Gasoline is pouring into a cylindrical tank of radius 4 feet. When the depth of the gasoline is 6 feet, the depth is increasing at 0.3 ft/sec. How fast is the volume of gasoline changing at that instant?

Round your answer to 3 decimal places!

The volume is changing at a rate of _______ feet^3 per second.

@ganeshie8 do you get this one? :/

Question

Gasoline is pouring into a cylindrical tank of radius 4 feet. When the depth of the gasoline is 6 feet, the depth is increasing at 0.3 ft/sec. How fast is the volume of gasoline changing at that instant?

Round your answer to 3 decimal places!

The volume is changing at a rate of _______ feet^3 per second.

@ganeshie8 do you get this one? :/

ganeshie8 · Answer

start by writing the Volume of cylinder formula..

ganeshie8 · Answer

$V = \pi r^2 h  =  \pi \times 4^2\times h  \ ~~~= 16\pi h$

anonymous · Answer

umm is it this? :/

πr^2h = v ?

anonymous · Answer

ohh okay haha :P

ganeshie8 · Answer

yes :) differentiate both sides with.respect.to $t$:

$V =   16\pi h$

$\dfrac{dV}{dt} =   ?$

anonymous · Answer

ummm 16pi ?

ganeshie8 · Answer

Note that we're NOT differentiating with.respect.to "h",

ganeshie8 · Answer

we're differentiating with.respect.to "t"

ganeshie8 · Answer

so you need to use chain rule

ganeshie8 · Answer

$V =   16\pi h$

$\dfrac{dV}{dt} =   16 \pi  \dfrac{dh}{dt}$

anonymous · Answer

ahh not sure how to do that here :(

so what happens after that? :/

ganeshie8 · Answer

And you're given `the depth is increasing at 0.3 ft/sec. `

ganeshie8 · Answer

that means $\dfrac{dh}{dt} =   0.3$

ganeshie8 · Answer

plug that in and simplify

anonymous · Answer

ahh okay 
so 16pi(0.3) = 15.07964474 ?

so rounded to three decimal places, we get

15.101  ? 

did i round that correctly ? :/ or should it be 15.081?  :/

ganeshie8 · Answer

http://prntscr.com/3cuij0

anonymous · Answer

whoah awesome! thank you!! :D

ganeshie8 · Answer

np :)