Question Statement

The volume $v$ of a sphere of radius $r$ is $v = (\frac{4 π}{3}) r^{3}$ . Find the surface area $s$ of the sphere if $s$ is the instantaneous rate of change of the volume with respect to the radius.

Background and Explanation

This problem connects the geometric formulas for a sphere through calculus. The "instantaneous rate of change" indicates we need to find the derivative of the volume function with respect to the radius. Recall that the power rule for differentiation states $\frac{d}{d x} (x^{n}) = n x^{n - 1}$ .

Solution

We are given the volume of a sphere as a function of its radius:

$v = \frac{4 π}{3} r^{3}$

To find the surface area $s$ , we calculate the instantaneous rate of change of volume with respect to the radius. This means we differentiate $v$ with respect to $r$ :

$\frac{d v}{d r} = \frac{d}{d r} (\frac{4 π}{3} r^{3})$

Using the power rule for differentiation, we bring down the exponent 3 and reduce the power by 1:

$\frac{d v}{d r} = \frac{4 π}{3} \cdot (3 r^{2})$

Simplifying the expression, the 3 in the numerator and denominator cancel:

$\frac{d v}{d r} = 4 π r^{2}$

Since $s$ is defined as the instantaneous rate of change of volume with respect to radius, we have:

$s = \frac{d v}{d r} = 4 π r^{2}$

Thus, the surface area of the sphere is $s = 4 π r^{2}$ .

Key Formulas or Methods Used

Volume of a sphere: $v = \frac{4}{3} π r^{3}$
Power rule for differentiation: $\frac{d}{d r} (r^{n}) = n r^{n - 1}$
Surface area of a sphere (result): $s = 4 π r^{2}$

Summary of Steps

Write down the given volume formula: $v = \frac{4 π}{3} r^{3}$
Differentiate with respect to $r$ using the power rule: $\frac{d v}{d r} = \frac{4 π}{3} \cdot 3 r^{2}$
Simplify to obtain the surface area: $s = 4 π r^{2}$

Question Statement

The volume $v$ of a sphere of radius $r$ is $v = (\frac{4 π}{3}) r^{3}$ . Find the surface area $s$ of the sphere if $s$ is the instantaneous rate of change of the volume with respect to the radius.

Background and Explanation

Solution

We are given the volume of a sphere as a function of its radius:

$v = \frac{4 π}{3} r^{3}$

To find the surface area $s$ , we calculate the instantaneous rate of change of volume with respect to the radius. This means we differentiate $v$ with respect to $r$ :

$\frac{d v}{d r} = \frac{d}{d r} (\frac{4 π}{3} r^{3})$

Using the power rule for differentiation, we bring down the exponent 3 and reduce the power by 1:

$\frac{d v}{d r} = \frac{4 π}{3} \cdot (3 r^{2})$

Simplifying the expression, the 3 in the numerator and denominator cancel:

$\frac{d v}{d r} = 4 π r^{2}$

Since $s$ is defined as the instantaneous rate of change of volume with respect to radius, we have:

$s = \frac{d v}{d r} = 4 π r^{2}$

Thus, the surface area of the sphere is $s = 4 π r^{2}$ .

Key Formulas or Methods Used

Volume of a sphere: $v = \frac{4}{3} π r^{3}$
Power rule for differentiation: $\frac{d}{d r} (r^{n}) = n r^{n - 1}$
Surface area of a sphere (result): $s = 4 π r^{2}$

Summary of Steps

Write down the given volume formula: $v = \frac{4 π}{3} r^{3}$
Differentiate with respect to $r$ using the power rule: $\frac{d v}{d r} = \frac{4 π}{3} \cdot 3 r^{2}$
Simplify to obtain the surface area: $s = 4 π r^{2}$

2.10 Q-3

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps

2.10 Q-3

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps