errors in related rates

Question

You are wondering how inaccuracies in the radius or height of your can effect the volume of the can. Use differentials to answer the questions below. 

An error of p percent in the radius will cause an error of approximately 

?????

 percent in the volume of the can. 

An error of p percent in the height will cause an error of approximately 

?????

 percent in the volume of the can. 

Note that although the sentences above contain the word "approximately", the answers are actually determined precisely by the procedure with which we compute them. For example for the dependence on the radius you consider h constant, and you use the differentials 

dV=V′(r)dr.

The percentage of the error in v is100dV/V and you know that p=100dr/r and dV=(dV/dr)dr.

Your answer will depend on p. Proceed similarly for the dependence of the error on h.

Answer 1

The volume of a cylinder with radius r and height h is

V = πr²h (1)

Next we differentiate (1) with respect to r and h.

(Strictly speaking we should be using partial derivatives, but I will follow the procedure outlined in the

question.)

dV/dr = 2πrh

dV/dh = πr²

Rewrite both as differentials

dV = 2πrhdr

dV = πr²dh

Divide both by (1)

dV/V = 2πrhdr/πr²h = 2dr/r (2)

dV/V = πr²dh/πr²h = dh/r (3)

Consider equation (2)

dr/r represents relative error in r.

dV/V represents relative error in V when h is held constant,

and it is twice the relative error in r.

Consider equation (3)

dh/h represents relative error in h.

dV/V represents relative error in V when r is held constant,

and it is equal the relative error in h.

The question talks about percentage error.

For that we multiply the results by 100 to arrive at an equivalent conclusion

100dV/V = 2(100dr/d)

100dV/V = 100dh/h

In the terminology of the question:

An error of p percent in the radius will cause an error of approximately 2p percent in the volume of the can.

An error of p percent in the height will cause an error of approximately p percent in the volume of the can.