about the intervals.
• f is:
   ♦ Increasing in (0, π/2) and (3π/2, 2π)
   ♦ Decreasing in (π/2, 3π/2)

8. Fig.22.76 below shows the graph.

Fig.22.76

• Note that:
   ♦ π/2 = 1.57
   ♦ 3π/2 = 4.71

Solved example 22.80
Find the intervals in which the function f given by
$f(x)\,=\,x^3 + \frac{1}{x^3}$
is (i) increasing (ii) decreasing.
Solution:
1. First we write the derivative:
$f'(x)\,=\,3x^2 \,-\, \frac{3}{x^4}$

2. Next, we equate the above result to zero:
$3x^2 \,-\, \frac{3}{x^4}~=~0$
⇒ $x^2 \,-\, \frac{1}{x^4}~=~0$
⇒ $x^2 ~=~\frac{1}{x^4}$
⇒ $x^6 ~=~1$
⇒ $x\,=\,1~\text{and}~x\,=\,-1$

3. So the domain can be divided into three intervals:
(−∞, −1), (−1, 1) and (1, ∞)

4. Consider the first interval (−∞, −1)
• −2 is a convenient point in this interval.
$f'(-2)\,=\,3(-2)^2 \,-\, \frac{3}{(-2)^4}$
= $12\,-\, \frac{3}{16}$
• So f '(x) is +ve in this interval. That means, f(x) is increasing in this interval.

5. Consider the second interval (−1, 1)
• 0.5 is a convenient point in this interval.
$f'(0.5)\,=\,3(0.5)^2 \,-\, \frac{3}{(0.5)^4}$
= $\frac{3}{4}\,-\, \frac{30000}{625}$
• So f '(x) is −ve in this interval. That means, f(x) is decreasing in this interval.

6. Consider the third interval (1, ∞)
• 2 is a convenient point in this interval.
$f'(2)\,=\,3(2)^2 \,-\, \frac{3}{(2)^4}$
= $12\,-\, \frac{3}{16}$
• So f '(x) is +ve in this interval. That means, f(x) is increasing in this interval.

7. Now we can write about the intervals.
• f is:
   ♦ Increasing in (−∞, −1) (1, ∞)
   ♦ Decreasing in (−1, 1)

Solved example 22.81
Let f be a function defined on [a,b] such that f '(x) > 0 for all x ∈ (a,b). Then prove that f is an increasing function on (a,b).
Solution:
1. Let x₁, x₂ ∈ (a,b) such that x₁ < x₂.
2. f is defined on [a,b]. Also, [x₁,x₂] is a subset of [a,b].
So f is continuous and differentiable in [x₁,x₂]
3. By mean value theorem,
There exists a number c in (x₁,x₂) such that:
$f'(c)\,=\,\frac{f(x_2) - f(x_1)}{x_2 - x_1}$
4. Given that, f '(x) > 0 for all x ∈ (a,b).
• So f '(c) > 0
⇒ $\frac{f(x_2) - f(x_1)}{x_2 - x_1}~>~0$
⇒ f(x₂) − f(x₁) > 0
⇒ f(x₂) > f(x₁)
5. So we can write:
When x₂ > x₁, we will obtain: f(x₂) > f(x₁)
• That means, f is an increasing function on (a,b).

---

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

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22.20 - Miscellaneous Examples on Applications of Derivatives - Part 2

In the previous section, we saw some miscellaneous examples. In this section, we will see a few more miscellaneous examples.

Solved example 22.70
Find the equation of the normal to the curve x² = 4y which passes through the point (1,2).
Solution:
1. In the fig.22.71 below, the curve x² = 4y is drawn in red color.

