Question

Help with rates of change.

A conical tank (with vertex down) is 9 ft. across the top and 16 ft. deep. If the water is flowing into the tank at a rate of 11 cubic ft. per minute, find the rate of change of depth of water when the tank is 8 ft. deep.

Answer 1

There are 7.48 gallons in cubic foot.

Answer 2

Ok let's make sure I have everything drawn correctly.
I think this is going to involve using similar triangles.

We were given \(\large V'=11\).
They tell us to find \(\large h'\) when \(\large h=8\).

Answer 3

So we'll want to use the formula for Volume of a Cone.\[\large V=\frac{1}{3}\pi r^2h\]

Hmm

Answer 4

So we'll need to take a derivative.
But before we do that, I think we need to get our volume function in terms of \(\large h\) alone.

We don't want to differentiate it while it has both \(\large h\) and \(\large r\) in it.
Hmmmm

Answer 5

Mmmm I think what we want to do is this...We can rewrite our \(\large r\) in terms of \(\large h\) using similar triangles.
Refer back to the previous picture I drew so you understand where these side lengths are coming from.

Ignore the fact that I called it \(\large r_2\) in the first picture, I shoulda just called it \(\large r\).

Answer 6

Relating the sides of these triangles like so,
\[\large \frac{h}{16}=\frac{r}{4.5}\]Rearranging things a tad, to solve for \(\large r\),\[\large r=\frac{9}{32}h\]

Answer 7

Hmm I hope I'm on the right track with this.. :p

Are you even here or no? XD

Answer 8

\[\large \color{orangered}{r=\frac{9}{32}h}\]
\[\large V=\frac{1}{3}\pi \color{orangered}{r}^2h \qquad \rightarrow \qquad V=\frac{1}{3}\pi \left(\color{orangered}{\frac{9}{32}h}\right)^2h\]

Answer 9

@zepdrix -- sorry, I stepped out for a bit; I'm here now, though. :)

Answer 10

Ok :o go back and check out what's been worked out so far. I think we're on the right track.

We've successfully written our Volume function in terms of \(\large h\) alone.
You should probably simplify it a tad before taking the derivative.

Then, take the derivative (with respect to time) and solve for h'.

Answer 11

All right. So, we would have take the derivative of that crazy V(h) function?/ :3

Answer 12

It's not so hard-- just the chain rule, product rule, constant multiple(I think) rule.

Answer 13

Let me see if I can do that. :D

Answer 14

Yah it shouldn't be too tough :O Let's simplify it first.
I think you get something like this before taking a derivative. Lemme know what you come up with.\[\large V=\frac{1}{3}\pi \left(\frac{9}{32}h\right)^2h \qquad \rightarrow \qquad V=\frac{81}{3072}\pi h^3\]

Answer 15

If you simplify it first, you don't have to worry about the product or chain rule.

Answer 16

Oh, I thought I would come up with a wrong answer if I simplified too early.

Answer 17

Hmm, I got something different. ;o \[V=\frac{ 27 }{ 1024 }h^3\]

Answer 18

Yah that's the same as mine :) I forgot to simplify the fraction.

Answer 19

Oh, missing the pi:\[V=\pi \frac{ 27 }{ 1024 }h^3\]

Answer 20

Oh, lol. Okay. :D

Answer 21

So, I now I can take the derivative? :3

Answer 22

Do you have an answer key by chance? :O There's a possibility that I did this problem entirely wrong. Not a big chance, but it may have happened lol.
Would be nice if we can check out work :D

Yes, now derivative.

Answer 23

\[V'(h) = \pi \frac{ 27 }{ 1024 }*3h^2\] I hope...lol

Answer 24

I do have the answer to the problem, yes. :3

Answer 25

When we take the derivative with respect to `time`, it not only produces a V' on the left, it should produce an h' on the right.\[\large V'=\pi \frac{ 27 }{ 1024 }*3h^2h'\]

Answer 26

From here, just plug in the information they gave us,\[\large V'=11\]\[\large h=8\]And solve for \(\large h'\)

Answer 27

So, it would be:\[11=\pi \frac{ 27 }{ 1024 }*3(8)^2*h'\]

Answer 28

true story

Answer 29

I got h' = 0.69

Answer 30

me too. Is that what we're looking for? :O

Answer 31

And checking the answer key; the answer is correct! :D

Answer 32

Yayyy team \c:/

So as the volume increases by 11 cubic feet per minute,
when the height of the water is 8 ft,
the rate of change of the height of water is 0.69 ft/min.

yay

Answer 33

I just have to remember this long crazy process... for tomorrow. Calc. Final. DX

Answer 34

Yah this one is a little trickier than other related rates problems due to the similar triangles nonsense :(