Related Rates

Question

Label the diagram and given information, model with an equation, differentiate and evaluate. Then answer the question with correct units. A coffee pot that has the shape of a circular cylinder of radius 4 inches is being filled with water flowing at a constant rate. At what rate is the water flowing into the coffee pot when the water level in the pot is rising at a rate of 0.3 in/sec? Volume of a cylinder: V = pir^2h

Answer 1

The key to solving any related rates problem is to write a function containing only one time dependent variable. Let me explain using the situation you have presented. 

 

We are looking for the rate at which the volume, V, is changing. That is, we want to find dV/dt (change in Volume over a  change in time). We are given the rate at which the height, h, is changing (dh/dt = 0.3 in/sec, change in height over change in time). Therefore, in order to solve the problem we need to write an equation for V as a function of h then take the derivative of both sides. Well, the problem already states the relationship for us.

 

Volume of a cylinder is: V = π r² h = Area of circle times h

 

So now, we need to think about what parts of this relationship are changing with time.

 

Is volume changing with time, that is, is volume changing as the height of water increases? Yes. And that is the rate we are looking for.

 

Is height changing with time? yes. and that rate is given as 0.3 in/sec.

 

Now, is the radius of the cylinder (of the circle) changing with time? No. the cylinder is not getting bigger or smaller, it is just being filled with water. So the radius is constant, and it is given in the problem (r = 4 inches).

 

So we can think about the relationship in this way:

V(t) = π r² h(t)

 

Now, we take the derivative of both sides:

 

dV/dt = π r² dh/dt  (remember, r is a constant, so we just move it in front of the derivative, just like we do for π)

 

Now, plug in what we know (dh/dt = 0.3 inches/sec and r = 4 inches)

 

dV/dt = π (4 inches)² (0.3 inches/sec)

 

And we evaluate to find:

 

dV/dt = 3.77 inches³/sec

The units make since because we have inches squared times inches, so that gives us inches cubed

 

It is important to note, this problem is simple because the radius does not change with time. If it did, we would need to find a way to relate r and h (write an equation for r as a function of h). That way, we could substitute in for r and have h as our only time dependent variable in our function.