Using Differentials to find Error and Percent Error

grapeape
Messages
10
Reaction score
0

Homework Statement


One side of a right triangle is known to be 20 cm long and the opposite angle is measured as 30(degrees), with a possible error of +/- 1 degree.
a) Use differentials to estimate the error in computing the length of the hypotenuse
b) what is the percentage error.


Homework Equations





The Attempt at a Solution


Well, using the given data I found that the hypotenuse when x=30(degrees) is 40 cm. The equation I used was h(x)=20/sin(x). I know that the change in (h) is equal to error x h'(x). When finding h'(x) I got -20cos(x)/sin(x)^2. I'm not sure if this is correct. My book doesn't do a great job at explaining anything so any help will be greatly appreciated!
 
Physics news on Phys.org
You're almost there. Expressed in terms of differentials, h' = dh/dx. So write dh in terms of dx. dx will represent the error in angle (be sure to express dx in radians, not degrees); the equation for dh will tell you the corresponding error in the hypotenuse.
 
so do I do ( -20cos(30)/sin(30)^2 ) x pi/180? if so I got -1.209. Can error be negative? Or since it's both +/- 1 degree, do you also do another equation multiplying by a negative pi/180. My real questions are, how is error represented (multiple numbers, one number, a continuum)? And, how do I go about converting this to a relative error, and then percent error.

Thanks!
 
grapeape said:
so do I do ( -20cos(30)/sin(30)^2 ) x pi/180? if so I got -1.209.
Looks good. The angle is 30 +/- 1 degrees, so the hypotenuse is 40 +/- 1.2 cm. You can express that as a percent error--what percent of 40 is 1.2?
 
I love this forum. Thanks for all the help!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top