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Is there an alternative method to solve this problem?

  • Thread starter asap9993
  • Start date
  • #1
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Homework Statement


Here's the problem:
Water is flowing into a water bottle at a rate of 16.5 cm^3/s. The diameter of the bottle varies with its height. How fast is the water level rising when the diameter is 6.30 cm?


Homework Equations


dV/dt = 16.5

radius = r = 3.15

dh/dt = ?

The Attempt at a Solution


I know it's a related rates problem, but there's no obvious geometric formula to take the derivative of. Or is there? So I pictured a regular Poland Spring water bottle. I noticed that most sections of the bottle come pretty close to a cylinder. So I guessed that I could approximate the rate with that of a cylinder.

V = pi(h)(r^2)

dV/dt = pi(r^2)(dh/dt)

the radius is the constant throughout a cylinder, so r^2 is a constant.

16.5 = pi(3.15^2)(dh/dt)

dh/dt = 0.529 cm/s

This is the correct answer, but is this the correct way to do this problem. I don't like taking risky guesses and approximations. Luckily it worked this time.
 

Answers and Replies

  • #2
Delphi51
Homework Helper
3,407
10
I don't care for the V = pi(h)(r^2).
Better to picture an infinitesimally thick layer of water of height dh, which has volume dV = πr²*dh. That gives you an expression for dh/dV that you can put together with the given value for dV/dt to get dh/dt. Same answer.
 
  • #3
tiny-tim
Science Advisor
Homework Helper
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hi asap9993! :smile:

(have a pi: π and try using the X2 tag just above the Reply box :wink:)

here's a reasonably systematic way …

the basic formula for volume is V = ∫x=0h A dx (A = area, h = height).

So dV/dh = … ? :smile:
 
  • #4
19
0
To tiny-tim,

I don't quite understand why we are concentrating on the rate of change of the volume with respect to height instead of time. If you define volume as that integral then,

dV/dh = A = pi(r^2) = pi(3.15^2) = 31.17 cm^2

I guess we could find how the height is changing with respect to time by the chain rule:

dh/dt = (dV/dt)(dh/dV)

dh/dt = (16.5)(1/31.17) = 0.529 cm/s

Is that it?
 
  • #5
tiny-tim
Science Advisor
Homework Helper
25,832
250
hi asap9993! :wink:
I don't quite understand why we are concentrating on the rate of change of the volume with respect to height instead of time.

I guess we could find how the height is changing with respect to time by the chain rule:

Is that it?
yes, that's right :smile:

do the obvious first (volume vs height), then use the chain rule to adapt it to the question asked :wink:
 

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