How to Solve a Conical Tank Calculus Problem

[punjabi_monster]

I have a calculus question and was wondering if some could help me see what I am doing wrong in this question. Thank you

A conical tank with an altitude of 10m and whose base has a radius of 4m is mounted with its vertex down. The tank is full of water which is draining through the vertex at the rate of 5 m^3/min. how fast is the level of the water dropping when the radius is 3?

the answer is 5pi/9

this is how i did it:

r/h=4/10
h=5pi/2

V=[(pi)(r^2)(h)]/3
V=[(pi)(r^2)(5r/2)]/3
V=[pi5r^3]/6
dv/dt=[15pi(r^2)]/6] * dr/dt
5=[(15pi(3^2)]/6] * dr/dt
dr/dt= (2pi)/9

 

[Data]

Firstly, the answer is actually \frac{5}{9\pi}\frac{\mbox{m}}{\mbox{min}}, not \frac{5\pi}{9}\frac{\mbox{m}}{\mbox{min}}.

Secondly, are you sure \frac{d r}{d t} is what the question is looking for? Re-read it.

 

[erik05]

I have a calculus question and was wondering if some could help me see what I am doing wrong in this question. Thank you

A conical tank with an altitude of 10m and whose base has a radius of 4m is mounted with its vertex down. The tank is full of water which is draining through the vertex at the rate of 5 m^3/min. how fast is the level of the water dropping when the radius is 3?
the answer is 5pi/9

this is how i did it:

r/h=4/10
h=5pi/2

V=[(pi)(r^2)(h)]/3
V=[(pi)(r^2)(5r/2)]/3
V=[pi5r^3]/6
dv/dt=[15pi(r^2)]/6] * dr/dt
5=[(15pi(3^2)]/6] * dr/dt
dr/dt= (2pi)/9

Notice how the question asks for how fast the level of the water is dropping. This would mean a change in height with respect to time, not the radius. Since it's asking for the rate of change when the radius is 3 then:

r=\frac{3h}{10}
V=\frac{1}{3} \pi (\frac{3h}{10})^2 h
V= \frac {3}{100} \pi h^3
\frac {dV}{dt}= \frac{9}{100} \pi h^2 \frac{dh}{dt}

Solving for dh/dt should get you the answer \frac{5}{9\pi}\frac{\mbox{m}}{\mbox{min}} as stated above by Data.

 

[Data]

You actually need to use r = \frac{2h}{5}. But other than that you're completely right.

 

[punjabi_monster]

thanks for your help.