# NCERT Solutions for Class 12 Maths Chapter 6 - Application of Derivatives - Exercise 6.1

### Access Exercises of Class 12 Maths Chapter 6 Application of Derivatives

Exercise 6.1 Solutions: 18 Questions (10 Long, 6 Short, 2 MCQs)

Exercise 6.2 Solutions: 19 Questions (10 Long, 7 Short, 2 MCQs)

Exercise 6.3 Solutions: 27 Questions (14 Long, 11 Short, 2 MCQs)

Exercise 6.4 Solutions: 9 Questions (7 Short, 2 MCQs)

Exercise 6.5 Solutions: 29 Questions (15 Long, 11 Short, 3 MCQs)

Miscellaneous Exercise Solutions: 24 Questions (14 Long, 4 Short, 6 MCQs)

**Exercise 6.1**

**1. Find the rate of change of the area of a circle with respect of its radius r when**

**(a) r = 3 cm (b) r = 4 cm**

**Sol.** Let A denotes the area of the circle when its radius is r, then A = *π*r^{2}.

Now , the rate of change of the area with respect to its radius is given by,

$$\frac{\text{dA}}{\text{dr}}=\frac{d}{dr}(\pi r^2)=2\pi r$$

(a) When r = 3 cm,

$$\bigg(\frac{\text{dA}}{dr}\bigg)_{r=3}=2\pi(3)$$

= 6*π* cm^{2} per cm

Hence, the area of the circle is changing at the rate of 6 cm^{2}/cm when its radius is 3 cm.

(b) When r = 4 cm,

$$\bigg(\frac{dA}{dr}\bigg)_{r=4}=\text{2}\pi(4)\\=8\pi\space\text{cm}^2\text{per cm}$$

Hence, the area of the circle is changing at the rate of 8*π* cm^{2}/cm when its radius is 4 cm.

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

**Sol.** Let x be the length of a side (edge), V be the volume and S be the surface area of the cube.

V = x^{3} and S = 6x^{2}

(∵ Cube has six square faces, each of side x.)

where, x is a function of time t. It is given that

$$\frac{dV}{dt}= 8\space\text{cm}^3/s. $$

Then, by using the chain rule, we have

$$8=\frac{dV}{dt}=\frac{d}{dt}(x^3)\\=3x^2.\frac{dx}{dt}\\\Rarr\space\frac{dx}{dt}=\frac{8}{3x^2}\space\text{...(i)}\\\text{Now,\space}\frac{dS}{dt}=\frac{d}{dt}(6x^2)\\=12x\frac{dx}{dt}\\=12x\bigg(\frac{8}{3x^2}\bigg)=\frac{32}{x}$$

[From equation (i)]

Thus, when x = 12 cm,

$$\frac{dS}{dt}=\frac{32}{12}\space\text{cm}^2/s\\=\frac{8}{3}\space\text{cm}^2/s.$$

Hence, if the length of the edge of a cube is 12 cm, then the surface area is increasing at the rate of

$$\frac{8}{3}\space\text{cm}^2/s.$$

**3. The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.**

**Sol.** Let r be the radius of circle and A be the area of circle at any time t. Area of circle A = *π*r^{2}.

Now , the rate of change of area A with respect to time t is given by

$$\frac{\text{dA}}{\text{dt}}=\frac{d}{dt}(\pi r^2)= 2\pi r\space\frac{dr}{dt}\\\text{(By chain rule)}$$

It is given that radius of circle is increasing at the rate of 3 cm/s,

$$\text{so}\space\frac{dr}{dt}=\text{3 cm/s}.\\\therefore\space\frac{dA}{dt}=2\pi r(3)=6\pi r\space\text{cm}^2/s$$

Thus, when r = 10 cm,

$$\frac{\text{dA}}{dt}=6\pi(10)= 60\pi\space\text{cm}^2/s.$$

Hence, the rate at which the area of the circle is increasing, when the radius is 10 cm, is 60*π* cm^{2}/s.

**4. An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?**

**Sol.** Let x be the length of an edge of cube and V be the volume of the cube. Then,

V = x^{3}

∴ Rate of change of volume w.r.t. time

$$\frac{dV}{dt}=\frac{d}{dt}(x^3)=3x^2\frac{dx}{dt}$$

(By chain ru le)

It is given that edge of the cube is increasing at the rate of 3 cm/s,

$$\text{so}\space\frac{dx}{dt}=3\space\text{cm/s}.$$

$$\therefore\space\frac{dV}{dt}= 3x^2(3)=9x^2\space\text{cm}^3/s$$

Thus, when x = 10 cm,

$$\frac{dV}{dt}=9(10)^2=900\space\text{cm}^3/s$$

Hence, the v olume of the cube is increasing at the rate of 900 cm^{3}/s when the edge is 10 cm long.

