Quick question folks

[Daniel_Feldman]

Yes, I know, I'm usually helping people with calculus stuff. However, I have a question of my own, as we've just started dfferentials in my Calc BC class, and did not cover them at all in AB.

The question: If the radius of the sphere is measured with a +/-6% error, the approximate relative error in volume would be _______.


A. +/-12%
B. +/-15%
C. +/-18%
D. +/-19.1%
E. None of These


I know that relative error is the ration dV/V, and dV=4pir^2dr. V=(4/3)pir^3, so

dV/V=(4pir^2dr)/((4/3)pir^3)=3dr/r. dr=+/-0.06. However, we do not know r. If someone can help me finish this, that would be awesome.

 

[pka]

Usually in such problems one is given a particular radius.
However, absent that I think the best you can do is \(\displaystyle \frac{{ \pm 0.18}}{r}\\).

 

[soroban]

Hello, Daniel!

If the radius of the sphere is measured with a $\displaystyle \pm6\%$ error,
the approximate relative error in volume would be _______.

$\displaystyle (A)\;\pm12\%\qquad(B)\;\pm15\%\qquad(C)\;\pm18\%\qquad(D)\;\pm19.1\%\qquad(E)\;\text{None of these}$

Your work is correct . . . you have a misunderstanding of one quantity.

The volume of a sphere is: .$\displaystyle V \:= \:\frac{4}{3}\pi r^3$ . . . Then: .$\displaystyle dV \:= \:4\pi r^2\ dr$

We have: .$\displaystyle \displaystyle{\frac{dV}{V} \:= \:\frac{4\pi r^2\ dr}{\frac{4}{3}\pi r^3} } \;= \;3\left(\frac{dr}{r}\right)$

. . We are given: .$\displaystyle \displaystyle{\frac{dr}{r} \:= \:\pm6\% }$ *

Therefore: .\(\displaystyle \displaystyle{\frac{dV}{V} \;= \;3(\pm6\%) \;= \;\pm 18\%\) . . . answer (C)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

*
NOTE . They gave us a <u>percent</u> of error: . $\displaystyle \frac{\text{error in }r}{r}$

 

[Daniel_Feldman]

Gotcha. Thanks guys.