Fig.22.70

2. We have to draw a normal to this curve. That normal should pass through (1,2).
• Substituting x = 1 and y = 2 in the equation x² = 4y, we find that, (1,2) does not lie in the given curve.
• We cannot draw "any line" from (1,2) towards the curve. The line must satisfy the following 2 conditions:
(i) The line must intersect the curve at (h,k)
(ii) The tangent of the curve at (h,k) should be perpendicular to the line.
3. So our first aim is to find h and k.
• The given equation is x² = 4y. This can be written as y = x² / 4
• So dy/dx = x/2. That means, the slope of tangent at any point x on the curve can be obtained using the equation:
slope = x/2
• Then the slope at (h,k) will be h/2
⇒ Slope of the normal passing through (1,2) and (h,k) will be -(2/h)
4. Now, slope of line passing through (1,2) and (h,k) is:
$\rm{\frac{k - 2}{h - 1}}$
5. Equating the results in (3) and (4), we get:
$\rm{\frac{k - 2}{h - 1}~=~\frac{-2}{h}}$
6. In the above step, there are two unknowns h and k. But there is only one equation.
• A second equation can be obtained based on the equation of the curve. Since (h,k) is a point on the curve x² = 4y, we get: h² = 4k
• Substituting this in (5), we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{k - 2}{h – 1}}    & {~=~}    &{\frac{-2}{h}}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{(h^2 / 4) - 2}{h – 1}}    & {~=~}    &{\frac{-2}{h}}    \\
{~\color{magenta}    3    }    &{\implies}    &{h^3 / 4 ~-~ 2h}    & {~=~}    &{-2h + 2}    \\
{~\color{magenta}    4    }    &{\implies}    &{\frac{h^3}{4}}    & {~=~}    &{2}    \\
{~\color{magenta}    5    }    &{\implies}    &{h^3}    & {~=~}    &{8}    \\
{~\color{magenta}    6    }    &{\implies}    &{h}    & {~=~}    &{2}    \\
\end{array}$                           
7. Using the equation x² = 4y again, we get:
2² = 4k ⇒ k = 1
8. So (h,k) is (2,1)
• Also, the slope of the normal = −(2/h) = −(2/2) = −1
9. Now, the normal passes through (h,k) and has a slope of −(2/h). The equation of such a line can be obtained as:
y − k = −(2/h)(x − h)
⇒ y − 1 = −1(x − 2)
⇒ y − 1 = −x + 2
⇒ x + y = 3

Solved example 22.71
Find the equation of tangents to the curve
y = cos (x+y), −2π ≤ x ≤ 2π that are parallel to the line x + 2y = 0.
Solution:
1. First we differentiate the given equation

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{\cos(x+y)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{-\sin(x+y)[1 + \frac{dy}{dx}]}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{-\sin(x+y) ~-~ \frac{dy}{dx} \sin(x+y)}    \\
{~\color{magenta}    4    }    &{\implies}    &{\frac{dy}{dx}~+~\frac{dy}{dx} \sin(x+y)}    & {~=~}    &{-\sin(x+y)}    \\
{~\color{magenta}    5    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{-\sin(x+y)}{1 ~+~ \sin(x+y)}}    \\
\end{array}$

2. The above result in (1) can be used to find the slope of tangent at any point we want.
• We want those points where slope is same as the slope of the line x + 2y = 0.
    ♦ Slope of this line is -1/2.
• Equating this to the result in (1), we get:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{-\sin(x+y)}{1 ~+~ \sin(x+y)}}    & {~=~}    &{\frac{-1}{2}}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{\sin(x+y)}{1 ~+~ \sin(x+y)}}    & {~=~}    &{\frac{1}{2}}    \\
{~\color{magenta}    3    }    &{\implies}    &{2 \sin(x+y)}    & {~=~}    &{1 ~+~ \sin(x+y)}    \\
{~\color{magenta}    4    }    &{\implies}    &{\sin(x+y)}    & {~=~}    &{1}    \\
\end{array}$

3. So we need to solve the equation sin(x+y) = 1
We can apply the first theorem (Details here)
• We know that sin(π/2) = 1.
• So we can write: $\rm{x+y~=~n\pi\,+\,(-1)^n \frac{\pi}{2}}$, where n is any integer.
• This is same as: $\rm{x+y~=~\left(n\,+\,(-1)^n \frac{1}{2}\right)\pi~=~u \pi}$
• Some of the possible values of u are tabulated below:
n = −4 ⇒ u = −3.5
n = −3 ⇒ u = −3.5
n = −2 ⇒ u = −1.5
n = −1 ⇒ u = −1.5
n =    0 ⇒ u = 0.5
n =    1 ⇒ u = 0.5
n =    2 ⇒ u = 2.5
n =    3 ⇒ u = 2.5
n =    4 ⇒ u = 4.5

• The given domain is [−2π,2π]. So the acceptable values of u are: −1.5 and 0.5.
• That means, (x+y) can have two values:
    ♦ −1.5π = −(3/2)π
    ♦    0.5π = (1/2)π.
• We can write two facts:
(i) At the point (x,y), where (x+y) = −(3/2)π, the tangent will be parallel to the given line.
(ii) At the point (x,y), where (x+y) = (1/2)π also, the tangent will be parallel to the given line.

