Partial Derivative Application Help Needed

rhm95

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Joined
Mar 8, 2015
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Prompt: Use differentials to estimate the amount of tin in a closed tin can with diameter 6 cm and height 19 cm if the tin is 0.04 cm thick. (Round your answer to two decimal places.) *answer is in cm^3

my work:

v=pi*r2h
dr = .04

I assumed that dv would be the amount of tin in the closed can where, (the fractions are partial derivatives)
dv = dv/dr *dr + dv * dv/dh *dh
then
dv/dr =2pi*r*h
dv/dh = pi*r^2

how i found dh,
h/r = 19/3
h=19/3 *r
so,
dh = 19/3 dr (and we know dr so gouda)

plug n chug,
dv = 2pi*r*h*dr + pi*r^2*19/3*dr
dv = 21.49

This however is not the correct answer. I have tried 4 or 5 other methods, this one seems the best and it still is not correct. What did I do wrong?
 
Prompt: Use differentials to estimate the amount of tin in a closed tin can with diameter 6 cm and height 19 cm if the tin is 0.04 cm thick. (Round your answer to two decimal places.) *answer is in cm^3

my work:

v=pi*r2h
dr = .04

I assumed that dv would be the amount of tin in the closed can where, (the fractions are partial derivatives)
dv = dv/dr *dr + dv * dv/dh *dh
then
dv/dr =2pi*r*h
dv/dh = pi*r^2

how i found dh,
h/r = 19/3
h=19/3 *r
so,
dh = 19/3 dr (and we know dr so gouda)
Sorry, but what is meant by "so gouda"?

plug n chug,
dv = 2pi*r*h*dr + pi*r^2*19/3*dr
dv = 21.49

This however is not the correct answer.
What is "the correct answer"?

By the way, this page may contain some useful information. ;)
 
Sorry, but what is meant by "so gouda"?

What is "the correct answer"?

By the way, this page may contain some useful information. ;)

Thanks for the link, I got it figured out. I assumed that the change in height is related to the change in radius which it's not.
 
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