Rbse Solutions Class 12 Maths Chapter 6 Exercise 6.1 Rates of Change

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Complete solutions for Class 12 Maths (NCERT) Exercise 6.1. Master the application of derivatives to solve Rates of Change problems, including calculating rates for area, volume, surface area, marginal cost, and marginal revenue. Includes ladder, cone, and spherical balloon examples.

This exercise focuses on Rates of Change, which involves using the derivative to find how one quantity changes in relation to another, often time.

Rbse Solutions Class 12 Maths Chapter 6 Exercise 6.1 Rates of Change

1. Rate of change of area of a circle w.r.t. radius $r$

The area of a circle is $A = \pi r^2$. The rate of change of area w.r.t. $r$ is $\frac{dA}{dr}$.

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

(a) When $r = 3$ cm:

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

(b) When $r = 4$ cm:

$$\frac{dA}{dr} = 2\pi (4) = \mathbf{8\pi \text{ cm}^2/\text{cm}}$$

2. Rate of change of surface area of a cube

Given: $\frac{dV}{dt} = 8 \text{ cm}^3/\text{s}$. Edge length $x = 12 \text{ cm}$. Find $\frac{dS}{dt}$.

  1. Volume ($V$) and Surface Area ($S$) formulas:$$V = x^3 \quad S = 6x^2$$
  2. Differentiate $V$ w.r.t. $t$ to find $\frac{dx}{dt}$:$$\frac{dV}{dt} = 3x^2 \frac{dx}{dt}$$$$8 = 3(12)^2 \frac{dx}{dt} \implies 8 = 3(144) \frac{dx}{dt} \implies \frac{dx}{dt} = \frac{8}{432} = \frac{1}{54} \text{ cm/s}$$
  3. Differentiate $S$ w.r.t. $t$ to find $\frac{dS}{dt}$:$$\frac{dS}{dt} = 12x \frac{dx}{dt}$$Substitute $x=12$ and $\frac{dx}{dt} = \frac{1}{54}$:$$\frac{dS}{dt} = 12(12) \left(\frac{1}{54}\right) = \frac{144}{54}$$Simplify by dividing by 18:$$\frac{dS}{dt} = \frac{8}{3}$$The surface area is increasing at the rate of $\mathbf{\frac{8}{3} \text{ cm}^2/\text{s}}$.

3. Rate of increase of the area of a circle

Given: $\frac{dr}{dt} = 3 \text{ cm/s}$. Find $\frac{dA}{dt}$ when $r = 10 \text{ cm}$.

  1. Area formula: $A = \pi r^2$
  2. Differentiate $A$ w.r.t. $t$:$$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$$
  3. Substitute the given values:$$\frac{dA}{dt} = 2\pi (10) (3) = 60\pi$$The area of the circle is increasing at the rate of $\mathbf{60\pi \text{ cm}^2/\text{s}}$.

4. Rate of increase of the volume of a cube

Given: $\frac{dx}{dt} = 3 \text{ cm/s}$. Find $\frac{dV}{dt}$ when $x = 10 \text{ cm}$.

  1. Volume formula: $V = x^3$
  2. Differentiate $V$ w.r.t. $t$:$$\frac{dV}{dt} = 3x^2 \frac{dx}{dt}$$
  3. Substitute the given values:$$\frac{dV}{dt} = 3(10)^2 (3) = 3(100)(3) = 900$$The volume of the cube is increasing at the rate of $\mathbf{900 \text{ cm}^3/\text{s}}$.

5. Rate of increase of enclosed area of a circular wave

The speed of the wave is the rate at which the radius increases: $\frac{dr}{dt} = 5 \text{ cm/s}$. Find $\frac{dA}{dt}$ when $r = 8 \text{ cm}$.

  1. Area formula: $A = \pi r^2$
  2. Differentiate $A$ w.r.t. $t$:$$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$$
  3. Substitute the given values:$$\frac{dA}{dt} = 2\pi (8) (5) = 80\pi$$The enclosed area is increasing at the rate of $\mathbf{80\pi \text{ cm}^2/\text{s}}$.

