# Thread: Differentials: estimate maximum percentage error

1. ## Differentials: estimate maximum percentage error

I'm having such a hard time with this one :/

A model for the surface area of a human body is given by $\displaystyle S = 0.1091w^{0.425}h^{0.725}$, where w is the weight (in pounds), h is the height (in inches), and S is measured in square feet. If the errors in measurement of w and h are at most 2%, use differentials to estimate the maximum percentage error in the calculated surface area.

2. Hi angel.white
Originally Posted by angel.white
A model for the surface area of a human body is given by $\displaystyle S = 0.1091w^{0.425}h^{0.725}$, where w is the weight (in pounds), h is the height (in inches), and S is measured in square feet. If the errors in measurement of w and h are at most 2%, use differentials to estimate the maximum percentage error in the calculated surface area.

The hint to is to work with $\displaystyle \ln S$ : $\displaystyle S=0.1091w^{0.425}h^{0.725} \implies \ln S =0.425\ln w +0.725\ln h+\ln 0.1091$

thus $\displaystyle \mathrm{d}\ln S=0.425\times\mathrm{d}\ln w + 0.725\times \mathrm{d}\ln h \Longleftrightarrow \frac{\mathrm{d}S}{S}=0.425\frac{\mathrm{d}w}{w} + 0.725\frac{\mathrm{d}h}{h}$

The maximum percentage error is the value of $\displaystyle \frac{\mathrm{d}S}{S}$ for $\displaystyle \frac{\mathrm{d}h}{h}=\frac{\mathrm{d}w}{w}=0.02$ and this is 2.3% as expected.

3. i know this is old, but i had that problem in my calculus homework today and i'm sure other people will in the future.

i didn't use lns or anything -
if you just try to find dS/S, it works out.
for ds, you need dS/dw + dS/dh
you end up with dS/S = ((.425)(.1091)(w^(.425-1))(h^7.25)dw + (.725)(.1091)(w^.425)(h^(.725-1))dh) / ((.1091)(w^.425)(h^.725))
do some cancellation and you get dS/S = .425dw/w + .725dh/h and then plug in .02 for (dw/w) and (dh/h)
it's essentially the same thing i guess, but i understand it better that way.