CBSE Class 12 Maths Chapter 6 Application of Derivatives NCERT Solutions

Solution:

Let V be the volume and S be the surface area of the spherical ball of salt at any instant, and let r be its radius. We are given that the rate of decrease of the volume is proportional to the surface area.

The volume of a sphere is given by $V\; =\; \backslash frac\{4\}\{3\}\; \backslash pi\; r^3$.

The surface area of a sphere is given by $S\; =\; 4\backslash pi\; r^2$.

According to the problem statement, the rate of decrease of volume is proportional to the surface area:

$$\backslash frac\{dV\}\{dt\}\; \backslash propto\; S$$

This can be written as:

$$\backslash frac\{dV\}\{dt\}\; =\; -kS$$

where k is a positive constant of proportionality (the negative sign indicates a decrease in volume).

Now, we differentiate the volume V with respect to time t:

$$\backslash frac\{dV\}\{dt\}\; =\; \backslash frac\{d\}\{dt\}\; \backslash left(\; \backslash frac\{4\}\{3\}\; \backslash pi\; r^3\; \backslash right)$$

$$\backslash frac\{dV\}\{dt\}\; =\; \backslash frac\{4\}\{3\}\; \backslash pi\; \backslash cdot\; 3r^2\; \backslash cdot\; \backslash frac\{dr\}\{dt\}$$

$$\backslash frac\{dV\}\{dt\}\; =\; 4\backslash pi\; r^2\; \backslash frac\{dr\}\{dt\}$$

Substitute this expression for $\backslash frac\{dV\}\{dt\}$ and the formula for S into the proportionality equation:

$$4\backslash pi\; r^2\; \backslash frac\{dr\}\{dt\}\; =\; -k(4\backslash pi\; r^2)$$

Now, we can solve for $\backslash frac\{dr\}\{dt\}$. Divide both sides by $4\backslash pi\; r^2$ (assuming $r\; \backslash neq\; 0$):

$$\backslash frac\{dr\}\{dt\}\; =\; \backslash frac\{-k(4\backslash pi\; r^2)\}\{4\backslash pi\; r^2\}$$

$$\backslash frac\{dr\}\{dt\}\; =\; -k$$

Since k is a positive constant, $-k$ is a negative constant. This shows that the rate of change of the radius, $\backslash frac\{dr\}\{dt\}$, is a constant. The negative sign indicates that the radius is decreasing.

Hence, the radius is decreasing at a constant rate.

Solution:

Let r be the radius of the circle and A be its area. Let P be its perimeter.

The area of the circle is given by $A\; =\; \backslash pi\; r^2$.

The perimeter of the circle is given by $P\; =\; 2\backslash pi\; r$.

We are given that the area of the circle increases at a uniform rate. This means that the rate of change of area with respect to time (t) is a constant. Let this constant rate be k.

$$\backslash frac\{dA\}\{dt\}\; =\; k$$

Now, differentiate the area formula with respect to time t:

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

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

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

Equating the two expressions for $\backslash frac\{dA\}\{dt\}$:

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

Now, we can express the rate of change of the radius $\backslash frac\{dr\}\{dt\}$ in terms of r:

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

Next, let's find the rate of change of the perimeter P with respect to time t. Differentiate the perimeter formula $P\; =\; 2\backslash pi\; r$ with respect to t:

$$\backslash frac\{dP\}\{dt\}\; =\; \backslash frac\{d\}\{dt\}(2\backslash pi\; r)$$

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

Substitute the expression for $\backslash frac\{dr\}\{dt\}$ we found earlier:

$$\backslash frac\{dP\}\{dt\}\; =\; 2\backslash pi\; \backslash left(\; \backslash frac\{k\}\{2\backslash pi\; r\}\; \backslash right)$$

$$\backslash frac\{dP\}\{dt\}\; =\; \backslash frac\{k\}\{r\}$$

This equation shows that $\backslash frac\{dP\}\{dt\}$ is proportional to $\backslash frac\{1\}\{r\}$, which means the rate of change of the perimeter varies inversely as the radius.

Hence, the perimeter varies inversely as the radius.

Solution:

Let the height of the kite be denoted by CD and the height of the boy be denoted by AB. Let the horizontal distance of the kite from the boy be DB, which is equal to AE. Let the length of the string be AC, denoted by y.

Given:

• Height of the kite, $CD\; =\; 151.5\; \backslash text\{\; m\}$.
• Height of the boy, $AB\; =\; 1.5\; \backslash text\{\; m\}$.
• The kite is moving horizontally, so its height is constant. The speed of the kite is the rate at which the horizontal distance changes, $\backslash frac\{dx\}\{dt\}\; =\; 10\; \backslash text\{\; m/s\}$.
• The distance of the kite from the boy along the string is $AC\; =\; y\; =\; 250\; \backslash text\{\; m\}$ at the instant we are interested in.

The effective height of the kite above the boy's eye level is $EC\; =\; CD\; -\; AB\; =\; 151.5\; -\; 1.5\; =\; 150\; \backslash text\{\; m\}$.

Let the horizontal distance of the kite from the boy at any time t be $AE\; =\; x$. The length of the string is $AC\; =\; y$.

We have a right-angled triangle $\backslash triangle\; CEA$, where $AE\; =\; x$, $EC\; =\; 150\; \backslash text\{\; m\}$, and $AC\; =\; y$.

By the Pythagorean theorem:

$$AE^2\; +\; EC^2\; =\; AC^2$$

$$x^2\; +\; (150)^2\; =\; y^2$$

We need to find how fast the string is being let out, which is $\backslash frac\{dy\}\{dt\}$, when $y\; =\; 250\; \backslash text\{\; m\}$.

First, let's find the value of x when $y\; =\; 250\; \backslash text\{\; m\}$:

$$x^2\; +\; (150)^2\; =\; (250)^2$$

$$x^2\; =\; (250)^2\; -\; (150)^2$$

$$x^2\; =\; (250\; -\; 150)(250\; +\; 150)$$

$$x^2\; =\; (100)(400)$$

$$x^2\; =\; 40000$$

$$x\; =\; \backslash sqrt\{40000\}\; =\; 200\; \backslash text\{\; m\}$$

Now, differentiate the equation $x^2\; +\; (150)^2\; =\; y^2$ with respect to time t:

$$\backslash frac\{d\}\{dt\}(x^2)\; +\; \backslash frac\{d\}\{dt\}((150)^2)\; =\; \backslash frac\{d\}\{dt\}(y^2)$$

$$2x\; \backslash frac\{dx\}\{dt\}\; +\; 0\; =\; 2y\; \backslash frac\{dy\}\{dt\}$$

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

$$\backslash frac\{dy\}\{dt\}\; =\; \backslash frac\{x\}\{y\}\; \backslash frac\{dx\}\{dt\}$$

Substitute the known values at the instant in question: $x\; =\; 200\; \backslash text\{\; m\}$, $y\; =\; 250\; \backslash text\{\; m\}$, and $\backslash frac\{dx\}\{dt\}\; =\; 10\; \backslash text\{\; m/s\}$:

$$\backslash frac\{dy\}\{dt\}\; =\; \backslash frac\{200\}\{250\}\; \backslash times\; 10$$

$$\backslash frac\{dy\}\{dt\}\; =\; \backslash frac\{4\}\{5\}\; \backslash times\; 10$$

$$\backslash frac\{dy\}\{dt\}\; =\; 8\; \backslash text\{\; m/s\}$$

Therefore, the string is being let out at a rate of 8 m/s.