Question

PLEASE HELP!! A water tank has the shape of an inverted circular cone with base radius 2m and height 4m. If water is being pumped into the tank at a rate of 2m^3/min, find the rate at which the water level is rising when the water is 3m deep.
Formula= V=1/3(PI)r^2h

Answer 1

rate of change defines a derivative

Answer 2

since the radius will also pop out a rate of change; we need to formulate a relation between h and r with to account for it as well

Answer 3

i just plugged stuff in idk if thats right

Answer 4

nah, simply "plugging stuff in" is not advisable until we have the proper design in which to plug them into

Answer 5

at the moment we have something that needs to be cooked up first before we plug things into it

Answer 6

okay my bad i just thought 2 was r and 3 was h

Answer 7

\[V=\frac{pi}{3}r^2h\]

this formula gives us the volume at any given r and h; its a position and not a speed (not a rate of change); to find out how fast its changing we take its derivative.

\[V'=\frac{pi}{3}r' ^2h+\frac{pi}{3}r^2h'\]

\[V'=\frac{2pi}{3}rr'h+\frac{pi}{3}r^2h'\]

this is the formula that will tell us how fast the volume is changing at any given position but it still needs some touching up since we dont know r' yet

Answer 8

okay makes sense

Answer 9

Lets see what we can gather from the information:

A water tank has the shape of an inverted circular cone with
base radius 2m and
height 4m.

If water is being pumped into the tank at a rate of
V' = 2m^3/min,

find the rate at which the water level is rising (h') when the water is 3m (h) deep.

V' = 2
h = 3

h' = find this part :)

the rest needs to be gotten from a picture