- #1
asd1249jf
Hello.
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 bizarre 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!
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 bizarre 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!