CBSE Class 12 Mathematics: Applications of Derivatives NCERT Solutions

The area of a circle, denoted by A, with radius r is given by the formula $A\; =\; \backslash pi\; r^2$. To find the rate of change of the area with respect to its radius, we need to calculate the derivative of A with respect to r, $\backslash frac\{dA\}\{dr\}$.

Using the power rule for differentiation, we get:

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

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

1. When $r\; =\; 3$ cm:

   $$\backslash frac\{dA\}\{dr\}\; \backslash bigg|\_\{r=3\}\; =\; 2\backslash pi\; (3)\; =\; 6\backslash pi$$

   Therefore, the area of the circle is changing at a rate of $6\backslash pi$ cm²/cm when the radius is 3 cm.

2. When $r\; =\; 4$ cm:

   $$\backslash frac\{dA\}\{dr\}\; \backslash bigg|\_\{r=4\}\; =\; 2\backslash pi\; (4)\; =\; 8\backslash pi$$

   Therefore, the area of the circle is changing at a rate of $8\backslash pi$ cm²/cm when the radius is 4 cm.

Let x be the length of an edge of the cube. The volume V of the cube is given by $V\; =\; x^3$, and the surface area S is given by $S\; =\; 6x^2$. Both V and S are functions of time t.

We are given that the volume is increasing at a rate of 8 cm³/s, which means $\backslash frac\{dV\}\{dt\}\; =\; 8$ cm³/s.

We need to find how fast the surface area is increasing, $\backslash frac\{dS\}\{dt\}$, when the edge length $x\; =\; 12$ cm.

First, let's find the rate at which the edge length is changing, $\backslash frac\{dx\}\{dt\}$. We differentiate the volume formula with respect to time t using the chain rule:

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

Substitute the given value of $\backslash frac\{dV\}\{dt\}$:

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

Now, solve for $\backslash frac\{dx\}\{dt\}$:

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

Next, we find the rate of change of the surface area by differentiating S with respect to time t, again using the chain rule:

$$\backslash frac\{dS\}\{dt\}\; =\; \backslash frac\{d\}\{dt\}(6x^2)\; =\; \backslash frac\{d\}\{dx\}(6x^2)\; \backslash cdot\; \backslash frac\{dx\}\{dt\}\; =\; 12x\; \backslash frac\{dx\}\{dt\}$$

Now, substitute the expression for $\backslash frac\{dx\}\{dt\}$ we found earlier:

$$\backslash frac\{dS\}\{dt\}\; =\; 12x\; \backslash left(\; \backslash frac\{8\}\{3x^2\}\; \backslash right)\; =\; \backslash frac\{96x\}\{3x^2\}\; =\; \backslash frac\{32\}\{x\}$$

Finally, we need to find $\backslash frac\{dS\}\{dt\}$ when the edge length is $x\; =\; 12$ cm:

$$\backslash frac\{dS\}\{dt\}\; \backslash bigg|\_\{x=12\}\; =\; \backslash frac\{32\}\{12\}\; =\; \backslash frac\{8\}\{3\}$$

Thus, the surface area is increasing at the rate of $\backslash frac\{8\}\{3\}$ cm²/s when the length of the edge is 12 cm.

Let r be the radius of the circle and A be its area. The formula for the area of a circle is $A\; =\; \backslash pi\; r^2$. We are given that the radius is increasing at a uniform rate of 3 cm/s, which means $\backslash frac\{dr\}\{dt\}\; =\; 3$ cm/s.

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

To find $\backslash frac\{dA\}\{dt\}$, we differentiate the area formula with respect to time t, using the chain rule:

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

Calculating the derivative of $\backslash pi\; r^2$ with respect to r gives $2\backslash pi\; r$. So, the equation becomes:

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

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

$$\backslash frac\{dA\}\{dt\}\; =\; 2\backslash pi\; (10)\; (3)\; =\; 60\backslash pi$$

Therefore, the area of the circle is increasing at the rate of $60\backslash pi$ cm²/s when the radius is 10 cm.