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Another differential problem I just can't close the deal on...
- Thread starter MrJoe2000
- Start date
D
Deleted member 4993
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A conical cistern is 12ft across the top and 12ft deep. Water is being poured into the cistern at 2cu. ft. per second. How fast is the area of the top surface increasing when the water is 6 feet deep?
Start with:
1) Drawing a sketch of the cone as a isoscales triangle.
2) Draw a water level at height h.
3) what is the radius of the top area at height h? what is the surface area(A)? Can you write an equation for the volume of the fluid (V) - within the cone - under the top surface, as a function of A
4) Now differentiate wrt to time.
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I've completed the problem up to the point where I get a dx/dt term which represents the changing of water level with time. I'm not sure how to find this rate.
So basically from the volume equation I get:
6/pi= 9*dx/dt + 36 dr/dt
how do I figure out this dx/dt term?
So basically from the volume equation I get:
6/pi= 9*dx/dt + 36 dr/dt
how do I figure out this dx/dt term?
D
Deleted member 4993
Guest
I've completed the problem up to the point where I get a dx/dt term which represents the changing of water level with time. I'm not sure how to find this rate.
So basically from the volume equation I get:
6/pi= 9*dx/dt + 36 dr/dt
how do I figure out this dx/dt term?
What is 'x'?
What is 'r'?
How did you get to the equation you wrote?
Where did you begin?
V= pi/3*r^2*x, where r is the radius and x is the height of the water. They're asking about the area when x=6. When x=6, r=3
A= pi*r^2
then you take the derivative of V: dV/dt= pi/3*[(r^2 *dx/dt) + (2xr * dr/dt)]
now you take the derivative of A to find the change in surface area with respect to time: dA/dt= 2pi*r*dr/dt
Then you find dr/dt from the dV/dt equation and then plug it back into the dA/dt equation to get the solution.
The problem I'm having is that I cannot come up with a value for dx/dt (which as I said before is the change in water height with respect to time)
A= pi*r^2
then you take the derivative of V: dV/dt= pi/3*[(r^2 *dx/dt) + (2xr * dr/dt)]
now you take the derivative of A to find the change in surface area with respect to time: dA/dt= 2pi*r*dr/dt
Then you find dr/dt from the dV/dt equation and then plug it back into the dA/dt equation to get the solution.
The problem I'm having is that I cannot come up with a value for dx/dt (which as I said before is the change in water height with respect to time)
D
Deleted member 4993
Guest
V= pi/3*r^2*x, where r is the radius and x is the height of the water. They're asking about the area when x=6. When x=6, r=3
A= pi*r^2
then you take the derivative of V: dV/dt= pi/3*[(r^2 *dx/dt) + (2xr * dr/dt)]
now you take the derivative of A to find the change in surface area with respect to time: dA/dt= 2pi*r*dr/dt
Then you find dr/dt from the dV/dt equation and then plug it back into the dA/dt equation to get the solution.
The problem I'm having is that I cannot come up with a value for dx/dt (which as I said before is the change in water height with respect to time)
If the cone is filled to height 'x' - the top surface of the fill line would be circle of radius 'r'
From similar triangle:
x/r = 12/6 → r = x/2
Area of the top surface A = π*r2 → r = √(A/π) → x = 2√(A/π)
The volume upto the fill-line:
V = π/3 * r2 * x = π/3 * (A/π) * 2√(A/π) = 2/(3√π) * A3/2 = = =
Now differentiate........
D
Deleted member 4993
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Okay so the frustrating part is that I'm getting the calculus right but I can't remember my simple geometry. Can you tell me where you got the x=2*r from?
.
D
Deleted member 4993
Guest
Is that relationship defined because there is a right triangle or just because the radius at the top of the cistern is 6 and the height is 12? I don't think I would have come up with that relationship on my own.
Did you draw a sketch - as I had suggested?