6. Rate of increase of the circumference of a circle

Given: $\frac{dr}{dt} = 0.7 \text{ cm/s}$. Find $\frac{dC}{dt}$.

  1. Circumference formula: $C = 2\pi r$
  2. Differentiate $C$ w.r.t. $t$:$$\frac{dC}{dt} = 2\pi \frac{dr}{dt}$$
  3. Substitute the given value:$$\frac{dC}{dt} = 2\pi (0.7) = 1.4\pi$$The rate of increase of the circumference is $\mathbf{1.4\pi \text{ cm/s}}$.

7. Rate of change of perimeter and area of a rectangle

Given: $x = 8 \text{ cm}$, $y = 6 \text{ cm}$.

$\frac{dx}{dt} = -5 \text{ cm/min}$ (decreasing).

$\frac{dy}{dt} = 4 \text{ cm/min}$ (increasing).

(a) Rate of change of the Perimeter ($P$):

  1. Perimeter formula: $P = 2(x + y)$
  2. Differentiate $P$ w.r.t. $t$:$$\frac{dP}{dt} = 2 \left(\frac{dx}{dt} + \frac{dy}{dt}\right)$$
  3. Substitute the given rates:$$\frac{dP}{dt} = 2 (-5 + 4) = 2(-1) = -2$$The perimeter is decreasing at the rate of $\mathbf{2 \text{ cm/min}}$.

(b) Rate of change of the Area ($A$):

  1. Area formula: $A = xy$
  2. Differentiate $A$ w.r.t. $t$ (Product Rule):$$\frac{dA}{dt} = \frac{dx}{dt} y + x \frac{dy}{dt}$$
  3. Substitute the given values:$$\frac{dA}{dt} = (-5)(6) + (8)(4) = -30 + 32 = 2$$The area is increasing at the rate of $\mathbf{2 \text{ cm}^2/\text{min}}$.

8. Rate of increase of the radius of a spherical balloon

Given: $\frac{dV}{dt} = 900 \text{ cm}^3/\text{s}$. Find $\frac{dr}{dt}$ when $r = 15 \text{ cm}$.

  1. Volume formula for a sphere: $V = \frac{4}{3} \pi r^3$
  2. Differentiate $V$ w.r.t. $t$:$$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$$
  3. Substitute the given values:$$900 = 4\pi (15)^2 \frac{dr}{dt}$$$$900 = 4\pi (225) \frac{dr}{dt}$$$$900 = 900\pi \frac{dr}{dt}$$$$\frac{dr}{dt} = \frac{900}{900\pi} = \frac{1}{\pi}$$The radius of the balloon increases at the rate of $\mathbf{\frac{1}{\pi} \text{ cm/s}}$.

9. Rate of increase of volume of a sphere w.r.t. radius

Find $\frac{dV}{dr}$ when $r = 10 \text{ cm}$.

  1. Volume formula: $V = \frac{4}{3} \pi r^3$
  2. Differentiate $V$ w.r.t. $r$:$$\frac{dV}{dr} = 4\pi r^2$$
  3. Substitute the given value:$$\frac{dV}{dr} = 4\pi (10)^2 = 400\pi$$The rate of increase of volume with the radius is $\mathbf{400\pi \text{ cm}^3/\text{cm}}$.

10. Rate of decrease of the height of a ladder

Let $x$ be the distance of the foot of the ladder from the wall and $y$ be the height of the ladder on the wall. The length of the ladder is $L=5 \text{ m}$.

Given: $L=5 \text{ m}$. $\frac{dx}{dt} = 2 \text{ cm/s} = 0.02 \text{ m/s}$. $x = 4 \text{ m}$. Find $\frac{dy}{dt}$.

