# Homework Help: Linear approximation and errors

1. Oct 23, 2007

### anthonym44

[SOLVED] Linear approximation

1. The problem statement, all variables and given/known data
Juan measures the circumference C of a spherical ball at 40cm and computes the ball's volume V. Estimate the maximum possible error in V if the error in C is as most 2cm. Recall that C=2(pi)r and V=(4/3)pi(r)

2. Relevant equations
deltaf - f'(a)h, a= 40, V=(4/3)pi(C/2pi)^3

3. The attempt at a solution

im not sure exactly how to compute to find the final answer but i believe that a = 40cm + or - 2cm and that somehow solving for r using circumference and then plugging it into the volume equation. if you can point me in the right direction i would appreciate it, thanks.

2. Oct 23, 2007

### EnumaElish

The easiest way is to assume maximum error (plus, then minus). Then take the difference between the calculated volume and true volume.

Or you could try expressing V as a function of C.

3. Oct 23, 2007

### anthonym44

thanks, that actually helps me a lot.

4. Oct 23, 2007

### anthonym44

after solving for the volume when the radius is (21/pi) which is a +2cm error i got the answer 1251.1cm^3. when i solved for the volume when the radius was (19/pi) a -2cm error i got 926.6cm^3. last i solved for the actual measured raius which was 40 and found the radius to be (20/pi) when i solved for this i got 1080.8. i then subtracted 926.6 (-2cm error) from this answer and got 154.2. when i subtracted 1080.8 from 1251.1 (+2cm error minus actual) i got the answer 170.3. I'm guessing this means that the maximum error is 170.3cm^3. can anyone verify this?

5. Oct 24, 2007

### HallsofIvy

V=(4/3)pi(C/2pi)^3= C^3/(6pi^2) is exactly what you want. What is the derivative of V with respect to C?

6. Oct 24, 2007

### anthonym44

4pi(c/2pi)^2

7. Oct 24, 2007

### anthonym44

i think i finally got the answer to be 324.22pi by plugging in 40 to the above equation for C and then multipliing by delta x which in this case is 2