Related rates: dh/dt given time

[virulent]

Homework Statement

A solution is being poured into a conical filter in a chemistry experiment at a rate of 5cm^3/min. The filter is 15 cm high with a diameter of 10 cm at the top. The solution is dropping out of the filter at a rate of 1 cm^3/min. Determine the rate at which the height of the solution is increasing 3 minutes in this process.

h = 15 cm
r = \frac{10cm}{2} = 5 cm
\frac{dV}{dt} = \frac{5cm^3}{min} - \frac{4cm^3}{min} = \frac{1cm^3}{min}
t = 3 minutes
\frac{dh}{dt} = ?

Homework Equations

V = \frac{1}{3}\pi r^2h
\frac{r}{h} = \frac{5cm}{15cm} => r = \frac{h}{3}

The Attempt at a Solution

Substitute.

V = \frac{1}{3}\pi (\frac{h}{3})^2h
...
V = \frac{\pi}{27} h^3

Differentiate.

\frac{dV}{dt} = \frac{\pi}{27} 3h^2 \frac{dh}{dt}


At this point I feel I went wrong somewhere because I have not incorporated t so I am unsure of where to go from there. Countless searches brought me no results which is even more confusing.

I feel I have to get a formula for h given t but that is also a problem, I think?

Thanks in advance!

 

[tiny-tim]

welcome to pf!

hi virulent! welcome to pf!

\frac{dV}{dt} = \frac{\pi}{27} 3h^2 \frac{dh}{dt}

that looks fine

but after all that work, you've lost the plot …

you've forgotten that dV/dt is known! ​