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?
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?