Assume that the radius $$ r $$ of a sphere is expanding at a rate of $$ 15 \mathrm{in} . / \mathrm{min} $$. The volume of a sphere is $$ V=\frac{4}{3} \pi r^{3} $$. Determine the rate at which the volume is changing with respect to time when $$ r=13 \mathrm{in} $$. The volume is changing at a rate of $$ \square \mathrm{in}^{3} / \mathrm{min} $$.

Question

UpStudy AI · Accepted Answer

We start with the formula for the volume of a sphere:
$$
V = \frac{4}{3} \pi r^3.
$$

Differentiate both sides with respect to time $$ t $$ to find the rate of change of the volume:
$$
\frac{dV}{dt} = \frac{d}{dt} \left(\frac{4}{3} \pi r^3\right).
$$

Using the chain rule, we obtain:
$$
\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}.
$$

Given that $$ \frac{dr}{dt} = 15 $$ in/min and $$ r = 13 $$ in, substitute these values into the equation:
$$
\frac{dV}{dt} = 4 \pi (13)^2 \times 15.
$$

Calculate $$ 13^2 $$:
$$
13^2 = 169.
$$

Then the expression becomes:
$$
\frac{dV}{dt} = 4 \pi \times 169 \times 15.
$$

Multiply the constants:
$$
4 \times 169 = 676,
$$
$$
676 \times 15 = 10140.
$$

Thus, the rate of change of the volume is:
$$
\frac{dV}{dt} = 10140 \pi \, \text{in}^3/\text{min}.
$$

The volume is changing at a rate of $$ 10140 \pi \, \text{in}^3/\text{min} $$.
The volume is changing at a rate of $$ 10140 \pi \, \text{in}^3/\text{min} $$.

Assume that the radius of a sphere is expanding at a rate of . The volume of a sphere is .
Determine the rate at which the volume is changing with respect to time when .
The volume is changing at a rate of .

Solución de inteligencia artificial de Upstudy

Responder

Solución

Beyond the Answer

preguntas relacionadas

Latest Calculus Questions

Assume that the radius of a sphere is expanding at a rate of . The volume of a sphere is . Determine the rate at which the volume is changing with respect to time when . The volume is changing at a rate of .