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

@jigglypuff314

Answer 2

factor out V = 1/3(pi)(h)^2(60 - h)
so = 20pih^2 - (1/3)pi h^3
then take derivative of that...

Answer 3

dV/dt = (dV/dh)(dh/dt) need to take derivatives.

If not then find the volume V at h=10 cm and then at 0.01 s later, V' for h=10.04cm, and take the numerical rate of change dV/dt= (V'-V))/0.01s

Answer 4

20(pi)(h)^2-(pi)(h)^3/3

Answer 5

take the derivative of that :)

Answer 6

40(pi)(h)-(pi)(h)^2

Answer 7

close :)
don't forget that when taking derivatives based on time you have to include dh/dt
at the ends
like 40(pi)(h)(dh/dt) - (pi)(h)^2(dh/dt)

before you didn't have to do that because when taking the derivative d/dx
it came out as dx/dx = 1
but now since your doing it as d/dt you'll have to keep the derivative parts :)

Answer 8

you were given that dh/dt = 4 and h = 10
so plug in so that
V' = 40(pi)(10)(4) - (3/2)(pi)(10)^2(4)
=

Answer 9

1000(pi)

Answer 10

yup :) that's what I got :)

Answer 11

one sec

Answer 12

http://answers.yahoo.com/question/index?qid=20140131174131AAGNOpX

Answer 13

1000/pie was incorrect

Answer 14

i did a txt this time

Answer 15

it's not 1000/pi
it's 1000pi

Answer 16

as for the yahoo answers
"MechEng2030" added an extra 0 to 200liters
"Michael" got what we got :)
and "Keith A" didn't convert radius to cm

Answer 17

we need cm^3/day

Answer 18

200 liters/day = 200000 cm^3/day

Answer 19

oh wait, do you mean for this question?

Answer 20

no, actually 1000p is not it either

Answer 21

I get \(1200\pi\) when I do the arithmetic

Answer 22

\[\frac{dV}{dh} = 40\pi h - \pi h^2\]\[\frac{dh}{dt} = 4\]\[\frac{dV}{dt} = \frac{dV}{dh}*\frac{dh}{dt} = 4(40\pi h - \pi h^2) = 4*(40*10\pi - 10*10*pi) = 4(300\pi)\]

Answer 23

did you say that for this problem she wanted cm^3/day and Not cm^3/sec ?

Answer 24

cm^3/sec

Answer 25

oh yeah, I see my mistake
@whpalmer4 was right about the 1200 cm^3/day :)

Answer 26

Answer to the original problem is \(1200\pi\).

Answer 27

incorrect

Answer 28

No, it is correct. 1200pi cm^3/s is the rate of change of the volume when the water is 10cm deep and rising 4 cm/s. If the answer your system expects is different, it is incorrect, or the problem has been transmitted incorrectly.

Answer 29

my head.........

Answer 30

basically, once again we agree that 1200 cm^3/day is correct
while you once again tell us that your teacher says that it is incorrect >,<

Answer 31

1200 just?

Answer 32

no, \(1200\pi \approx 3769.9111843077518862\)

Answer 33

yeah sorry, typo :P

Answer 34

darn...

Answer 35

Here's a check: at the precise moment where the height is 10 cm, the volume is

\[V = \frac{1}{3}\pi(60-10)(10^2) \approx 5235.99\]
At 1/1000 of a second later, the height has gone up by 1/1000*4 cm/s = 0.004 cm
h = 10.004 cm
\[V = \frac{1}{3}\pi(60-10.004)(10.004^2) \approx 5239.76\]\[\frac{dV}{dt} \approx \frac{\Delta V}{\Delta t} = \frac{5239.76-5235.99}{0.001} = 3770.41\]
\[1200\pi \approx 3769.91\]\[\text{error} = \frac{3770.41-3769.91}{3769.91} = 0.013\%\]

Answer 36

wow nice @whpalmer4 !!!

Answer 37

Even after all these years of doing related rates problems, I still like to do such a calculation to check my work :-)

Answer 38

ughhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

Answer 39

mmm just stick with these answers
and when possible, have your teacher give you her explanations for these problems
then come back aye? :P