4. We obtained the sum (x+y). If we can find x or y, we will be able to calculate the other.
• Let us calculate y. It can be done in 3 steps:
(i) Consider the two points that we determined in (3). At those two points, cos(x+y) will be zero. This is because, (x+y) is a multiple of (1/2)π.
(ii) So from the given equation of the curve, we get:
y = cos (x+y) = 0
(iii) That means, at both the points determined in (3), y will be zero

5. Now we can calculate the values of x:
• When (x + y) = −(3/2)π, we get (x+0) = −(3/2)π
⇒ x = −(3/2)π 
• When (x + y) = (1/2)π, we get (x+0) = (1/2)π
⇒ x = (1/2)π

6. So the two points are: [−(3/2)π, 0] and [(1/2)π, 0].
We can easily write the equation of the line passing through each of those points. The slope is −(1/2).
• Line through [−(3/2)π, 0]:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y-0}    & {~=~}    &{\frac{-1}{2} \left(x - (-3/2)\pi \right)}    \\
{~\color{magenta}    2    }    &{\implies}    &{y}    & {~=~}    &{\frac{-1}{2} \left(x + \frac{3 \pi}{2} \right)}    \\
{~\color{magenta}    3    }    &{\implies}    &{2y}    & {~=~}    &{-x - \frac{3 \pi}{2}}    \\
{~\color{magenta}    4    }    &{\implies}    &{4y}    & {~=~}    &{-2x - 3 \pi}    \\
{~\color{magenta}    5    }    &{\implies}    &{2x + 4y + 3 \pi}    & {~=~}    &{0}    \\
\end{array}$

• Line through [(1/2)π, 0]:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y-0}    & {~=~}    &{\frac{-1}{2} \left(x – (1/2)\pi \right)}    \\
{~\color{magenta}    2    }    &{\implies}    &{y}    & {~=~}    &{\frac{-1}{2} \left(x + \frac{\pi}{2} \right)}    \\
{~\color{magenta}    3    }    &{\implies}    &{2y}    & {~=~}    &{-x - \frac{\pi}{2}}    \\
{~\color{magenta}    4    }    &{\implies}    &{4y}    & {~=~}    &{-2x - \pi}    \\
{~\color{magenta}    5    }    &{\implies}    &{2x + 4y + \pi}    & {~=~}    &{0}    \\
\end{array}$

7. The graph is shown below:

Fig.22.71

• The given curve is drawn in red color.
• The given line is drawn in green color.
• We see that:
The two tangents (magenta color) are parallel to the green line.

Solved example 22.72
Show that the normal at any point 𝜃 to the curve
x = a cos 𝜃 + a 𝜃 sin 𝜃 , y = a sin 𝜃 − a 𝜃 cos 𝜃
is at a constant distance from the origin.
Solution:
1. First we differentiate the given equation

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x}    & {~=~}    &{a \cos \theta + a \theta \sin \theta}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{dx}{d \theta}}    & {~=~}    &{a (-\sin \theta) + a \theta (\cos \theta) + a (1) \sin \theta}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{a \theta \cos \theta}    \\
{~\color{magenta}    4    }    &{\implies}    &{y}    & {~=~}    &{a \sin \theta - a \theta \cos \theta}    \\
{~\color{magenta}    5    }    &{\implies}    &{\frac{dy}{d \theta}}    & {~=~}    &{a (\cos \theta) - [a \theta (-\sin \theta) + a (1) \cos \theta]}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{a \cos \theta + a \theta \sin \theta - a \cos \theta}    \\
{~\color{magenta}    7    }    &{{}}    &{{}}    & {~=~}    &{a \theta \sin \theta}    \\
{~\color{magenta}    8    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{dy}{d \theta} \div \frac{dx}{d \theta}}    \\
{~\color{magenta}    9    }    &{{}}    &{{}}    & {~=~}    &{\frac{a \theta \sin \theta}{a \theta \cos \theta}}    \\
{~\color{magenta}    10    }    &{{}}    &{{}}    & {~=~}    &{\tan \theta}    \\
\end{array}$

