Wednesday, January 1, 2025

22.22 - Miscellaneous Examples on Applications of Derivatives - Part 4

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

Solved example 22.82
A circular disc of radius 3 cm is being heated. Due to expansion, it's radius increases at the rate of 0.05 cm/s. Find the rate at which its area is increasing when radius is 3.2 cm.
Solution:
1. We have the formula for area of a circle: A = πr2.
2. The radius increases at the rate of 0.05 cm/s.
• So radius at time t will be 0.05t
4. So area A at time t = π(0.05t)2 = 0.0025πt2 cm2
• Now the rate of change of area w.r.t radius can be obtained as:
dA/dt = 0.0025π(2t) = 0.005 πt cm2/s
• This result can be used to find the rate of change of area at any instant.
5. We want the rate when radius is 3.2 cm. So we want the instant at which the radius is 3.2 cm. For that, we can use the result in (2). We get:
3.2 = 0.05t
⇒ t = 64
6. So from (4) we get:
(dA/dt)t=64 = 0.005π(64) = 0.320π cm2/s

Solved example 22.83
An open topped box is to be constructed by removing equal squares from each corner of a 3 metre by 8 metre rectangular sheet of aluminium and folding up the sides. Find the volume of the largest such box.
Solution:
Step I: Writing the problem as a function
1. In fig.22.77 below, the original rectangular sheet is shown in yellow color. Red squares are cut off from the corners.

Fig.22.77

• So when the flaps are folded up, we will get a box with:
   ♦ base a rectangle (8 − 2x) × (3 − 2x) m2
   ♦ height x m
• Then the volume of the box = x(8 − 2x)(3 − 2x) m3

2. It is clear that, V depends on x. That means, V is a function of x. So we can write:
V = f(x) = x(8 − 2x)(3 − 2x) m3

3. We want V to be as large as possible. That means, we want the largest of all 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:
f '(x) =12x2 − 44x + 24

f ''(x) = 24x − 44

2. Equate the first derivative to zero and solve for x:
12x2 − 44x + 24 = 0
⇒ x = 2/3, x = 3 

3. So the only two points in category I are: x = 2/3 and x = 3
4. We obtained f '(x) = 12x2 − 44x + 24
• 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 = 2/3 and x = 3

Step III: Find f ''(x) at the critical points
1. f ''(2/3) = 24(2/3) − 44 = −28
2. f ''(3) =  24(3) − 44 = 28

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

Step V: Finding the actual maximum value:
• The local maximum value at x = 2/3 is given by:
f(2/3) = (2/3)[8 − 2(2/3)][3 − 2(2/3)] = 200/27 m3.

Solved example 22.84
Manufacturer can sell x items at a price of rupees $\rm{5 - \frac{x}{100}}$ each. The cost price of x items is Rs $\rm{\frac{x}{5} + 500}$. Find the number of items he should sell to earn maximum profit.
Solution:
Step I: Writing the problem as a function
1. When the manufacturer sells x items, he will get an income of: Rs $\rm{x\left(5 - \frac{x}{100} \right)}$
2. To produce those x items, he would have spent:
Rs $\rm{\frac{x}{5} + 500}$
3. So the profit P can be obtained as:
$\rm{P\,=\, x\left(5 - \frac{x}{100} \right)~-~\frac{x}{5} + 500}$
4. It is clear that, P depends on x. That means, P is a function of x. So we can write:
$\rm{P\,=\, f(x)\,=\,x\left(5 - \frac{x}{100} \right)~-~\frac{x}{5} + 500}$
5. We want P to be as large as possible. That means, we want the largest of all 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:
f '(x) = $\rm{\frac{-x}{50} + \frac{24}{5}}$

f ''(x) = $\rm{\frac{-1}{50}}$

2. Equate the first derivative to zero and solve for x:
$\rm{\frac{-x}{50} + \frac{25}{4} \,=\, 0}$
⇒ x = 240

3. So the only one point in category I is: x = 240
4. We obtained f '(x) = $\rm{\frac{-x}{50} + \frac{24}{5}}$
• 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 = 240

Step III: Find f ''(x) at the critical points
f ''(x) is a constant: $\rm{\frac{-1}{50}}$

Step IV: Applying the second derivative test
• Since f ''(240) is −ve, there is a local maximum at x = 240

Step V: Final result:
• For maximum profit, the manufacturer must sell 240 items.