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

**Sol.** Let r be the radius of the circular wave and A be the area, then

A = *π*r^{2}

Therefore, the rate of change of area (A) with respect to time (t) is given by

$$\frac{dA}{dt}=\frac{d}{dt}(\pi r^2)\\\Rarr\space\frac{dA}{dt}=2\pi r\frac{dr}{dt}\space\\\text{(\text{By chain rule})}$$

It is given that waves move in circles at the speed of 5 cm/s.

$$\text{So,}\space\frac{dr}{dt}=\text{5 cm/s}\\\therefore\space\frac{dA}{dt}=2\pi r×5=10\pi r\space\text{cm}^2/s $$

Thus, when r = 8 cm,

$$\frac{dA}{dt}=10\pi(8)=80\pi\space\text{cm}^2/s$$

Hence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate of 80*π* cm^{2}/s.

**6. The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?**

**Sol.** At any instant of time t let the radius of the circle be r and its circumference be C, then C = 2*π*r

⇒ Rate of increase of circumference

$$\frac{dC}{dt}=2\pi\frac{dr}{dt}$$

(Differentiating w.r.t. t, by chain rule)

$$\text{where}\space\frac{dr}{dt}\space\text{is rate of increase of radius,}\\ \text{so}\space\frac{dr}{dt}=0.7\space\text{cm/s.}\\\therefore\space\frac{dC}{dt}=2\pi(0.7)\text{cm/s}=1.4\pi\space\text{cm/s}\\\bigg(\because\space\frac{dr}{dt}=0.7\space\text{cm}/s\space\text{given}\bigg)$$

Hence, rate of increase of circumference is 1.4*π* cm/s.

**7. The length x of a rectangle is decreasing at the rate of 5 cm/min and the width y is increasing at the rate of 4 cm/min. When x = 8 cm, y = 6 cm, find the rate of change of**

**(a) the perimeter**

**(b) the area of the rectangle.**

**Sol.** At any instant of time t, let length, breadth, perimeter and area of the rectangle are x, y, P and A respectively, then

P = 2(x + y) and A = xy ...(i)

Perimeter of rectangle = (length + breadth) and area = length × breadth

$$\text{It is given that}\space\frac{dx}{dt}=-5\space\text{cm/min}\\\text{and}\space\frac{dy}{dx}=4\space\text{cm/min}$$

(–ve sign shows that the length is decreasing)

(a) Now , P = 2(x + y). On differentiating w.r.t. t, we get

Rate of change of perimeter

$$\frac{dP}{dt}=2\bigg(\frac{dx}{dt}+\frac{dy}{dt}\bigg)\\=2[-5+4]\space\text{cm/min}=-2\text{cm/min}\\\bigg[\because\space\frac{dx}{dt}=-5\space\text{and}\space\frac{dy}{dt}=4\bigg]$$

Hence, perimeter of the rectangle is decreasing (– ve sign) at the rate of 2 cm/min.

(b) Here, area of rectangle A = xy. On differentiating w.r.t. t, we get

Rate of change

$$\frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dy}{dt}$$

= 8 × 4 + 6 × (– 5)

$$\bigg(\because\space \frac{dx}{dt}=-5\space\text{and}\space\frac{dy}{dt}=4\bigg)$$

= 32 – 30 = 2 cm^{2}/min

Hence, area of the rectangle is increasing at the rate of 2 cm^{2}/min.

**Note :** If rate of change is increasing, we take positive sign and if rate of change is decreasing, then we take negative sign.

**8. A balloon, which always remains spherical on inflation, is being inflated by pumping in ****900 cubic cm of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.**

**Sol.** At any instant of time t let the radius of the balloon be r and its volume be V, then volume of balloon V = (4/3)*π* r^{3}.

The balloon is being inflated at 900 cubic cm/s i.e., then rate of change of volume with respect to time is 900 cm^{2}/s.

