Question

related rates: problem 1 (1 point) if the radius of a sphere is increasing at a constant rate of 2 cm/sec. then the volume is increasing at a rate of cm³/sec when the radius is 2 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.

Explanation:

Step1: Differentiate volume formula

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

Step2: Substitute given values

We know that $\frac{dr}{dt}=2$ cm/sec and $r = 2$ cm. Substitute these values into the derivative formula: $\frac{dV}{dt}=4\pi(2)^{2}\times2$.

Step3: Calculate the result

$\frac{dV}{dt}=4\pi\times4\times2=32\pi$ $cm^{3}/sec$.

Answer:

$32\pi$