How Accurately Can We Estimate Volume Error Using Tangent Line Approximation?

[gh_pluvilias]

Homework Statement

You measure a cubic container and find it to be about 10 cm on each side. From this, you conclude that it holds 1000 cc. However, your measurement is accurate only to within ±0.1 cm. (So you can be sure that the side is between 9.9 and 10.1 cm.)

What are x, f(x) , and a in this problem?
Use the tangent line approximation to estimate the error in your value of 1000 for the volume.
Give the percent error in your measurement of the length of the side.
Give the percent error in your estimate of the volume.

Homework Equations

f(a)+f'(a)(x-a)

The Attempt at a Solution

I just don't get this problem... this is all I've got figured out:
the width, w of the it is measured to be 10 cm with a error of plus or minus 0.1. V=10^3=1000. hence the 1000cc. I think I could use differentials to estimate max error in the calculated volume, but I don't know how to assign f(x), a and x... anyway this is what i have -- V=w^3 and V'=3w^2. So the differentials dV and dw I think would be related in dV=3w^2dw. Setting w to 10 yields dV=300dw

...and that's all I got.

Even if I am on the right track here I don't know where my equation fits into everything and how to answer the questions I'm given to begin with.

Please help, I need this soon!

 

[Dick]

Yes, dV=300cm^2*dw. dw=0.1cm or -0.1cm. Because those are the differences between 10cm and 9.9cm and 10.1cm. That gives you an estimate for the change in V, which is estimated by dV, yes?

 

[gh_pluvilias]

I'm not sure how to work this into the entire problem though, especially in regards to TLA

 

[Dick]

I'm not sure how to work this into the entire problem though, especially in regards to TLA

dV=300cm^2*dw IS the TLA. It's not quite the same as the true error, it's an approximation. But it's close. The true errors are the difference between 1000cm^3 and (9.9cm^3) and (10.1cm)^3.