Question

Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 16cm and a height of 8cm, at the rate of 4 cm3/min. At what rate is the depth of the water changing at the instant when the water in the tank is 6 cm deep? Show all work and include units in your answer. Give an exact answer showing all work and include units in your answer.

Answer 1

ap calc ab problem

Answer 2

start with the volume equation

V = (1/3)pi*r^2*h

Answer 3

then we want to try to eliminate r by using the proportion

if r = 16 and h = 8, we know that r = 2h

so we get

V = (1/3)*pi*(2h)^2*h

= (1/3)*pi*(4)h^3

Answer 4

then we differentiate w/ respect to time

Answer 5

V = (4)(1/3)*(pi)h^3

dV/dt = (4/3)(pi)(3)(h^2) * dh/dt

Answer 6

then it's just plug and chug from there

Answer 7

So the ultimate goal after you plug in is to find dh/dt?

Answer 8

yup! dh/dt is the rate at which the depth/height is changing

Answer 9

we already know dV/dt (4) and h

Answer 10

Wait when did we find h?

Answer 11

"At what rate is the depth of the water changing at the instant when the water in the tank is 6 cm deep?"

Answer 12

h = 6

Answer 13

Ohhh ok my bad. so then this is what I got:
4 = (4/3)(pi)(3)(6^2) * dh/dt
4= 452.16 * dh/dt
dh/dt = .0096 this sounds very wrong

Answer 14

well, it makes sense since the cone is very wide compared to its height

Answer 15

let me just check the numbers real quick

Answer 16

actually i got .0088

Answer 17

@DanJS @Directrix would you mind checking my work? might have made a mistake but I don't see where

Answer 18

`Give an exact answer showing all work and include units in your answer.`
0.0088 sounds correct! :)
But that is an approximation.
Oh, and maybe throw a negative in front to show that the water is height is decreasing.\[\large\rm h'=-\frac{1}{36\pi}\frac{cm}{\min}\]

Answer 19

Oh okay it was so small it scared me for a second haha! Thanks for your help both of you!