Question

The volume of water in a bowl is given by V = 1/3(pi)(h)^2(60 - h) cm^3, where h is the depth of the water in centimeters. If the depth is increasing at the rate of 4 cm/sec when the water is 10 cm deep, how fast is the volume increasing at the instant?

Answer 1

V = (1/3)pi(60h^2 - h^3)
dh/dt(at h=10) = 4

Find dV/dt (at h=10)

Answer 2

Can you solve the question now ??

Answer 3

Let me see. One second

Answer 4

5000pi/3 right?

Answer 5

mine was 1200 pi :(

Answer 6

Oh, how did you get it? I must have simplified! :D

Answer 7

Well. where do we go after finding 1200pi?

Answer 8

Would you mind giving me your solution ?

Answer 9

I'm on a study guide. I thought 5000pi/3 (the wrong one) was the final answer but I think we might need to do implicit diffs, right?

Answer 10

Post your solution for this question..I'll correct it :)

Answer 11

That's the only solution I had, 5000pi/3 :)
This is a real confusin section :P

Answer 12

I mean; show me your steps..how you reached the answer..

Answer 13

Well for the wrong one i just simplified (1/3)(pi)(60(10)^2-10^3)

Answer 14

You need to find dV/dt at h = 10
So differentiate first..then substitute h

Answer 15

ok let me see if i can

Answer 16

so
20pi^2-pi^2(h)/3
20pi^2-pi^2(10)/3
50pi^2/3

Answer 17

pi^2 ??

Answer 18

ya, im a little confused. my other answer still had p^2 in it? am i off?

Answer 19

pi is a constant..

Answer 20

so are you differentiating a constant ??

Answer 21

i might be. I'm kinda lost

Answer 22

V = (1/3)pi(60h^2 - h^3)
dv/dt = (1/3)pi(120h - 3h^2) * dh/dt
dv/dt = 1200 pi

Does that makes sense ??

Answer 23

ok a couple questions. where does th120 and 3 come from?

Answer 24

Do you know differentiation ??

Answer 25

finding derivative right?

Answer 26

Yea..what's the derivative of x^n ??

Answer 27

nx^n-1

Answer 28

Now, do you know Chain Rule ??

Answer 29

i think. Should i use it in this problem?

Answer 30

Whats the derivative of x^n with respect to t ?

Answer 31

I'm not sure

Answer 32

d(x^n)/dt = [d(x^n)/dx] * [dx/dt] = nx^(n-1) * dx/dt

Chain Rule ^^ :)

Answer 33

ok. so am i going to apply it to this one? I'm not all that sure about where to go after 1200pi

Answer 34

1200*pi is the rate at which volume is increasing at h=10

Answer 35

so, the final is 1200pi?

Answer 36

yep..

Answer 37

one second

Answer 38

I don't think because it needs to be in cm^3/sec

Answer 39

yea..it is in cm^3/sec

Answer 40

that's still not matching up. Let me write the equation out for you. holdon

Answer 41

1/3(pi)(h)^2(60 - h)
\[1/3pih^2(60 -h)\]

Answer 42

"...
V = (1/3)pi(60h^2 - h^3)
dv/dt = (1/3)pi(120h - 3h^2) * dh/dt
dv/dt = 1200 pi
..."

IDK what else to explain :(

Answer 43

Ok. 1 second let me see if i can verify it. @sourwing

Answer 44

@agent0smith Can you help here please..

Answer 45

thanks for calling someone. sour might be away

Answer 46

\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
\[V=\frac{1}{3}\pi h^2(60-h)\]
You're given \(\dfrac{dh}{dt}=4\), and you're to find \(\dfrac{dV}{dt}\) when \(h=10\).

Differentiate both sides with respect to \(t\) (I use the product rule, but nothing is stopping you from expanding the polynomial and just using the power rule term-by-term):
\[\frac{dV}{dt}=\frac{\pi}{3}\left[2h(60-h)\frac{dh}{dt}+h^2(-1)\frac{dh}{dt}\right]\]
Plugging in the given values, you have
\[\frac{dV}{dt}=\frac{\pi}{3}\left[2(10)(60-10)(4)+10^2(-1)(4)\right]=\frac{\pi}{3}(4000-400)=\color{red}{1200\pi}\frac{\text{cm}^3}{\text{s}}\]
\(\color{blue}{\text{End of Quote}}\)