CBSE Class 12 Mathematics Chapter 6: Applications of Derivatives NCERT Solutions

Let A be the area of the circle and r be its radius. The formula for the area of a circle is given by $A\; =\; \backslash pi\; r^2$.

To find the rate of change of the area with respect to its radius, we differentiate A with respect to r:

$$\backslash frac\{dA\}\{dr\}\; =\; \backslash frac\{d\}\{dr\}\; \backslash left(\; \backslash pi\; r^2\; \backslash right)$$

Using the power rule for differentiation, we get:

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

Now, we evaluate this rate at the given values of r:

(a) When $r\; =\; 3cm$:

Substitute r = 3 into the expression for $\backslash frac\{dA\}\{dr\}$:

$$\backslash frac\{dA\}\{dr\}\; =\; 2\backslash pi\; \backslash left(3\backslash right)\; =\; 6\backslash pi$$

Thus, the rate of change of the area of the circle with respect to its radius when r = 3 cm is $6\backslash pi$ cm²/cm.

(b) When $r\; =\; 4cm$:

Substitute r = 4 into the expression for $\backslash frac\{dA\}\{dr\}$:

$$\backslash frac\{dA\}\{dr\}\; =\; 2\backslash pi\; \backslash left(4\backslash right)\; =\; 8\backslash pi$$

Thus, the rate of change of the area of the circle with respect to its radius when r = 4 cm is $8\backslash pi$ cm²/cm.

Let x be the length of the edge of the cube. Let V be the volume and S be the surface area of the cube. We are given that the volume is increasing at a rate of $8cm^3/s$, which means $\backslash frac\{dV\}\{dt\}\; =\; 8$.

The formulas for the volume and surface area of a cube are:

$V\; =\; x^3$

$S\; =\; 6x^2$

We need to find $\backslash frac\{dS\}\{dt\}$ when $x\; =\; 12cm$.

First, let's find the rate of change of the edge length, $\backslash frac\{dx\}\{dt\}$, by differentiating the volume formula with respect to time t:

$$\backslash frac\{dV\}\{dt\}\; =\; \backslash frac\{d\}\{dt\}\; \backslash left(\; x^3\; \backslash right)$$

Using the chain rule, $\backslash frac\{dV\}\{dt\}\; =\; \backslash frac\{d\}\{dx\}\; \backslash left(\; x^3\; \backslash right)\; \backslash frac\{dx\}\{dt\}$:

$$8\; =\; 3x^2\; \backslash frac\{dx\}\{dt\}$$

Now, we can express $\backslash frac\{dx\}\{dt\}$ in terms of x:

$$\backslash frac\{dx\}\{dt\}\; =\; \backslash frac\{8\}\{3x^2\}\; \backslash qquad\; \backslash dots\; (1)$$

Next, let's find the rate of change of the surface area, $\backslash frac\{dS\}\{dt\}$, by differentiating the surface area formula with respect to time t:

$$\backslash frac\{dS\}\{dt\}\; =\; \backslash frac\{d\}\{dt\}\; (6x^2)$$

Using the chain rule, $\backslash frac\{dS\}\{dt\}\; =\; \backslash frac\{d\}\{dx\}\; (6x^2)\; \backslash frac\{dx\}\{dt\}$:

$$\backslash frac\{dS\}\{dt\}\; =\; 12x\; \backslash frac\{dx\}\{dt\}$$

Now, substitute the expression for $\backslash frac\{dx\}\{dt\}$ from equation (1) into this equation:

$$\backslash frac\{dS\}\{dt\}\; =\; 12x\; \backslash left(\; \backslash frac\{8\}\{3x^2\}\; \backslash right)$$

Simplify the expression:

$$\backslash frac\{dS\}\{dt\}\; =\; \backslash frac\{12x\; \backslash times\; 8\}\{3x^2\}\; =\; \backslash frac\{96x\}\{3x^2\}\; =\; \backslash frac\{32\}\{x\}$$

Finally, we need to find the rate of increase of the surface area when the edge length $x\; =\; 12cm$. Substitute $x\; =\; 12$ into the expression for $\backslash frac\{dS\}\{dt\}$:

$$\backslash frac\{dS\}\{dt\}\; =\; \backslash frac\{32\}\{12\}$$

Simplify the fraction:

$$\backslash frac\{dS\}\{dt\}\; =\; \backslash frac\{8\}\{3\}$$

Therefore, the surface area of the cube is increasing at the rate of $\backslash frac\{8\}\{3\}\; cm^2/s$ when the length of its edge is 12 cm.

Let A be the area of the circle and r be its radius. We are given that the radius is increasing at a rate of 3 cm/s, so $\backslash frac\{dr\}\{dt\}\; =\; 3$ cm/s.

The formula for the area of a circle is $A\; =\; \backslash pi\; r^2$.

We want to find the rate at which the area is increasing, which is $\backslash frac\{dA\}\{dt\}$, when the radius $r\; =\; 10cm$.

Differentiate the area formula with respect to time t, using the chain rule:

$$\backslash frac\{dA\}\{dt\}\; =\; \backslash frac\{d\}\{dt\}\; \backslash left(\; \backslash pi\; r^2\; \backslash right)$$

$$\backslash frac\{dA\}\{dt\}\; =\; \backslash frac\{d\}\{dr\}\; \backslash left(\; \backslash pi\; r^2\; \backslash right)\; \backslash frac\{dr\}\{dt\}$$

$$\backslash frac\{dA\}\{dt\}\; =\; 2\backslash pi\; r\; \backslash frac\{dr\}\{dt\}$$

Now, substitute the given values: $r\; =\; 10cm$ and $\backslash frac\{dr\}\{dt\}\; =\; 3cm/s$:

$$\backslash frac\{dA\}\{dt\}\; =\; 2\backslash pi\; (10)\; (3)$$

$$\backslash frac\{dA\}\{dt\}\; =\; 60\backslash pi$$

Therefore, the area of the circle is increasing at the rate of $60\backslash pi\; cm^2/s$ when the radius is 10 cm.