Question

Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 6cm and a height of 12cm, at the rate of 3 cm3/min. At what rate is the depth of the water changing at the instant when the water in the tank is 9 cm deep? Give an exact answer showing all work and include units in your answer.

Answer 1

@jim_thompson5910

Answer 2

What do you have so far?

Answer 3

r = 6cm
h = 12cm
r = 3cm^3/min
\[v = \pi r^2h\]

Answer 4

`r = 3cm^3/min` is incorrect

I think you meant to say dV/dt = -3 to represent the fact that the volume is decreasing by 3 cubic cm

Answer 5

also, you have the wrong volume formula

Answer 6

volume of a cone

\[\Large V = \frac{1}{3}\pi*r^2*h\]

Answer 7

Oh... So then you take the derivative, yes?

Answer 8

yeah but first you have to somehow get everything in terms of h

you have to find a connection between r and h and do a substitution

Answer 9

a picture might help

Answer 10

r = 1/2h?

Answer 11

yeah that looks good, r = h/2

Answer 12

plug that in, derive, then isolate dh/dt

Answer 13

\[\frac{ 2 }{ 3 } \pi (\frac{ h^2 }{ 2 })\]

Answer 14

When you plugged r = h/2, and simplified, did you get

\[\Large V = \frac{1}{12}\pi h^3\]

??

Answer 15

So you get \[\frac{ dv }{ dt } = \frac{ 1 }{ 4 } \pi h^2 \frac{ dh }{ dt }\]

Answer 16

good

Answer 17

dv/dt = -3
h = 9

solve for dh/dt

Answer 18

\[-3 = \frac{ 1 }{ 4 } \pi (9)^2 \frac{ dh }{ dt }\]
\[\frac{ -12 }{ 81 \pi } = \frac{ dh }{ dt }\]

Answer 19

make sure you reduce the fraction as much as possible

Answer 20

\[\frac{ -4 }{ 27 \pi } = \frac{ dh }{ dt }\]

Answer 21

yep

\[\Large \frac{dh}{dt} = -\frac{4}{27\pi} \approx -0.047157\]

the units for dh/dt are cm/min

the rate of the depth of the water is changing at roughly -0.047157 cm/min at exactly the depth of h = 9 cm

Answer 22

Thank you!!

Answer 23

glad to be of help