Hello.(adsbygoogle = window.adsbygoogle || []).push({});

I've encountered a problem when trying to attack this problem:

A right triangle has sides of length x, y and z. Measurements show that legs x and y have lengths of 119ft and 120ft, respectively. The percentage relative uncertainty in each measurement is 1%. For x, this means that 100 * dx / x = 1. (That 1 would be reported as 1%)

A)Use the pythagorean theorem to calculate the percentage relative uncertainty in z

B)Use the law of cosines and your result from part a to calculate the percentage relative uncertainty in the right angle.

So for part A, z = (x^2 + y^2)^.5

therefore,

dz = (.5(2x)^-.5)dx + (.5(2y)^-.5)dy

if I simply substitute the values in for x, dx, y and dy I get a bizzare answer for the percentage relative unceratainty, and 100*dz/z should be 1. Where am I going wrong with this problem?

and for part B)

z^2 = x^2 + y^2 - 2xycos(theta)

so the partial derivation would be

2z = -2xy(-sin(theta) * curlyd(theta)/curlydz)

but I am completely stuck where to go from here.

A help will be greatly appriciated!

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Help! Partial Derivation to calculate percentage relative uncertainty

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**