Question

A spherical snowball is melting in such a way that its diameter is decreasing at rate of 0.4 cm/min. At what rate is the volume of the snowball decreasing when the diameter is 12 cm?

Rate of change of volume = cm3/min

Answer 1

im confused about what the formula is for volume

Answer 2

@jim_thompson5910 can you help me out here?

Answer 3

?????? why u dont use the formula for the volume of a sphere?

Answer 4

so V = (4 pi a^3)/3

Answer 5

yes and u must get the derivative with respetc to time
i meant dV/dt

Answer 6

\[\frac{ dV }{ dt }=(4pir^2)\frac{ dr }{ dt }\]

Answer 7

but i need this in diameter

Answer 8

@julian25

Answer 9

hey Zepdrix

Answer 10

Hey c:

Answer 11

how have you been doing?

Answer 12

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

They told us that,\[\large d'=-0.4\]

Understand why this is negative?

Answer 13

Good good c:

Answer 14

because its decreasing

Answer 15

So they've told us to find \(\large V'\) when \(\large d=12\)

Answer 16

So we just need to rewrite our volume formula in terms of diameter right?

Answer 17

ya

Answer 18

How does diameter relate to radius? :D

Answer 19

d=2r
d d = 2 d r

Answer 20

d=2r

Answer 21

Hmm ok let's try using that, don't forget about the `cube` being applied to the entire substitution.

\[\large V=\frac{4}{3}\pi r^3 \qquad \rightarrow \qquad V=\frac{4}{3}\pi \left(\frac{1}{2}d\right)^3\]

Answer 22

oops its actually d d/dt = 2 dr/dt

Answer 23

I skipped a step in there, I hope that wasn't confusing.\[\large d=2r \qquad \rightarrow \qquad \frac{1}{2}d=r\]And then I made a substitution for the r.

Answer 24

na your good

Answer 25

V=4/3π(1/2d)^3 so dV/dt=4pi(1/2d)^2

Answer 26

So you should be able to simplify that down a bit.. I think it gives us something like this,\[\large V=\frac{1}{6}\pi d^3\]

I think we want to simplify before we take a derivative. Otherwise you'll need to remember to apply the chain rule. If you're comfortable with that, then it's fine.

Answer 27

Cause I can see in your dV/dt, you didn't apply the chain rule. A factor of 1/2 should be popping out from that 1/2d due to the chain rule.

Answer 28

ok, ya i suck at this stuff

Answer 29

so we would have 1/2pi(d)^2

Answer 30

Either way works fine. If we do it the way you did, we'll apply the chain rule to the inside.\[\large \frac{dV}{dt}=4\pi\left(\frac{1}{2}d\right)^2\frac{d}{dt}\left(\frac{1}{2}d\right)\]

Answer 31

Looks good, you're missing one really important piece of information though.

Answer 32

we should have called our diameter D, I'm realizing that now. lol

Answer 33

never to late lol

Answer 34

What happens when you take the derivative of diameter with respect to `time`.
The same thing that happens when you take the derivative of Y with respect to X.

Since it's in respect to a different variable, we need to multiply it by the dD/dt term.

Answer 35

I really really prefer prime notation, it makes it so much easier to read.
Let's try that real quick.

\[\large V=\frac{4}{3}\pi \left(\frac{1}{2}d\right)^3\]Taking the derivative we found to be,\[\large V'=4\pi\left(\frac{1}{2}d\right)^2\left(\frac{1}{2}d\right)'\]Which gives us,\[\large V'=2\pi\left(\frac{1}{2}d\right)^2d'\]

Answer 36

ok, but now where does -0.4 cm/min go?

Answer 37

Given:
\(\large d=12\)
\(\large d'=-0.4\)

Solve for \(\large V'\)

Just plug and chug!

Answer 38

ah ok got it

Answer 39

Ya you got it! c: too quick for me hah

Answer 40

so dV/dt=2π(1/2(12))^2(-.4)

Answer 41

Looks good.

Answer 42

ok tyvm, you got time for another?

Answer 43

?

Answer 44

Oh sorry was microwaving some soup c:
Sure