Linearization and Differentials and Percent Error

[SeekerOfDragons]

I have a Calc problem I'm having trouble getting started and figuring out the correct answer:

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The circumference of the equator of a sphere is measured as 10 cm with a possible error of .4 cm. This measurement is then used to calculate the radius. The radius is then used to calculate the surface area and volume of the sphere. Estimate the percentage errors in the calculated values of:
a) the Radius
b) the Surface Area
c) the Volume
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The equations that I have are:

Radius = Circumference / 2(pi)
Surface Area = 4(pi) Radius^2
Volume = (4/3)(pi) Radius^3

any help will be greatly appreciated.

 

[BigGlenntheHeavy]

\(\displaystyle Given: \ C \ = \ 10cm \ (at \ great \ circle), \ dC \ = \ \pm.4cm\)

\(\displaystyle Find \ the \ percentage \ errors \ in \ radius, \ surface \ area, \ and \ volume.\)

\(\displaystyle 10 \ = \ 2\pi r, \ then \ r \ = \ \frac{5}{\pi}.\)

\(\displaystyle C \ = \ 2\pi r, \ dC \ = \ 2\pi dr, \ \pm.4 \ = \ 2\pi dr, \ dr \ = \ \pm\frac{.2}{\pi}.\)

\(\displaystyle A \ = \ 4\pi r^{2}, \ dA \ = \ 8\pi rdr \ = \ 8\pi(\frac{5}{\pi})(\pm\frac{.2}{\pi}) \ = \ \pm\frac{8}{\pi}.\)

\(\displaystyle V \ = \ \frac{4\pi r^{3}}{3}, \ dV \ = \ 4\pi r^{2}dr, \ = \ 4\pi(\frac{25}{\pi^{2}})(\pm\frac{.2}{\pi}) \ = \ \pm(\frac{20}{\pi^{2}})\)

\(\displaystyle Now \ \frac{dr}{r} \ = \ \frac{\pm.2/\pi}{5/ \pi} \ = \ \pm.04 \ = \ relative \ error, \ 4 \% \ = \ percentage \ error.\)

\(\displaystyle \frac{dA}{A} \ = \ \frac{\pm8/\pi}{100/\pi} \ = \ \pm.08 \ = \ 8\% \ = \ percentage \ error\)

\(\displaystyle \frac{dV}{V} \ = \ \frac{\pm20/\pi^{2}}{500/3\pi^{2}} \ = \ \pm.12 \ = \ 12\% \ = \ percentage \ error.\)

\(\displaystyle Ergo, \ percentage \ errors: \ radius-4\%, \ surface \ area-8\%, \ and \ volume-12\%.\)

 

[SeekerOfDragons]

Thanks. That's what I was coming up with except for I was getting .8% for the Surface area. I found my error in my calculations though. Dropped a variable during the work that threw it all off by 10.