Solved example 22.85
Find the maximum area of an isosceles triangle inscribed in the ellipse $\rm{\frac{x^2}{a^2}\,+\,\frac{y^2}{b^2}\,=\,1}$ with it's vertex at one end of the major axis.
Solution:
Step I: Writing the problem as a function
1. The given ellipse is: $\rm{\frac{x^2}{a^2}\,+\,\frac{y^2}{b^2}\,=\,1}$
• So it is clear that:
    ♦ Major axis lies along the x-axis.
    ♦ Minor axis lies along the y-axis.
    ♦ Center of ellipse is at the origin O.
2. Thus we get fig.22.78 below:

Fig.22.78

• The vertex A of the isosceles triangle is at one end of the major axis.
3. Let the base of the isosceles triangle intersect the x-axis at D(h,0).
• Then the y-coordinates of B and C can be obtained as shown below:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x^2}{a^2}\,+\,\frac{y^2}{b^2}}    & {~=~}    &{1}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{h^2}{a^2}\,+\,\frac{y^2}{b^2}}    & {~=~}    &{1}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{y^2}{b^2}}    & {~=~}    &{1~-~\frac{h^2}{a^2}}    \\
{~\color{magenta}    4    }    &{\implies}    &{y^2}    & {~=~}    &{b^2 \left(1~-~\frac{h^2}{a^2} \right)}    \\
{~\color{magenta}    5    }    &{\implies}    &{y}    & {~=~}    &{\pm \sqrt{b^2 \left(1~-~\frac{h^2}{a^2} \right)}}    \\
\end{array}$

• So the y-coordinate of B is $\sqrt{b^2 \left(1~-~\frac{h^2}{a^2} \right)}$
• And the y-coordinate of C is $-\sqrt{b^2 \left(1~-~\frac{h^2}{a^2} \right)}$
• Then the distance BC will be $2 \sqrt{b^2 \left(1~-~\frac{h^2}{a^2} \right)}$
4. Next we want the coordinates of A. It can be calculated as shown below:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x^2}{a^2}\,+\,\frac{y^2}{b^2}}    & {~=~}    &{1}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{x^2}{a^2}\,+\,\frac{0^2}{b^2}}    & {~=~}    &{1}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{x^2}{a^2}}    & {~=~}    &{1}    \\
{~\color{magenta}    4    }    &{\implies}    &{x^2}    & {~=~}    &{a^2}    \\
{~\color{magenta}    5    }    &{\implies}    &{x}    & {~=~}    &{\pm a}    \\
\end{array}$

• So the coordinates of A are: (a,0).
• Then the altitude AD = (h+a)
5. Now we can write the expression for the area of triangle ABC:
Area = 1/2 × Base × Altitude = 1/2 × BC × AD
= $\frac{1}{2} × 2 \sqrt{b^2 \left(1~-~\frac{h^2}{a^2} \right)} × (h+a)$
• This can be simplified as:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\text{Area}}    & {~=~}    &{\frac{1}{2} × 2 \sqrt{b^2 \left(1~-~\frac{h^2}{a^2} \right)} × (h+a)}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{b \sqrt{\left(1~-~\frac{h^2}{a^2} \right)} × (h+a)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{b \sqrt{\left(\frac{a^2 – h^2}{a^2} \right)} × (h+a)}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{b}{a} \sqrt{\left(a^2 – h^2 \right)} × (h+a)}    \\
\end{array}$

6. It is clear that, Area A depends on h. That means, A is a function of h. So we can write:
$\rm{A\,=\, f(x)\,=\,\frac{b}{a} \sqrt{\left(a^2 – h^2 \right)} × (h+a)}$
5. We want A to be as large as possible. That means, we want the largest of all 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:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(h)}    & {~=~}    &{\frac{b}{a} \sqrt{\left(a^2 – h^2 \right)} × (h+a)}    \\
{~\color{magenta}    2    }    &{\implies}    &{f'(h)}    & {~=~}    &{\frac{b}{a} \left[\sqrt{\left(a^2 – h^2 \right)} × (1) ~+~(h+a) (1/2) \left(a^2 – h^2 \right)^{-1/2} (-2h) \right]}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{b}{a} \left[\sqrt{\left(a^2 – h^2 \right)}  ~-~ \frac{h(h+a)}{\sqrt{\left(a^2 – h^2 \right)}} \right]}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{b}{a} \left[\frac{a^2 – h^2 ~-~h(h+a)}{\sqrt{\left(a^2 – h^2 \right)}} \right]}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{b}{a} \left[\frac{a^2 – h^2 – h^2 – ha}{\sqrt{\left(a^2 – h^2 \right)}} \right]}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\frac{b}{a} \left[\frac{a^2 – 2 h^2  – ha}{\sqrt{\left(a^2 – h^2 \right)}} \right]}    \\
\end{array}$                           

• In this problem, calculation of f ''(h) will be a lengthy process. So we will use the first derivative test

2. Equate the first derivative to zero and solve for h:
$\rm{a^2 – 2 h^2  – ha \,=\, 0}$
⇒ $\rm{2 h^2 + ha - a^2 \,=\, 0}$
• Using quadratic formula, we get:
h = a/2 and h = −a
• h cannot be −a. Because, then in fig.22.78 above, D will coincide with the left end of the major axis. Parts of the triangle will fall outside the ellipse.

