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Calculus/Related Rates of a Cone

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Question
Water is leaking out of an inverted conical tank at a rate of 10,000 cm/min.  The tank has a height of 6m and the diameter at the top is 4m.  Find the rate at which the depth of the water in the tank is changing when the height of the water is 2m.  How do I go about solving this problem?

Answer

drawing
Ok William ,let's define some variable based on the drawing in the
attachment :

R=4m   (" The radius of the inverted base ")
H=6m   (" The height of the Conical tank ")
h=h(t) (" The height of the water as a function of time ")
d=d(t) (" The depth of the water as a function of time ")
Γ=Γ(t) (" The amount of water remaining in the tank ")
A=A(t) (" The anount of water droped ")

Here are some know facts :
1. d(t)+h(t)=H=6m
2. The volume of the tank is Vo=(1/3)πHR^2=32π
3. The volume of the water remaining in the tank is Γ(t)=Vo-A(t)
4. A(t)=10000t (" Because A'(t)=10000 ")
5. Γ(t)=(1/3)πh(t)r^2(t)

Now, considering the left sketch of the drawing, we claim that :
h/H=r/R -> h/6=r/4 -> h=(3/2)r -> d=6-h -> d(t)=6-(3/2)r(t).
What we need now to find is the r(t) :
Γ(t)=(1/3)πh(t)r^(t) & Γ(t)=Vo-A(t)=32π-10000t , so
(1/3)πh(t)r^2(t)=32π-10000t
(1/3)π(3/2)r(t)r^2(t)=32π-10000t
2πr^3(t)=32π-10000t -> r(t)=[16-(50000/π)t]^(1/3)
This gives us :
d(t)=6-(3/2)[16-(50000/π)t]^(1/3) We found the depth changing as function of time. To calculate the rate , we derive :
d'(t)=-(3/2)*[-50000/π][16-(50000/π)t]^(-2/3)
d'(t)=[7500/π][16-(50000/π)t]^(-2/3). Now we need to know at which
time the height of the water is 2m..
If h(to)=2m then d(to)=4m, so we will solve this equation :
4=6-(3/2)[16-(50000/π)to]^(1/3) (" to is the time the water is 2m hight")
Once we calculated to, we plug it into d'(t) & we calculate d'(to).
I'll leave it to you to do it.

Alon.  

Alon Mandes

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