Question 1

Find the rate of change of the area of a circle with respect to its radius \(r\) when (a) \(r = 3 \text{ cm}\) (b) \(r = 4 \text{ cm}\).

Solution:

The area \(A\) of a circle with radius \(r\) is given by \(A = \pi r^2\).

The rate of change of area with respect to its radius is: \[ \frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2\pi r \]

(a) When \(r = 3 \text{ cm}\): \[ \frac{dA}{dr} = 2\pi(3) = 6\pi \text{ cm}^2/\text{cm} \]

(b) When \(r = 4 \text{ cm}\): \[ \frac{dA}{dr} = 2\pi(4) = 8\pi \text{ cm}^2/\text{cm} \]

Question 2

The volume of a cube is increasing at the rate of \(8 \text{ cm}^3/\text{s}\). How fast is the surface area increasing when the length of an edge is \(12 \text{ cm}\)?

Solution:

Let \(x\) be the length of the side, \(V\) be the volume, and \(S\) be the surface area of the cube.

We have \(V = x^3\) and \(S = 6x^2\).

Given: \(\frac{dV}{dt} = 8 \text{ cm}^3/\text{s}\).

Differentiating \(V = x^3\) with respect to time \(t\): \[ \frac{dV}{dt} = 3x^2 \frac{dx}{dt} \implies 8 = 3x^2 \frac{dx}{dt} \implies \frac{dx}{dt} = \frac{8}{3x^2} \]

Now, differentiate \(S = 6x^2\) with respect to \(t\): \[ \frac{dS}{dt} = 12x \frac{dx}{dt} \]

Substituting the value of \(\frac{dx}{dt}\): \[ \frac{dS}{dt} = 12x \left(\frac{8}{3x^2}\right) = \frac{32}{x} \]

When \(x = 12 \text{ cm}\): \[ \frac{dS}{dt} = \frac{32}{12} = \frac{8}{3} \text{ cm}^2/\text{s} \]

Question 3

The radius of a circle is increasing uniformly at the rate of \(3 \text{ cm}/\text{s}\). Find the rate at which the area of the circle is increasing when the radius is \(10 \text{ cm}\).

Solution:

Area \(A = \pi r^2\). Given: \(\frac{dr}{dt} = 3 \text{ cm}/\text{s}\).

Rate of change of area with respect to time: \[ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \frac{dr}{dt} \]

When \(r = 10 \text{ cm}\): \[ \frac{dA}{dt} = 2\pi(10)(3) = 60\pi \text{ cm}^2/\text{s} \]

Question 4

An edge of a variable cube is increasing at the rate of \(3 \text{ cm}/\text{s}\). How fast is the volume of the cube increasing when the edge is \(10 \text{ cm}\) long?

Solution:

Let \(x\) be the edge of the cube. Given: \(\frac{dx}{dt} = 3 \text{ cm}/\text{s}\).

Volume \(V = x^3\). Rate of change of volume: \[ \frac{dV}{dt} = 3x^2 \frac{dx}{dt} \]

When \(x = 10 \text{ cm}\): \[ \frac{dV}{dt} = 3(10)^2(3) = 900 \text{ cm}^3/\text{s} \]

Question 5

A stone is dropped into a quiet lake and waves move in circles at the speed of \(5 \text{ cm}/\text{s}\). At the instant when the radius of the circular wave is \(8 \text{ cm}\), how fast is the enclosed area increasing?

Solution:

Given: \(\frac{dr}{dt} = 5 \text{ cm}/\text{s}\). Area \(A = \pi r^2\).

Rate of change of area: \[ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \]

When \(r = 8 \text{ cm}\): \[ \frac{dA}{dt} = 2\pi(8)(5) = 80\pi \text{ cm}^2/\text{s} \]

Question 6

The radius of a circle is increasing at the rate of \(0.7 \text{ cm}/\text{s}\). What is the rate of increase of its circumference?

Solution:

Circumference \(C = 2\pi r\). Given: \(\frac{dr}{dt} = 0.7 \text{ cm}/\text{s}\).

