Question

Related rates problem

Answer 1

@ganeshie8 Hey ganeshie can you just check this for me, it's been a while since I've done related rates and as hoping you can just see if I'm doing it right haha.

Answer 2

Problem: If a snowball melts so that its surface area decreases at a rate of 2 cm^2/ min, find the rate at which the diameter decreases when the diameter is 6m.

Answer 3

So I'll let A be the surface area, and let y be the diameter.

dA/dt = -2cm^2/min

And I'm looking for dy/dt...
(diameter = 2r)

\[A = 4 \pi r^2 = (4 \pi \frac{ y }{ 2 })^2~~~~ r = \frac{ y }{ 2 }\]

A = pi*y^2

\[A' = ( \pi y^2)'\]
\[\frac{ dA }{ dt } = \pi(y^2)' = \pi 2y y' = \pi ^2 y \frac{ dy }{ dt }\]

So, \[\large \frac{ dy }{ dt } = \frac{ \frac{ dA }{ dt } }{ \pi 2 y } = \frac{ -2 cm^2/\min }{ \pi*2*6 }=-\frac{ 1 }{ 6\pi } cm/\min \]

Answer 4

that looks good to me :)

Answer 5

except for typoes here and there *

Answer 6

Alright cool, I was just looking at this problem thought I'd give it a shot since it's been a while since I've done these kinds of problem lol, and yeah haha >.<

Answer 7

just a small observation :
you could have done this w/o substituting r = d/2 early

Answer 8

D = 2r

D' = 2 dr/dt

Answer 9

so diameter changes twice as fast as the radius ^

Answer 10

Oh right, don't know how I missed that lol

Answer 11

how are you getting A = pi*y^2 ?

Answer 12

Oh I see how you got it.

A = 4*pi*r^2

A = 4pi * ( (y/2) )^2

A = 4pi * ( y^2)/(4)

A = pi*y^2

So nvm that bit