3. So the only one point in category I is: x = a/2
4. We obtained f '(h) = $\frac{b}{a} \left[\frac{a^2 – 2 h^2  – ha}{\sqrt{\left(a^2 – h^2 \right)}} \right]$
• This function is not defined when h = a. But when h = a, the function itself becomes invalid because there will be no inscribed triangle.
• Therefore, there will be no points in category II.
5. So the only one critical point is: h = a/2
6. The domain is (−a,a).
• Based on the critical point, the domain can be divided into two intervals: (−a, a/2) and (a/2, a)

Step III: Analyzing the signs of f '(x) in the two intervals
1. Consider the first interval. Zero is a convenient point.
• We get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f'(h)}    & {~=~}    &{\frac{b}{a} \left[\frac{a^2 – 2 h^2  – ha}{\sqrt{\left(a^2 – h^2 \right)}} \right]}    \\
{~\color{magenta}    2    }    &{\implies}    &{f'(0)}    & {~=~}    &{\frac{b}{a} \left[\frac{a^2 – 2 (0)^2  – (0)a}{\sqrt{\left(a^2 – (0)^2 \right)}} \right]}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{b}{a} \left[\frac{a^2}{\sqrt{\left(a^2 \right)}} \right]}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{b}    \\
\end{array}$

• So in the first interval. f '(x) is +ve.

2. Consider the second interval. $\frac{3a}{4}$  is a convenient point.
• We get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f'(h)}    & {~=~}    &{\frac{b}{a} \left[\frac{a^2 – 2 h^2  – ha}{\sqrt{\left(a^2 – h^2 \right)}} \right]}    \\
{~\color{magenta}    2    }    &{\implies}    &{f'(3a/4)}    & {~=~}    &{\frac{b}{a} \left[\frac{a^2 – 2 (3a/4)^2  – (3a/4)a}{\sqrt{\left(a^2 – (3a/4)^2 \right)}} \right]}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{b}{a} \left[\frac{a^2 – 2 \left(\frac{9 a^2}{16} \right)  – \frac{3 a^2}{4}}{\sqrt{\left(a^2 – \frac{9 a^2}{16} \right)}} \right]}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{b}{a} \left[\frac{\frac{16 a^2 - 18 a^2 – 12 a^2}{16}}{\sqrt{\left(\frac{16 a^2 – 9 a^2}{16} \right)}} \right]}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{b}{a} \left[\frac{\frac{16 a^2 - 30 a^2}{16}}{\sqrt{\left(\frac{7 a^2}{16} \right)}} \right]}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\frac{b}{a} \left[\frac{\frac{- 14 a^2}{16}}{\sqrt{\left(\frac{7 a^2}{16} \right)}} \right]}    \\
\end{array}$

• So in the second interval. f '(x) is −ve.

Step IV: Applying the first derivative test
• Since f '(x) changes sign from +ve to −ve at the critical point, we have a maximum at that critical point.

Step V: Final result:
• For maximum area, h = a/2
So from step I, we get:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\text{Area}}    & {~=~}    &{\frac{1}{2} × 2 \sqrt{b^2 \left(1~-~\frac{h^2}{a^2} \right)} × (h+a)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\text{Area (max)}}    & {~=~}    &{\sqrt{b^2 \left(1~-~\frac{(a/2)^2}{a^2} \right)} × ((a/2)+a)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\sqrt{b^2 \left(1~-~\frac{a^2 / 4}{a^2} \right)} × (3a/2)}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\sqrt{3 b^2 / 4} × (3a/2)}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{\sqrt{3} b}{2} × \frac{3a}{2}}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\frac{3 \sqrt{3} \,ab}{4}}    \\
\end{array}$


The link below gives a few more miscellaneous examples:

Miscellaneous Exercise


In the next section, we will see Maxima and Minima.

Previous

Contents


Copyright©2024 Higher secondary mathematics.blogspot.com

Sunday, December 29, 2024

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.

Previous

Contents

Next

Copyright©2024 Higher secondary mathematics.blogspot.com

Wednesday, December 25, 2024

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.

Previous

Contents

Next

Copyright©2024 Higher secondary mathematics.blogspot.com