Related Rates Conical Tank

[jagman1990]

I am doing some test corrections and I am having a problem with one.

Water runs into a conical tank at the rate of 9ft^3/min. How fast is the water level rising when the water is 6 feet deep, the radius is 2 feet and dr/dt = .16ft/min?

I have done a different version of this problem using similar triangles where dr/dt was not provided. We were instructed to not use similar triangles for this problem.

V=(π/3)r^2h

dV/dt=9ft^3/min

I took the derivative dV/dt = (2π/3) * r *dr/dt *dh/dt

9 = (2π/3) * 2 *.16 *dh/dt

dh/dt = 9/((2π/3) * 2 *.16)

I got dh/dt = 13.43 ft/min which was marked incorrect.

Can someone please throw some help my way as to where I may have set the problem up incorrectly.

Thank you!

 

[DrPhil]

I am doing some test corrections and I am having a problem with one.

Water runs into a conical tank at the rate of 9ft^3/min. How fast is the water level rising when the water is 6 feet deep, the radius is 2 feet and dr/dt = .16ft/min?

I have done a different version of this problem using similar triangles where dr/dt was not provided. We were instructed to not use similar triangles for this problem.

V=(π/3)r^2h

dV/dt=9ft^3/min

I took the derivative dV/dt = (2π/3) * r *dr/dt *dh/dt...Doesn't have correct units - can't be right.

9 = (2π/3) * 2 *.16 *dh/dt

dh/dt = 9/((2π/3) * 2 *.16)

I got dh/dt = 13.43 ft/min which was marked incorrect.

Can someone please throw some help my way as to where I may have set the problem up incorrectly.

Thank you!

You don't need the similar triangles because you are already given r and dr/dt.

Look at V as a product of r^2 and h, and differentiate using the product rule.

\(\displaystyle \displaystyle V = \dfrac{\pi}{3}\ r^2\ h\)

\(\displaystyle \displaystyle \dfrac{dV}{dt} = \dfrac{\pi}{3}\ \left[h\ (2r)\ \dfrac{dr}{dt} + r^2\ \dfrac{dh}{dt} \right]\)

Looks like the only unknown in that equation is dh/dt.

 

[jagman1990]

You don't need the similar triangles because you are already given r and dr/dt.

Look at V as a product of r^2 and h, and differentiate using the product rule.

\(\displaystyle \displaystyle V = \dfrac{\pi}{3}\ r^2\ h\)

\(\displaystyle \displaystyle \dfrac{dV}{dt} = \dfrac{\pi}{3}\ \left[h\ (2r)\ \dfrac{dr}{dt} + r^2\ \dfrac{dh}{dt} \right]\)

Looks like the only unknown in that equation is dh/dt.

I can't believe I overlooked that and ignored the product rule. Stupid mistake and I appreciate the help.