Differentiation application problem

  • #1

Homework Statement


An inverted cone has a depth of 10cm and a base radius of 5cm. Water is poured into it at the rate of 1.5 cm^3/ min. Find the rate at which level of water in the cone is rising, when the depth of water is 4cm.

Homework Equations




The Attempt at a Solution

[/B]

I know that this question requires the application of chain rule. However having only recently studied it, I am yet to fully understand it's application. Therefore looking for an answer which gives a correct approach to such questions.

Thanks.
 

Answers and Replies

  • #2
SteamKing
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Homework Statement


An inverted cone has a depth of 10cm and a base radius of 5cm. Water is poured into it at the rate of 1.5 cm^3/ min. Find the rate at which level of water in the cone is rising, when the depth of water is 4cm.

Homework Equations




The Attempt at a Solution

[/B]

I know that this question requires the application of chain rule. However having only recently studied it, I am yet to fully understand it's application. Therefore looking for an answer which gives a correct approach to such questions.

Thanks.
Well, PF doesn't provide answers to members, only guidance. You must try harder to understand the question.

For example, how would you calculate the volume of water in this cone, given the depth of the water?
 
  • #3
Ok, so what I've done is this -

In the ques. h=2r or r=h/2
So the formula for volume of cone is written in terms of 'h'.

Now,
dv/dt = 1.5 (given)
And dv/dh is calculated.

Then after applying chain rule, dh/dt comes out to be 3/8π

Now could you please tell me if I'm right?
Thanks
 
  • #4
SteamKing
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Ok, so what I've done is this -

In the ques. h=2r or r=h/2
So the formula for volume of cone is written in terms of 'h'.

Now,
dv/dt = 1.5 (given)
And dv/dh is calculated.

Then after applying chain rule, dh/dt comes out to be 3/8π
Is this (3/8)π or 3/(8π)?
 
  • #5
haruspex
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Then after applying chain rule, dh/dt comes out to be 3/8π
As SteamKing notes, that is technically (3/8)π, but maybe you did not mean that. Either way, I get a different answer. Please post your working.
 
  • #6
Volume of cone = 1/3π r2h
r=5 cm. h = 10 cm
r= h/2

1/3π (h2/4) h

πh3/12

dv/dt = 1.5
dh/dt = ?

dv/dt = dh/dt * dv/dh
1.5 = dh/dt * πh2/4

dh/dt = 6/πh2

Now putting h=4,
I got the answer 3/(8π)
 
  • #7
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The height of the cone is not 4 cm, that is the starting depth of the water, and not the height of the cone.
 
  • #8
SteamKing
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Volume of cone = 1/3π r2h
r=5 cm. h = 10 cm
r= h/2

1/3π (h2/4) h

πh3/12

dv/dt = 1.5
dh/dt = ?

dv/dt = dh/dt * dv/dh
1.5 = dh/dt * πh2/4

dh/dt = 6/πh2

Now putting h=4,
I got the answer 3/(8π)
You should write complete equations in your work, rather than isolated expressions.

For example, write

V = πh3/12 rather than just πh3/12

It makes your thoughts much easier to follow in extended calculations.
 

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