Question

The volume of a sphere with radius r is V(r)=(4/3)pir^3. If a child is adding snow to a perfectly spherical snowball at a rate of 5cm^3/second, at what rate is the radius of the snowball increasing when the radius is 10cm?

Answer 1

Is this a calculus question?

Answer 2

@whpalmer4 yes, related rates; given \(V=\frac43\pi r^3\) and \(\frac{dV}{dt}=5\); what is \(\frac{dr}{dt}\) when \(r=10\)?

Answer 3

Yeah, just didn't want to suggest a calculus solution if it wasn't a calculus class!

Answer 4

yeah its calculus

Answer 5

@heradog do you know how to find \(\dfrac{dr}{dV}\)?

Answer 6

\[dr/dt=dV/dt*dr/dV\] Right? but not sure how to go about it no

Answer 7

Can you solve the volume equation for r?

Answer 8

4188.79

Answer 9

no, solve for r in terms of V

Answer 10

This looks like rate of change.

Answer 11

no or now?

Answer 12

no. we aren't ready for numbers yet.

Answer 13

this is why I'm so stupid

Answer 14

Nah, you just don't understand it yet, big difference between that and stupid!

Answer 15

We need to find an expression for r in terms of V. When we have that, we will differentiate it with respect to V to give us dr/dV, as a formula. We'll multiply that by our known value of dr/dt and evaluate it at r = 10...

Answer 16

okay so instead of v= I need r=?

Answer 17

Okay, \[V=\frac{4}{3}\pi r^3\]\[\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}\](by implicit differentiation)
Solve that for \(\frac{dr}{dt}\) and we get
\[\frac{dr}{dt} = \frac{dV}{dt}*\frac{1}{4\pi r^2}\] and now we plug in our numbers:
\[\frac{dr}{dt} = (5 cm^3/s)*\frac{1}{4\pi (10 cm)^2} = \frac{5}{400\pi}\frac{cm}{s} \approx 0.00398\frac{cm}{s}\]

Answer 18

If we want to check, we can evaluate the volume at the time r = 10 cm:

\[V = \frac{4}{3}\pi(10)^3 \approx 4188.79 \]And now after 1 second passes, find the new radius:
\[V = \frac{4}{3}\pi r^3\]\[\sqrt[3]\frac{3V}{4\pi} = r\]\[\sqrt[3]{\frac{3*(4188.79+5)}{4\pi}} \approx 10.004\]so our answer looks reasonable.

Answer 19

Okay cool