2. So the slope of tangent at any point 𝜃 is tan 𝜃.
⇒ slope of normal at any point is −(1/tan 𝜃)

3. For any point , the coordinates in terms of x and y can be written as:
[(a cos 𝜃 + a 𝜃 sin 𝜃), (a sin 𝜃 − a 𝜃 cos 𝜃)]

4. Now we have point and slope. The equation of the normal can be obtained as:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y – y_1}    & {~=~}    &{m(x – x_1)}    \\
{~\color{magenta}    2    }    &{\implies}    &{y – (a \sin \theta – a \theta \cos \theta)}    & {~=~}    &{\frac{-1}{\tan \theta} \left(x - ( a \cos \theta + a \theta \sin \theta) \right)}    \\
{~\color{magenta}    3    }    &{\implies}    &{y \tan \theta – (a \sin \theta – a \theta \cos \theta) \tan \theta}    & {~=~}    &{\left(- x + ( a \cos \theta + a \theta \sin \theta) \right)}    \\
{~\color{magenta}    4    }    &{\implies}    &{x + y \tan \theta – (a \sin \theta – a \theta \cos \theta) \tan \theta - ( a \cos \theta + a \theta \sin \theta)}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{\implies}    &{x + y \tan \theta – \left[(a \sin \theta – a \theta \cos \theta) \tan \theta + ( a \cos \theta + a \theta \sin \theta) \right]}    & {~=~}    &{0}    \\
\end{array}$

• This is in the form Ax + By + C = 0
Where, A = 1, B = tan 𝜃 and
C = a sin 𝜃 − a 𝜃 cos 𝜃 + a cos 𝜃 + a 𝜃 sin 𝜃 , y =

5. Now we can write the distance:
• Distance d of any point (x₁,y₁) from Ax + By + C = 0, is given by:
$d~=~\frac{\left|A x_1~+~B y_1~+~C \right|}{\sqrt{A^2~+~B^2}}$
• So distance d of origin from Ax + By + C = 0, is given by:
$d~=~\frac{\left|C \right|}{\sqrt{A^2~+~B^2}}$

• Thus we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{d}    & {~=~}    &{\frac{(a \sin \theta – a \theta \cos \theta) \tan \theta + ( a \cos \theta + a \theta \sin \theta)}{\sqrt{1 + \tan^2 \theta}}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{a \frac{\sin^2 \theta}{\cos \theta} – a \theta \sin \theta + a \cos \theta + a \theta \sin \theta}{\sqrt{\sec^2 \theta}}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{a \frac{\sin^2 \theta}{\cos \theta}+ a \cos \theta }{\sec \theta}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{\frac{a \sin^2 \theta ~+~ a \cos^2 \theta}{\cos \theta}}{\sec \theta}}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{a \sin^2 \theta ~+~ a \cos^2 \theta}{\cos \theta} \left(\frac{1}{\sec \theta} \right)}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\frac{a (\sin^2 \theta ~+~ \cos^2 \theta)}{1}}    \\
{~\color{magenta}    7    }    &{{}}    &{{}}    & {~=~}    &{a}    \\
\end{array}$

• So the distance is a constant.

6. The graph of the given function is shown in fig.22.72 below. Value of a is assumed as 30.

Fig.22.72

• The graph is drawn in pink color. Two random points are marked in yellow color. The normal at those points are drawn in green color.
• Both the normals have the same distance of 30 from the origin.

Solved example 22.73
The slope of the tangent to the curve
x = t² + 3t −8, y = 2t² − 2t − 5
at the point (2,−1) is
(A) 22/7    (B) 6/7    (C) 7/6    (D) −6/7.
Solution:
1. First we differentiate the given equation
• x = t² + 3t −8
⇒ dx/dt = 2t + 3
• y = 2t² − 2t − 5
⇒ dy/dt = 4t − 2
dy/dx = (dy/dt)(dt/dx) = (4t − 2)/(2t + 3)

2. Given that x = t² + 3t −8
• So for the point (2, −1), we can write:
2 = t² + 3t −8
⇒ t² + 3t −10 = 0
Solving this quadratic equation, we get: t = 2 and t = −5

3. Let us check using y values:
Given that y = 2t² − 2t − 5
• So for the point (2, −1), we can write:
−1 = 2t² − 2t − 5
⇒ 2t² − 2t − 4 = 0
⇒ t² − t − 2 = 0
Solving this quadratic equation, we get: t = 2 and t = −1

4. (t = 2) is common for both x and y. So we can write:
The point (2, −1) is obtained when t = 2

5. So the slope at t = 2 can be written as:
(dy/dx)_{t = 2} = (4(2) − 2)/(2(2) + 3) = (8 − 2)/(4 + 3) = 6/7

So the correct option is (B)

Solved example 22.74
The line y = mx + 1 is a tangent to the curve y² = 4x if the value of m is
(A) 1    (B) 2    (C) 3    (D) 1/2.
Solution:
1. Fig.22.73 below, shows the rough sketch of the curve y² = 4x
• We assume that, the line y = mx + 1 touches the curve at (h,k).