  1. Pythagorean theorem: $x^2 + y^2 = 5^2$
  2. Find $y$ when $x=4$:$$4^2 + y^2 = 25 \implies 16 + y^2 = 25 \implies y^2 = 9 \implies y = 3 \text{ m}$$
  3. Differentiate the Pythagorean equation w.r.t. $t$:$$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$$$$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$$
  4. Substitute the known values:$$4 (0.02) + 3 \frac{dy}{dt} = 0$$$$0.08 + 3 \frac{dy}{dt} = 0$$$$\frac{dy}{dt} = -\frac{0.08}{3} = -\frac{8}{300} = -\frac{2}{75} \text{ m/s}$$To convert to cm/s, multiply by 100:$$\frac{dy}{dt} = -\frac{2}{75} \times 100 = -\frac{200}{75}$$Simplify by dividing by 25:$$\frac{dy}{dt} = -\frac{8}{3} \text{ cm/s}$$The height is decreasing at the rate of $\mathbf{\frac{8}{3} \text{ cm/s}}$.

11. Points where y-coordinate changes 8 times as fast as x-coordinate

Curve: $6y = x^3 + 2$.

Given: $\frac{dy}{dt} = 8 \frac{dx}{dt}$.

  1. Differentiate the curve equation w.r.t. $t$:$$6 \frac{dy}{dt} = 3x^2 \frac{dx}{dt}$$
  2. Substitute the given rate relationship:$$6 \left(8 \frac{dx}{dt}\right) = 3x^2 \frac{dx}{dt}$$
  3. Solve for $x$ (assuming $\frac{dx}{dt} \ne 0$):$$48 \frac{dx}{dt} = 3x^2 \frac{dx}{dt}$$$$48 = 3x^2 \implies x^2 = 16 \implies x = \pm 4$$
  4. Find the corresponding $y$-coordinates using $6y = x^3 + 2$:
    • If $x = 4$: $6y = (4)^3 + 2 = 64 + 2 = 66 \implies y = 11$. Point is $(4, 11)$.
    • If $x = -4$: $6y = (-4)^3 + 2 = -64 + 2 = -62 \implies y = -\frac{62}{6} = -\frac{31}{3}$. Point is $\left(-4, -\frac{31}{3}\right)$.The points are $\mathbf{(4, 11)}$ and $\mathbf{\left(-4, -\frac{31}{3}\right)}$.

12. Rate of increase of the volume of an air bubble

Given: $\frac{dr}{dt} = \frac{1}{2} \text{ cm/s}$. Find $\frac{dV}{dt}$ when $r = 1 \text{ cm}$.

  1. Volume formula: $V = \frac{4}{3} \pi r^3$
  2. Differentiate $V$ w.r.t. $t$:$$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$$
  3. Substitute the given values:$$\frac{dV}{dt} = 4\pi (1)^2 \left(\frac{1}{2}\right) = 2\pi$$The volume of the bubble is increasing at the rate of $\mathbf{2\pi \text{ cm}^3/\text{s}}$.

13. Rate of change of volume of a sphere w.r.t. $x$

Given: Diameter $D = \frac{3}{2}(2x + 1)$.

Find $\frac{dV}{dx}$.

  1. Radius in terms of $x$:$$r = \frac{D}{2} = \frac{1}{2} \cdot \frac{3}{2}(2x + 1) = \frac{3}{4}(2x + 1)$$
  2. Volume formula: $V = \frac{4}{3} \pi r^3$$$V = \frac{4}{3} \pi \left[\frac{3}{4}(2x + 1)\right]^3$$$$V = \frac{4}{3} \pi \cdot \frac{3^3}{4^3} (2x + 1)^3 = \frac{4}{3} \pi \cdot \frac{27}{64} (2x + 1)^3 = \frac{9\pi}{16} (2x + 1)^3$$
  3. Differentiate $V$ w.r.t. $x$:$$\frac{dV}{dx} = \frac{9\pi}{16} \cdot 3(2x + 1)^2 \cdot \frac{d}{dx}(2x + 1)$$$$\frac{dV}{dx} = \frac{27\pi}{16} (2x + 1)^2 \cdot 2$$$$\frac{dV}{dx} = \frac{27\pi}{8} (2x + 1)^2$$The rate of change of volume w.r.t. $x$ is $\mathbf{\frac{27\pi}{8} (2x + 1)^2}$.