Rate of change of circumference: \[ \frac{dC}{dt} = 2\pi \frac{dr}{dt} = 2\pi(0.7) = 1.4\pi \text{ cm}/\text{s} \]

Question 7

The length \(x\) of a rectangle is decreasing at the rate of \(5 \text{ cm}/\text{min}\) and the width \(y\) is increasing at the rate of \(4 \text{ cm}/\text{min}\). When \(x = 8 \text{ cm}\) and \(y = 6 \text{ cm}\), find the rates of change of (a) the perimeter, and (b) the area of the rectangle.

Solution:

Given: \(\frac{dx}{dt} = -5 \text{ cm}/\text{min}\) (negative because it is decreasing) and \(\frac{dy}{dt} = 4 \text{ cm}/\text{min}\).

(a) Perimeter \(P = 2(x + y)\). \[ \frac{dP}{dt} = 2\left(\frac{dx}{dt} + \frac{dy}{dt}\right) = 2(-5 + 4) = -2 \text{ cm}/\text{min} \]

(b) Area \(A = xy\). \[ \frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt} = (8)(4) + (6)(-5) = 32 - 30 = 2 \text{ cm}^2/\text{min} \]

Question 8

A balloon, which always remains spherical on inflation, is being inflated by pumping in \(900 \text{ cm}^3/\text{s}\) of gas. Find the rate at which the radius of the balloon increases when the radius is \(15 \text{ cm}\).

Solution:

Volume of a sphere \(V = \frac{4}{3}\pi r^3\). Given: \(\frac{dV}{dt} = 900 \text{ cm}^3/\text{s}\).

\[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \implies 900 = 4\pi r^2 \frac{dr}{dt} \]

When \(r = 15 \text{ cm}\): \[ 900 = 4\pi(15)^2 \frac{dr}{dt} = 4\pi(225) \frac{dr}{dt} = 900\pi \frac{dr}{dt} \]

\[ \frac{dr}{dt} = \frac{1}{\pi} \text{ cm}/\text{s} \]

Question 9

A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is \(10 \text{ cm}\).

Solution:

Volume \(V = \frac{4}{3}\pi r^3\). The rate of change of volume with respect to radius is: \[ \frac{dV}{dr} = 4\pi r^2 \]

When \(r = 10 \text{ cm}\): \[ \frac{dV}{dr} = 4\pi(10)^2 = 400\pi \text{ cm}^3/\text{cm} \]

Question 10

A ladder \(5 \text{ m}\) long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of \(2 \text{ cm}/\text{s}\). How fast is its height on the wall decreasing when the foot of the ladder is \(4 \text{ m}\) away from the wall?

Solution:

Let \(x\) be the distance of the foot from the wall and \(y\) be the height on the wall. By Pythagoras theorem: \(x^2 + y^2 = 5^2 = 25\). Given: \(\frac{dx}{dt} = 2 \text{ cm}/\text{s}\).

Differentiating wrt \(t\): \[ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \implies y\frac{dy}{dt} = -x\frac{dx}{dt} \]

When \(x = 4 \text{ m}\), \(y = \sqrt{25 - 4^2} = 3 \text{ m}\). Substituting values: \[ 3 \frac{dy}{dt} = -(4)(2) \implies \frac{dy}{dt} = -\frac{8}{3} \text{ cm}/\text{s} \]

The height is decreasing at the rate of \(\frac{8}{3} \text{ cm}/\text{s}\).

Question 11

A particle moves along the curve \(6y = x^3 + 2\). Find the points on the curve at which the \(y\)-coordinate is changing \(8\) times as fast as the \(x\)-coordinate.

Solution:

Given curve: \(6y = x^3 + 2\). Differentiating wrt \(t\): \[ 6 \frac{dy}{dt} = 3x^2 \frac{dx}{dt} \implies 2 \frac{dy}{dt} = x^2 \frac{dx}{dt} \]

Given condition: \(\frac{dy}{dt} = 8 \frac{dx}{dt}\). Substituting this: \[ 2(8 \frac{dx}{dt}) = x^2 \frac{dx}{dt} \implies 16 = x^2 \implies x = \pm 4 \]

When \(x = 4\), \(6y = 4^3 + 2 = 66 \implies y = 11\). Point: \((4, 11)\).

