# Chapter 3 - Section 3.1 - Tangents and the Derivative at a Point - Exercises - Page 118: 32

The rate of change of the volume $V$ of the ball with respect to $r=2$ is $16\pi$.

#### Work Step by Step

The function of the volume of the ball with a radius $r$ is $f(r)=(4/3)\pi r^3$. The rate of change of the volume $V$ of the ball with respect to $r=2$ is the derivative $f'(2)$, which can be calculated by $$f'(2)=\lim_{h\to0}\frac{f(h+2)-f(2)}{h}=\lim_{h\to0}\frac{\frac{4}{3}\pi(h+2)^3-\frac{4}{3}\pi \times2^3}{h}$$ $$f'(2)=\lim_{h\to0}\frac{\frac{4}{3}\pi(h^3+6h^2+12h+8)-\frac{4}{3}\pi\times8}{h}$$ $$f'(2)=\lim_{h\to0}\frac{\frac{4}{3}\pi h^3+8\pi h^2+16\pi h}{h}=\lim_{h\to0}\Big(\frac{4}{3}\pi h^2+8\pi h+16\pi\Big)$$ $$f'(2)=\frac{4}{3}\pi\times0^2+8\pi\times0+16\pi=16\pi$$ The rate of change of the volume $V$ of the ball with respect to $r=2$ is $16\pi$.

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