The diameter and height of a paper cup in the shape of a cone are both 4 inches and water is leaking out at the rate of .5 cubic inch per second. Find the rate at which the water level is dropping when the diameter of the surface is 2 inches.
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v= volume
r= radius
h= height
All at any time (t)
We're given that the height of the cone is 4 and the diameter of the cone is 4. Thus, the radius of the cone is 2.
The governing equation for volume of the cone is: v =1/3pir^2h, however we need to equate h and r (so that we only have one variable to deal with). 4/h = 2/r ===> h=2r
Plugging back into our governing equation for h we get v=1/3pir^2(2r) = 2/3pir^3.
Now we have an equation for v, we also have information for dv/dt! We know that dv/dt=1/2. So now we take the derivative of our governing equation and we get that dv/dt = 2pir^2*dr/dt.
Solving for dr/dt we get dr/dt=1/(4pir^2).
This solves in general, we now want to know what is happening exactly when the diameter is 2 (or as we've equated, our radius =1) essentially we need dr/dt when r=1.
Plugging into our general equation we get dr/dt when r=1 is 1/4pi.
r= radius
h= height
All at any time (t)
We're given that the height of the cone is 4 and the diameter of the cone is 4. Thus, the radius of the cone is 2.
The governing equation for volume of the cone is: v =1/3pir^2h, however we need to equate h and r (so that we only have one variable to deal with). 4/h = 2/r ===> h=2r
Plugging back into our governing equation for h we get v=1/3pir^2(2r) = 2/3pir^3.
Now we have an equation for v, we also have information for dv/dt! We know that dv/dt=1/2. So now we take the derivative of our governing equation and we get that dv/dt = 2pir^2*dr/dt.
Solving for dr/dt we get dr/dt=1/(4pir^2).
This solves in general, we now want to know what is happening exactly when the diameter is 2 (or as we've equated, our radius =1) essentially we need dr/dt when r=1.
Plugging into our general equation we get dr/dt when r=1 is 1/4pi.
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