When \(x = -4\), \(6y = (-4)^3 + 2 = -62 \implies y = -31/3\). Point: \((-4, -31/3)\).

Question 12

The radius of an air bubble is increasing at the rate of \(1/2 \text{ cm}/\text{s}\). At what rate is the volume of the bubble increasing when the radius is \(1 \text{ cm}\)?

Solution:

Given: \(\frac{dr}{dt} = \frac{1}{2} \text{ cm}/\text{s}\). Volume \(V = \frac{4}{3}\pi r^3\).

\[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \]

When \(r = 1 \text{ cm}\): \[ \frac{dV}{dt} = 4\pi(1)^2 \left(\frac{1}{2}\right) = 2\pi \text{ cm}^3/\text{s} \]

Question 13

A balloon, which always remains spherical, has a variable diameter \(\frac{3}{2}(2x + 1)\). Find the rate of change of its volume with respect to \(x\).

Solution:

Diameter \(D = \frac{3}{2}(2x + 1) \implies \text{radius } r = \frac{3}{4}(2x + 1)\).

Volume \(V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \left[ \frac{3}{4}(2x+1) \right]^3 = \frac{4}{3}\pi \cdot \frac{27}{64}(2x+1)^3 = \frac{9\pi}{16}(2x+1)^3\).

Rate of change wrt \(x\): \[ \frac{dV}{dx} = \frac{9\pi}{16} \cdot 3(2x+1)^2 \cdot 2 = \frac{27\pi}{8}(2x+1)^2 \]

Question 14

Sand is pouring from a pipe at the rate of \(12 \text{ cm}^3/\text{s}\). The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is \(4 \text{ cm}\)?

Solution:

Given: \(\frac{dV}{dt} = 12 \text{ cm}^3/\text{s}\) and \(h = \frac{1}{6}r \implies r = 6h\).

Volume of cone \(V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(6h)^2h = 12\pi h^3\).

Differentiating wrt \(t\): \[ \frac{dV}{dt} = 36\pi h^2 \frac{dh}{dt} \implies 12 = 36\pi h^2 \frac{dh}{dt} \]

When \(h = 4 \text{ cm}\): \[ 12 = 36\pi(4^2)\frac{dh}{dt} \implies 12 = 576\pi \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{1}{48\pi} \text{ cm}/\text{s} \]

Question 15

The total cost \(C(x)\) in Rupees associated with the production of \(x\) units of an item is given by \(C(x) = 0.007x^3 - 0.003x^2 + 15x + 4000\). Find the marginal cost when \(17\) units are produced.

Solution:

Marginal cost \(MC = \frac{dC}{dx} = 0.021x^2 - 0.006x + 15\).

When \(x = 17\): \[ MC = 0.021(17)^2 - 0.006(17) + 15 = 0.021(289) - 0.102 + 15 = 6.069 - 0.102 + 15 = 20.967 \]

Marginal Cost is ₹ 20.97 (approx).

Question 16

The total revenue in Rupees received from the sale of \(x\) units of a product is given by \(R(x) = 13x^2 + 26x + 15\). Find the marginal revenue when \(x = 7\).

Solution:

Marginal Revenue \(MR = \frac{dR}{dx} = 26x + 26\).

When \(x = 7\): \[ MR = 26(7) + 26 = 182 + 26 = 208 \]

Marginal Revenue is ₹ 208.

Question 17

The rate of change of the area of a circle with respect to its radius \(r\) at \(r = 6 \text{ cm}\) is: (A) \(10\pi\) (B) \(12\pi\) (C) \(8\pi\) (D) \(11\pi\).

Solution:

\[ \frac{dA}{dr} = 2\pi r = 2\pi(6) = 12\pi \]

Correct option is (B).

Question 18

The total revenue in Rupees received from the sale of \(x\) units of a product is given by \(R(x) = 3x^2 + 36x + 5\). The marginal revenue, when \(x = 15\) is: (A) \(116\) (B) \(96\) (C) \(90\) (D) \(126\).

Solution:

Marginal Revenue \(MR = 6x + 36\).

When \(x = 15\): \[ MR = 6(15) + 36 = 90 + 36 = 126 \]

Correct option is (D).