Hi Everyone, I'm hoping someone can help me solve this problem. It is one of those that you get at the end of all the problems in a chapter, less intuitive and tougher. This one comes from Calculus by Finney and Thomas (first edition) probelm #47. The answer in the back of the book is

| x - 3.385 | <= 0.002 but I don't know how to get to it. Here is the problem:

Grinding Engine Cylinders.

Before contracting to grind engine cylinders to a cross-section area of 9 in sqr, you want to know how much deviation from the ideal cylinder diameter of x0 = 3.385in.

You can allow and still have the area come within 0.01 in sqr of the required 9 in sqr.

To find out, you let A = pi (x/2)sqr and look for the interval in which you must hold x to make

|A - 9| <= 0.01.

What interval do you find?

Solution

Short of sustituting A = pi (x/2)sqr into |A - 9| <= 0.01 and solving for x to tget x <= 3.387 which i can see is the difference between the 2 amounts in the solution ( | x - 3.385 | <= 0.002 ).. I wouldn't don't know how to go about solving this.

And having just written this all out I've realised how to do it ( and so will not delete this in case it comes to use for somebody else....)

So I realised that since we are trying to find the values within which the diamter needs to be in, this would be formularised as such:

| x - diameter | <= Error

Where Error = my result of x <= 3.387 lessdiameter 3.385 = 0.002 , substituting gives:

| x - 3.385 | <= 0.002