Question

A spherical snowball is melting. Its radius is decreasing at 0.9 cm per hour when the radius is 15 cm. How fast is its volume decreasing at that time?

Enter the exact answer.

Answer 1

Please write the formula for the volume of a sphere, in terms of its radius.

Answer 2

\[V = \frac{ 4 }{ 3 }\pi r^3\]

Answer 3

Correct. You may notice that its volume will decrease \(V_{before}-V_{after}\). In this case, the radius before is 15, and the radius later is 15 - 0.9 = 14.1. Plug these into the formula for the volume

Answer 4

\(\large \Delta V= \frac {4\pi}{3}\ 15^3-\frac {4\pi}{3}\ 14.1^3\)

Answer 5

that's not the right answer, I put it in on my homework and it was wrong

Answer 6

We should be using differentials. Find \(\dfrac{dV}{dr}\)

Answer 7

Let's try tkhunny formula

Answer 8

Find the 1st derivative of Volume with respect to Radius.

Answer 9

\[V \prime = 4\pi r^2h\]

Answer 10

you should take the derivative with respect to time (understanding which variables are changing with time... in this case V and r)
so
\[ \frac{d}{dt} v = \frac{d}{dt} \frac{4}{3} \pi r^3 \\
\frac{dv}{dt}= 4 \pi r^2 \frac{dr}{dt}
\]

notice we have *rates* dv/dt = how fast is its volume decreasing at that time
and dr/dt = radius is decreasing at 0.9 cm per hour

Answer 11

notice we used the chain rule to do d/dt r^3 to get 3 r^2 dr/dt

Answer 12

They give you r= 15 cm, dr/dt = 0.9 cm/hr
so you can find dv/dt

Answer 13

What helps me understand partials is Leibnitz notation: \(\large \frac{\partial V}{\partial t} = \frac{\partial V}{\partial r}\cdot \frac{\partial r}{\partial t}\)