Question

the volume of a cantaloupe is approximated by ( v=\frac{4}{3}pi r^{3} ). the radius is growing at the rate of 0.8 cm/week, at a time when the radius is 7.9 cm. how fast is the volume changing at that moment?
the volume is changing at a rate of about
(round to one decimal place as needed.)

Explanation:

Step1: Differentiate the volume formula

The volume formula of a sphere is \(V=\frac{4}{3}\pi r^{3}\). Differentiate both sides with respect to time \(t\) using the chain - rule.
If \(y = f(x)\) and \(x = g(t)\), then \(\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}\).
Differentiating \(V\) with respect to \(t\): \(\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}\)

Step2: Substitute the given values

We are given that \(\frac{dr}{dt}=0.8\) cm/week and \(r = 7.9\) cm.
Substitute these values into the formula \(\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}\).
\(\frac{dV}{dt}=4\pi(7.9)^{2}(0.8)\)
First, calculate \((7.9)^{2}=62.41\)
Then \(4\pi\times62.41\times0.8=(4\times0.8)\pi\times62.41 = 3.2\pi\times62.41\)
\(3.2\times3.14\times62.41=10.048\times62.41\approx627.0\)

Answer:

\(627.0\)