Question

1. the volume v of a sphere of radius r changes over time t.

a. find an equation relating dv/dt to dr/dt.
b. at what rate is the volume changing if the radius increases at 2 in/min when the radius is 4 inches?
c. at what rate is the radius changing if the volume increases at 10 in³/min when the radius is 5 inches?

Explanation:

Step1: Volume formula of sphere

The volume formula of a sphere is \(V=\frac{4}{3}\pi r^{3}\).

Step2: Differentiate \(V\) with respect to \(t\)

Using the chain - rule \(\frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}\).
Differentiate \(V = \frac{4}{3}\pi r^{3}\) with respect to \(r\): \(\frac{dV}{dr}=4\pi r^{2}\).
So, \(\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}\).

Step3: Solve part b

Given \(\frac{dr}{dt} = 2\) in/min and \(r = 4\) inches.
Substitute into \(\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}\).
\(\frac{dV}{dt}=4\pi(4)^{2}\times2\)
\(=4\pi\times16\times 2\)
\(=128\pi\approx 402.12\) in³/min.

Step4: Solve part c

Given \(\frac{dV}{dt}=10\) in³/min and \(r = 5\) inches.
From \(\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}\), we can solve for \(\frac{dr}{dt}\).
\(\frac{dr}{dt}=\frac{\frac{dV}{dt}}{4\pi r^{2}}\)
Substitute \(\frac{dV}{dt}=10\) and \(r = 5\)
\(\frac{dr}{dt}=\frac{10}{4\pi(5)^{2}}=\frac{10}{100\pi}=\frac{1}{10\pi}\approx0.0318\) in/min.

Answer:

a. \(\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}\)
b. \(\frac{dV}{dt}\approx402.12\) in³/min
c. \(\frac{dr}{dt}\approx0.0318\) in/min