On differentiating w.r.t. t, we get

Rate of change of volume

$$\frac{dV}{dt}=\bigg(\frac{4}{3}\pi\bigg)\bigg(3r^2\space\frac{dr}{dt}\bigg)$$

Give r = 15 cm

$$\Rarr\space 900=\bigg(\frac{4}{3}\pi\bigg)\begin{Bmatrix}3 (15)^2\space\frac{dr}{dt}\end{Bmatrix}\\\Rarr\space\frac{dr}{dt}=\frac{900}{3×(15)^2}×\frac{3}{4\pi}$$

⇒ Rate of change of radius r,

$$\frac{dr}{dt}=\frac{900}{4\pi×(15)^2}\\=\frac{225}{\pi×225}\\=\frac{1}{\pi}\space\text{cm/s}$$

Hence, the rate at which the radius of the balloon increases when the radius is 15 cm is

$$\frac{1}{\pi}\space\text{cm/s.}$$

**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 cm.**

**Sol.** Let r be the radius of spherical balloon and V be its volume.

$$\text{Then,\space r=10 cm and V}=\frac{4}{3}\pi r^3$$

Rate of change of volume w.r.t. radius r,

$$\frac{dV}{dr}=\bigg(\frac{4}{3}\bigg)3r^2\\\text{(differentiating w.r.t. r)}$$

= 4*π*r^{2} = 4π(10)^{2} = 400π (∵ r = 10 cm)

Hence, the volume of the balloon is increasing at the rate of 400p cm^{3}/cm.

**10. A ladder 5 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 m/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?**

**Sol.** Let AB = 5 m be the ladder and y be the height of the wall at which the ladder touches. Also, let the foot of the ladder be at B whose distance from the wall is x.

Given that the bottom of ladder is pulled along the ground at 2 m/s,

$$\text{so}\space\frac{dx}{dt}=2\space\text{m/s}$$

As we know that ΔABC is right angled, so by Pythagoras theorem, we have

x^{2} + y^{2} = 5^{2 }...(i)

When x = 4, then

y^{2} = 5^{2} – 4^{2}

$$\Rarr\space y=\sqrt{25-16}$$

⇒ y = 3 m

Differentiating equation (i) w.r.t. time t on both sides, we get

$$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\\\Rarr\space x\frac{dy}{dt}+y\frac{dy}{dt}=0\\\Rarr\space 4×2×3×\frac{dy}{dt}=0\\\bigg[\because\space x=4\space\text{and}\space\frac{dx}{dt}=2\bigg]$$

⇒ The rate of fall of height on the wall

$$\frac{dy}{dt}=-\frac{8}{3}\space\text{cm/s}$$

[Negative sign shows that height of ladder on the wall is decreasing at the rate of

$$\frac{8}{3}\space\text{cm/s.}]$$

**11. A particle move 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.**

$$\textbf{Sol.}\space\text{6y = x}^3+2\space\text{and}\space\frac{dy}{dt}=8\frac{dx}{dt}$$

On differentiating w.r.t. t, we get

$$6\frac{dy}{dt}=3x^2\frac{dx}{dt}\\\Rarr\space 6×8\frac{dx}{dt}=3x^2\frac{dx}{dt}$$

⇒ 3x^{2} = 48

⇒ x^{2} = 16

⇒ x = ± 4

When x = 4, then

6y = (4)^{3} + 2

⇒ 6y = 64 + 2

$$\Rarr\space y=\frac{66}{6}=11$$

When x = – 4, then

6y = (– 4)^{3} + 2

⇒ 6y = – 64 + 2

$$\Rarr\space y=\frac{66}{6}=11$$

When x = – 4, then

6y = (– 4)^{3} + 2

⇒ 6y = – 64 + 2

$$\Rarr\space y=\frac{66}{6}=11$$

When x = – 4, then

6y = (– 4)^{3} + 2

⇒ 6y = – 64 + 2

$$\Rarr\space y=\frac{-62}{6}=\frac{31}{3}$$

Hence, the required points on the curv

e are (4, 11) are

$$\bigg(-4,\frac{-31}{3}\bigg).$$

**12. The radius of an air bubble is increasing at the rate of**

$$\frac{\textbf{1}}{\textbf{2}}\space\textbf{cm/s}.\space\textbf{At what rate is the volume of the}\space\\\textbf{ bubble increasing when the radius is 1 cm?}$$

**Sol.** The air bubble is in the shape of a sphere.

Let r be the radius of bubble and V be the volume of bubble at any time t.

$$\text{Then, rate of change of radius}\\\space\frac{dr}{dt}=\frac{1}{2}\space\text{cm/s and r = 1 cm (Given)}\\\text{Now , volume of the bubble V =}\frac{4}{3}\pi r^3$$

On differentiating w.r.t. t, we get

Rate of volume increasing

$$\frac{dV}{dt}=\bigg(\frac{4}{3}\pi\bigg)\bigg(3r^2\frac{dr}{dt}\bigg)\\=4\pi r^2\frac{dr}{dt}=4\pi(1)^2\frac{1}{2}\\\bigg(\text{On putting r=1}\space\text{and}\space\frac{dr}{dt}=\frac{1}{2}\bigg)$$

= 2*π* cm^{3}/s

Hence, the rate at which the volume of the bubble increases is 2*π* cm^{3}/s.

