# Thread: Relative Error through Differentials

1. ## Relative Error through Differentials

Hey, I'm not sure if I got this answer correct, I've been having a hard time with using differentials to find the various types of errors, I was hoping someone could look over this and point me in the right direction if I'm way off.

The question is
The area of a right triangle with hypotenuse $H$ is given by the formula
$A=\frac{1}{4}H^2sin(2\theta)$
where $\theta$ is one of the acute angles. Use differentials to estimate the relative errors of the area $A$ if $H=4 cm$ exactly and $\theta$ is measured to be $30^o$ with an error of measurement of $\pm15'$
So I know that the relative error is found through the quotient relation $\frac{\triangle A}{A}\approx\frac{dA}{A}=\frac{(A')(d\theta)}{A}$, so to solve this, I plug the differentiated function $A'$ and the given function $A$ in, as well as $d\theta=\frac{0.25}{30}=0.008333$ right? *I got my $d\theta$ value from the ratio between the number of erroneous minutes per whole degree by the total number of degrees.

Is this the correct path to take? My work to solve the problem is below.

$\frac{\triangle A}{A}\approx\frac{dA}{A}=\frac{(A')(d\theta)}{A}$
$\frac{(A')(d\theta)}{A}=\frac{[(\frac{1}{2}H)(\sin(2\theta))+(\frac{1}{4}H^2)(\co s(2\theta)2)](0.008333)}{\frac{1}{4}H^2\sin(2\theta)}$
$\frac{(A')(d\theta)}{A}=\frac{(2\sin(60)+4\cos(60) )(0.008333)}{4\sin(60)}$
$\frac{(A')(d\theta)}{A}=\frac{1.732+4}{3.464}$
$\frac{(A')(d\theta)}{A}=0.01373=\boxed{1.373\%}$

So the relative error is $1.373\%$?

Thanks for taking the time to look this over!
-Kasper

2. Originally Posted by Kasper
$\frac{\triangle A}{A}\approx\frac{dA}{A}=\frac{(A')(d\theta)}{A}$
$\frac{(A')(d\theta)}{A}=\frac{[(\frac{1}{2}H)(\sin(2\theta))+(\frac{1}{4}H^2)(\co s(2\theta)2)](0.008333)}{\frac{1}{4}H^2\sin(2\theta)}$
You have found wrong $A'$

$\frac{(A')(d\theta)}{A}=\frac{[(\frac{1}{4}H^2)(\cos(2\theta)2)](0.008333)}{\frac{1}{4}H^2\sin(2\theta)}$

Explanation:
$
A=\frac{1}{4}H^2sin(2\theta)
$

$
dA=(\frac{1}{2}H)(\sin(2\theta))dH+(\frac{1}{4}H^2 )(\cos(2\theta)2)d\theta
$

since dH=0, hence the expression used should as given by me above

3. Originally Posted by malaygoel
You have found wrong $A'$

$\frac{(A')(d\theta)}{A}=\frac{[(\frac{1}{4}H^2)(\cos(2\theta)2)](0.008333)}{\frac{1}{4}H^2\sin(2\theta)}$

Explanation:
$
A=\frac{1}{4}H^2sin(2\theta)
$

$
dA=(\frac{1}{2}H)(\sin(2\theta))dH+(\frac{1}{4}H^2 )(\cos(2\theta)2)d\theta
$

since dH=0, hence the expression used should as given by me above
Using that I got a relative error of $0.9622\%$, is that what you got? That does make perfect sense though, thanks!

I was misunderstanding, I thought for some reason that I wasn't to include the differential of H and $\theta$ in the differentiation of A. But to just multiply the differentiated function A' by the differential of $\theta$. Thanks again!