14. Rate of increase of the height of a sand cone

Given: $\frac{dV}{dt} = 12 \text{ cm}^3/\text{s}$. Height $h = \frac{1}{6} r \implies r = 6h$. Find $\frac{dh}{dt}$ when $h = 4 \text{ cm}$.

  1. Volume formula for a cone: $V = \frac{1}{3} \pi r^2 h$
  2. Express $V$ in terms of $h$ only (using $r = 6h$):$$V = \frac{1}{3} \pi (6h)^2 h = \frac{1}{3} \pi (36h^2) h = 12\pi h^3$$
  3. Differentiate $V$ w.r.t. $t$:$$\frac{dV}{dt} = 12\pi (3h^2) \frac{dh}{dt} = 36\pi h^2 \frac{dh}{dt}$$
  4. Substitute the given values ($\frac{dV}{dt} = 12$ and $h = 4$):$$12 = 36\pi (4)^2 \frac{dh}{dt}$$$$12 = 36\pi (16) \frac{dh}{dt}$$$$12 = 576\pi \frac{dh}{dt}$$$$\frac{dh}{dt} = \frac{12}{576\pi} = \frac{1}{48\pi}$$The height is increasing at the rate of $\mathbf{\frac{1}{48\pi} \text{ cm/s}}$.

💰 Marginal Cost and Revenue (Exercises 15-18)

Marginal Cost (MC) is the instantaneous rate of change of the total cost $C(x)$ w.r.t. the number of units produced $x$. $\text{MC} = C'(x)$.

Marginal Revenue (MR) is the instantaneous rate of change of the total revenue $R(x)$ w.r.t. the number of units sold $x$. $\text{MR} = R'(x)$.

15. Marginal Cost when $x = 17$

$$C(x) = 0.007x^3 – 0.003x^2 + 15x + 4000$$

  1. Find the Marginal Cost function $C'(x)$:$$C'(x) = 0.007(3x^2) – 0.003(2x) + 15$$$$C'(x) = 0.021x^2 – 0.006x + 15$$
  2. Evaluate $C'(17)$:$$C'(17) = 0.021(17)^2 – 0.006(17) + 15$$$$C'(17) = 0.021(289) – 0.102 + 15$$$$C'(17) = 6.069 – 0.102 + 15 = 20.967$$The marginal cost is $\mathbf{\text{₹} 20.967}$.

16. Marginal Revenue when $x = 7$

$$R(x) = 13x^2 + 26x + 15$$

  1. Find the Marginal Revenue function $R'(x)$:$$R'(x) = 26x + 26$$
  2. Evaluate $R'(7)$:$$R'(7) = 26(7) + 26 = 182 + 26 = 208$$The marginal revenue is $\mathbf{\text{₹} 208}$.

17. Multiple Choice: Rate of change of area of a circle at $r = 6$ cm

Area $A = \pi r^2$. Rate of change w.r.t. $r$ is $\frac{dA}{dr} = 2\pi r$.

At $r=6$: $\frac{dA}{dr} = 2\pi (6) = 12\pi$.

The correct answer is (B) $12\pi$.

18. Multiple Choice: Marginal Revenue when $x = 15$

$$R(x) = 3x^2 + 36x + 5$$

  1. Marginal Revenue $R'(x)$:$$R'(x) = 6x + 36$$
  2. Evaluate $R'(15)$:$$R'(15) = 6(15) + 36 = 90 + 36 = 126$$

The correct answer is (D) 126.

Ex 6.1
Ex 6.2
Ex 6.3
Ch 6 Misc.

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