Expert: Alon Mandes - 9/30/2008

Question
Hi, could you help me with this related rate problem?

a horizontal cylindrical oil tank (the length is parallel to the ground) of radius 2.00m is being emptied. Find how fast the width w of the oil surface is changing when the depth h is 0.500m and changing at the rate of 0.0500 m/min.

Answer
Ok, first of all let's find a relation between w & h . From simple
geometry & Pythagoras theorem you can find out that :
w=2*(4-h^2)^(1/2). Explanation : "My advise is to sketch a circle & create a triangle, which it's base is w & it's vertex is the center of the circle. That means that it's height is h, & so that the relation that I wrote stands ".
Now that we have the relation, we need to find h(t).
We know that h'(t)=1/20, so practically h(t)=4-t/20. " I performed
integration". Thus, let's find w(t) : w(t)=2*(4-h(t)^2)^(1/2).
w(t)=2*[4-(4-t/20)^2]^(1/2)=2*[4-(16-2t/5+t^2/400)]^(1/2)
w(t)=2*[(2/5)t-t^2/400-12]^(1/2). Great, so now we will have to
derive w(t) & calculate the derivative when h=1/2. So, first
let's calculate at which time to the height is 1/2 meter ??
We get back to our formula w=2*(4-h^2)^(1/2) & we substitute h=1/2
wo=2*(4-ho^2)^(1/2) --> wo=2*(4-1/4)^(1/2)=3.87 . All right,
Now we have to calculate w'(3.87). I will leave it to you as an
exercise....

Alon.