Fig.22.73

2. First we differentiate the equation y² = 4x
2y(dy/dx) = 4
⇒ dy/dx = 4/(2y) = 2/y
• So the slope at (h,k) will be: 2/k

3. The line y = mx + 1 is the tangent at (h,k).
So m = 2/k
• Therefore, the equation of the line becomes:
y = (2/k)x + 1

4. Points of intersection of the line and curve can be obtained by solving their equations:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y^2}    & {~=~}    &{4x}    \\
{~\color{magenta}    2    }    &{\implies}    &{\left(\frac{2x}{k}~+~1 \right)^2}    & {~=~}    &{4x}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{4x^2}{k^2}~+~\frac{4x}{k}~+~1}    & {~=~}    &{4x}    \\
{~\color{magenta}    4    }    &{\implies}    &{4x^2 + 4kx + k^2}    & {~=~}    &{4k^2 x}    \\
{~\color{magenta}    5    }    &{\implies}    &{4x^2 + 4kx – 4k^2 x + k^2}    & {~=~}    &{0}    \\
{~\color{magenta}    6    }    &{\implies}    &{4x^2 + 4k(1 – k)x + k^2}    & {~=~}    &{0}    \\
\end{array}$                           

• For finding the points of intersection, we need to solve the above quadratic equation.
• But since the line is a tangent, there will be only one point of intersection. That means, for the above quadratic equation, there will be only one solution.
• Recall that, if the quadratic equation ax² + bx + c has only one solution, the discriminant b² - 4ac will be zero.
• So for our present quadratic equation, we get:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{[4k(1 – k)]^2 ~-~ 4(4)k^2}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{\implies}    &{16 k^2 (1-k)^2 ~-~ 4(4)k^2}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{\implies}    &{16 k^2 (1-2k + k^2) ~-~ 4(4)k^2}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{\implies}    &{16 k^2 – 32 k^3 + 16 k^4 ~-~ 16 k^2}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{\implies}    &{– 32 k^3 + 16 k^4}    & {~=~}    &{0}    \\
{~\color{magenta}    6    }    &{\implies}    &{– 2 +  k}    & {~=~}    &{0}    \\
{~\color{magenta}    7    }    &{\implies}    &{k}    & {~=~}    &{2}    \\
\end{array}$

5. So from (3), we get:
m = 2/k = 2/2 = 1
• Thus the correct option is (A).

Solved example 22.75
The normal at the point (1,1) on the curve
2y + x² = 3 is
(A) x+y=0    (B) x−y=0    (C) x+y+1=0    (D) x−y=0.
Solution:
1. First we differentiate the equation of the curve. We get:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{2y + x^2}    & {~=~}    &{3}    \\
{~\color{magenta}    2    }    &{\implies}    &{2 \frac{dy}{dx} + 2x}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{dy}{dx} + x}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{-x}    \\
\end{array}$                           
• Using this result, we can find the slope of tangent at any point.

2. The slope of tangent  at (1,1) will be −1.
• So the slope of normal at (1,1) will be the −ve reciprocal, which is 1.

3. The normal has a slope of 1 and it passes through (1,1).
• So the equation of the normal can be obtained as:
y − 1 = 1(x − 1)
⇒ y − 1 = x − 1
⇒ x − y = 0
• So the correct option is (B)

Solved example 22.76
The points on the curve 9y² = x³, where the normal to the curve makes equal intercepts with the axes are
$\rm{(A)~\left(4,\,\pm \frac{8}{3} \right)~~~(B)~\left(4,\,- \frac{8}{3} \right)~~~(C)~\left(4,\,\pm \frac{3}{8} \right)~~~(D)~\left(\pm4,\, \frac{3}{8} \right)}$
Solution:
1. First we differentiate the equation of the curve. We get:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{9 y^2}    & {~=~}    &{x^3}    \\
{~\color{magenta}    2    }    &{\implies}    &{9(2y)\frac{dy}{dx}}    & {~=~}    &{3x^2}    \\
{~\color{magenta}    3    }    &{\implies}    &{(6y)\frac{dy}{dx}}    & {~=~}    &{x^2}    \\
{~\color{magenta}    4    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{x^2}{6y}}    \\
\end{array}$                           