**13. A balloon, which always remains spherical, has a variable diameter**

$$\frac{\textbf{3}}{\textbf{2}}\space\textbf{(2x+1)}.\space\textbf{Find the rate of change of}\\\textbf{its volume with respect to x.}$$

$$\textbf{Sol.}\space\text{Given, the diameter of the balloon =}\\\frac{3}{2}(2x+1)\\\therefore\space\text{Radius of the balloon =}\frac{\text{Diameter}}{2}\\=\frac{1}{2}\bigg[\frac{3}{2}(2x+1)\bigg]\\=\frac{3}{4}(2x+1)$$

For the volume V, the balloon is given by

$$\text{V}=\frac{4}{3}\space\pi(\text{radius})^3\\=\frac{4}{3}\pi\bigg[\frac{3}{4}(2x+1)\bigg]^3\\=\frac{9\pi}{16}(2x+1)^3$$

For the rate of change of volume, differentiate w.r.t. to x, we get

$$\frac{dV}{dx}=\frac{9\pi}{16}×3(2x+1)^2×2\\=\frac{27\pi}{8}(2x+1)^2\\\text{Thus, rate of change of volume is}\\\space\frac{27\pi}{8}(2x+1)^2.$$

**14. Sand is pouring from a pipe at the rate of ****12 cm ^{3}/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 cm?**

**Sol.** Let r be the radius, h be the height and V be the volume of sand cone at any time t.

$$\text{It is given that}\space\frac{dV}{dt}=12\space\text{cm}^3/s\\\text{and h}=\frac{1}{6}r\\\Rarr\space \text{ r = 6h }\\\therefore\space\text{V}=\frac{1}{3}\pi r^2h\\=\frac{1}{3}\pi(6h)^2h=12\pi h^3$$

On differentiating w.r.t. t, we get

$$\frac{dV}{dt}=(12\pi)\bigg(3h^2\frac{dh}{dt}\bigg)\\=36\pi h^2\space\frac{dh}{dt}\\\Rarr\space 12=36\pi(4)^2\space\frac{dh}{dt}\\\bigg(\because\space h=4\space \text{cm and }\frac{dV}{dt}=12\space\text{cm}^3/s\bigg)\\\Rarr\space\frac{dh}{dt}=\frac{12}{36\pi×16}=\frac{1}{48\pi}\space\text{cm/s}$$

Hence, when the height of the sand cone is 4 cm, its height is increasing at the rate of

$$\frac{1}{48\pi}\space\text{cm/s.}$$

**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.**

$$\textbf{Sol.}\space\therefore\space\text{Marginal Cost (MC) =}\frac{dC}{dt}$$

= 0.007(3x^{2}) – 0.003(2x) + 15

When x = 17

MC = 0.021(17)^{2} – 0.006(17) + 15

= 0.021(289) – 0.006(17) + 15

= 6.069 – 0.102 + 15 = 20.967

Hence, when 17 units are produced, the marginal cost is ₹ 20.967.

**16. The total revenue in Rupees received from the sale of x units of product is given by**

**R(x) = 13x ^{2} + 26x + 15**

**Find the marginal revenue when x = 7.**

$$\textbf{Sol.}\space\therefore\space\text{Marginal Revenue (MR) =}\frac{dR}{dx}\\=\frac{d}{dx}(13x^2+26x+15)$$

= 13 × 2x + 26

= 26x + 26

When x = 7, MR = 26(7) + 26 = 182 + 26 = 208

Hence, the required marginal revenue is ₹ 208.

**17. The rate of change of the area of a circle with respect to its radius r at r = 6 cm is**

**(a) 10 π (b) 12π (c) 8π (d) 11π**

**Sol.** (b) 12*π*

If A is the area of the circle corresponding to radius r, then A = *π*r^{2}

On differentiating w.r.t. r, we get

$$\frac{dA}{dr}=2\pi r$$

On putting r = 6, we get

$$\bigg(\frac{dA}{dr}\bigg)=2\pi(6)=12\pi\space\text{cm}^3/\text{cm} $$

Hence, the required rate of change of the area of a circle is 12*π* cm^{2} per cm.

**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**

**Sol.** (d) 126

Given, R(x) = 3x^{2} + 36x + 5

$$\therefore\space\text{Marginal Revenue (MR) =}\frac{dR}{dx}\\=\frac{d}{dx}(3x^2+36x+5)$$

= 6x + 36

When x = 15, MR = 6 × 15 + 36 = 126.

Hence, the required marginal revenue is ₹ 126.