# Estimate relative error using differentials

• Aug 8th 2010, 01:28 PM
tillymc
Estimate relative error using differentials
Area of right triangle with hypotenuse H is

A=(1/4)H^2sin(2theta)

where theta is one of the acute angles.

Use differentials to estimate the relative errors of the area A if H=4cm and theta is measured to be 30 degrees with an error of measurement of 15 minutes of arc.

note: a minute of arc, 1' is approximately equal to (1/60) of a degree.

I'm not quite sure what a measurement of arc is and whether i'm supposed to find the derivative or what? Im so lost, thanks in advance.
• Aug 8th 2010, 07:02 PM
skeeter
Quote:

Originally Posted by tillymc
Area of right triangle with hypotenuse H is

A=(1/4)H^2sin(2theta)

where theta is one of the acute angles.

Use differentials to estimate the relative errors of the area A if H=4cm and theta is measured to be 30 degrees with an error of measurement of 15 minutes of arc.

note: a minute of arc, 1' is approximately equal to (1/60) of a degree.

I'm not quite sure what a measurement of arc is and whether i'm supposed to find the derivative or what? Im so lost, thanks in advance.

if $\displaystyle \theta$ is equal to 30 degrees, then $\displaystyle A = 2\sqrt{3} \, cm^2$

note that $\displaystyle 15' = \left(\frac{15}{60}\right)^\circ = \frac{\pi}{720} \, rad$

radians need to be used since the derivative of sine = cosine only works in radians.

$\displaystyle A = \frac{H^2}{4} \sin(2\theta)$

$\displaystyle \frac{dA}{d\theta} = \frac{H^2}{2} \cos(2\theta)$

$\displaystyle dA = \frac{H^2}{2} \cos(2\theta) \, d\theta$

$\displaystyle dA = \frac{4^2}{2} \cos\left(\frac{\pi}{3}\right) \cdot \frac{\pi}{720}$

$\displaystyle dA = \frac{\pi}{180}$

$\displaystyle \frac{dA}{A} \approx 0.005$ ... area error is about 0.5%