Maximum Error in calculated surface area, volume of sphere

Jakotheshadows

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Jun 29, 2008
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47
"The circumference of a sphere was measured to be 84 cm with a possible error of .5 cm. a) Use differentials to estimate the maximum error in the calculated surface area. What is the relative error? b) Use differentials to estimate the maximum error in the calculated volume. What is the relative error?"

First, I use the circumference to find the radius. since C = 2*pi*r, 84 = 2*pi*r, 42 = pi * r, then r = 42 / pi..
a) dr = .5, A = 4*pi*r^2, dA = 8*pi*(42/pi)*.5 = 168 cm squared maximum error.
total surface area = 2245.99 cm squared = A. dA / A = .07 or 7%
b) V = (4/3)*pi*r^3, dV = (4*pi*(42/pi)^2) *.5 = 1122.99 cm cubed maximum error.
total volume = 10008.91 cm cubed = V. dV / V = .11 or 11%

The answers in the appendix give:
for a) 27 cm squared max, .012 relative
for b) 179 cm cubed max, .018 relative

Where is my error?
 

soroban

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Jan 28, 2005
Messages
5,588
Re: Maximum Error

Hello, Jakotheshadows!

The circumference of a sphere was measured to be 84 cm with a possible error of 0.5 cm.
a) Use differentials to estimate the maximum error in the calculated surface area.
What is the relative error?

b) Use differentials to estimate the maximum error in the calculated volume.
What is the relative error?

First, I use the circumference to find the radius.
\(\displaystyle \text{Since }C = 2\pi r,\;84 = 2\pi r \quad\Rightarrow\quad r = \frac{42}{\pi}\) . Right!

\(\displaystyle a)\;dr = 0.5\) . . . . no

The answers in the appendix give:
. . \(\displaystyle a)\;27\text{ cm}^2,\;\;0.012\)
. . \(\displaystyle b)\;179\text{ cm}^3,\;\;0.018\)

We have: .\(\displaystyle C \:=\:2\pi r \quad\Rightarrow\quad dC \:=\:2\pi\,dr\;\;[1]\)

When they measured the circumference, they found that \(\displaystyle C = 84\) cm
. . with a possible error of 0.5 cm in the circumference.
\(\displaystyle \text{That is: }\:dC = 0.5 = \tfrac{1}{2}\)

\(\displaystyle \text{So [1] becomes: }\;\tfrac{1}{2} \:=\:2\pi dr \quad\Rightarrow\quad dr \:=\:\tfrac{1}{4\pi}\)


Now you can give it another try . . .


 

Jakotheshadows

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Joined
Jun 29, 2008
Messages
47
Thank you. I should pay closer attention to the wording.
 
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