Question

related rates: problem 1 (1 point) if the radius of a sphere is increasing at a constant rate of 4 cm/sec, then the volume is increasing at a rate of cm³/sec when the radius is 5 cm. hint: $\frac{dv}{dt}=\frac{dv}{dr}cdot\frac{dr}{dt}$, and the volume of a sphere is $v = \frac{4}{3}pi r^{3}$. preview my answers submit answers you have attempted this problem 0 times. you have unlimited attempts remaining. email instructor

Explanation:

Step1: Differentiate volume formula

The volume of a sphere is $V=\frac{4}{3}\pi r^{3}$. Differentiating $V$ with respect to $r$ using the power - rule, we get $\frac{dV}{dr}=4\pi r^{2}$.

Step2: Use the chain - rule

We know that $\frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}$. Given $\frac{dr}{dt} = 4$ cm/sec and we need to find $\frac{dV}{dt}$ when $r = 5$ cm. Substitute $\frac{dV}{dr}=4\pi r^{2}$ and $\frac{dr}{dt}=4$ into the chain - rule formula: $\frac{dV}{dt}=(4\pi r^{2})\cdot4 = 16\pi r^{2}$.

Step3: Plug in the value of $r$

When $r = 5$ cm, $\frac{dV}{dt}=16\pi(5)^{2}=16\pi\times25 = 400\pi$ $cm^{3}/sec$.

Answer:

$400\pi$