• Using this result, we can find the slope of tangent at any point.
• Therefore, the slope of normal at any point will be given by the −ve reciprocal, which is: $\rm{\frac{-6y}{x^2}}$

2. Let (h,k) be the point on the curve at which, the normal is drawn.
• Then the slope of that normal will be: $\rm{\frac{-6k}{h^2}}$
• So we have a "point on the normal" and the "slope of the normal". Then the equation of the normal can be written as:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y-k}    & {~=~}    &{\frac{-6k}{h^2}(x-h)}    \\
{~\color{magenta}    2    }    &{\implies}    &{y-k}    & {~=~}    &{\frac{-6kx}{h^2} ~+~\frac{6k}{h}}    \\
{~\color{magenta}    3    }    &{\implies}    &{y}    & {~=~}    &{\frac{-6kx}{h^2} ~+~\frac{6k}{h} ~+~k}    \\
\end{array}$

• This equation is in the form y = mx + c
• So the y-intercept c is $\rm{\frac{6k}{h} ~+~k}$

3. The intercept form of any line is:
$\rm{\frac{x}{a} + \frac{y}{b} = 1}$
Where 'a' and 'b' are the x and y intercepts respectively.
• In our present case, the intercepts are equal. So the equation becomes:
$\rm{\frac{x}{a} + \frac{y}{a} = 1}$
• So we can write:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x}{a} + \frac{y}{b}}    & {~=~}    &{1}    \\
{~\color{magenta}    2    }    &{{}}    &{\frac{x}{\frac{6k}{h} ~+~k} + \frac{y}{\frac{6k}{h} ~+~k}}    & {~=~}    &{1}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{h}{\frac{6k}{h} ~+~k} + \frac{k}{\frac{6k}{h} ~+~k}}    & {~=~}    &{1}    \\
{~\color{magenta}    4    }    &{\implies}    &{\frac{h}{\frac{6k + kh}{h}} + \frac{k}{\frac{6k + kh}{h}}}    & {~=~}    &{1}    \\
{~\color{magenta}    5    }    &{\implies}    &{\frac{h^2}{6k + kh}~+~\frac{kh}{6k + kh}}    & {~=~}    &{1}    \\
{~\color{magenta}    6    }    &{\implies}    &{h^2 + kh}    & {~=~}    &{6k + kh}    \\
{~\color{magenta}    7    }    &{\implies}    &{h^2}    & {~=~}    &{6k}    \\
\end{array}$

4. We need one more equation connecting h and k. For that, we can use the equation of the curve. We get: 9k² = h³

5. So the two equations are:
   ♦ h² = 6k
   ♦ 9k² = h³
• We get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{h^2}    & {~=~}    &{6k}    \\
{~\color{magenta}    2    }    &{{}}    &{9 k^2}    & {~=~}    &{h^3}    \\
{~\color{magenta}    3    }    &{\implies}    &{k^2}    & {~=~}    &{\frac{h^3}{9} ~=~\left(\frac{h^2}{6} \right)^2}    \\
{~\color{magenta}    4    }    &{\implies}    &{\frac{h^3}{9}}    & {~=~}    &{\frac{h^4}{36}}    \\
{~\color{magenta}    5    }    &{\implies}    &{h}    & {~=~}    &{\frac{36}{9}}    \\
{~\color{magenta}    6    }    &{\implies}    &{h}    & {~=~}    &{4}    \\
\end{array}$

6. Substituting this value of h in the other equation, we get:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{9k^2}    & {~=~}    &{h^3}    \\
{~\color{magenta}    2    }    &{{}}    &{9 k^2}    & {~=~}    &{4^3}    \\
{~\color{magenta}    3    }    &{\implies}    &{k^2}    & {~=~}    &{\frac{16(4)}{9}}    \\
{~\color{magenta}    4    }    &{\implies}    &{k}    & {~=~}    &{\frac{\pm 4(\pm 2)}{\pm 3}}    \\
{~\color{magenta}    5    }    &{\implies}    &{k}    & {~=~}    &{\pm{\frac{8}{3}}}    \\
\end{array}$                           

7. So the point (h,k) is
option (A): $\rm{\left(4,\,\pm \frac{8}{3} \right)}$

8. Fig.22.74 below shows the graph:

Fig.22.74

• The curve is plotted in red color.
• The green lines are the normals at $\rm{\left(4,\,\pm \frac{8}{3} \right)}$
• We see that, both the green lines have equal x and y intercepts.

---

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

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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 maximum 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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