# Related Rates Conical Water Tank find rate of change of the water depth [duplicate]

An inverted conical water tank with a height of $20m$ and radius of $5m$ is drained through the hole in the vertex at a rate of $2m^3 /h$. What is the rate of change of the water depth when the water is $4m$ deep?

## marked as duplicate by user228113, user147263, John B, Pragabhava, user296602 Feb 25 '16 at 2:02

• Well I've drawn everything out and listed down what I know I have the equation for the volume of a cone and I think that the $2m^3 /h$ is the rate of change of the Volume over time ? but I'm not entirely sure the $/h$ in $2m^3 /h$ is throwing me off. – J2R5M3 Dec 14 '15 at 2:40
• @J2T5M3 Yes, here $h$ is most likely hours – mysatellite Dec 14 '15 at 2:42
• I know that I ultimately want to get $dh/dt$. – J2R5M3 Dec 14 '15 at 2:43
• @J2T5M3 correct, but perhaps you should assign a different variable for the height to avoid confusing the height $h$ with the time in hours $h$ – mysatellite Dec 14 '15 at 2:45
• given that $h = 3V/Pir^2$ – J2R5M3 Dec 14 '15 at 2:45

Step 1: Find an equation that relates the quantities. In this question, it is the relationship between the height h and the volume v. Find an equation that gets the volume from the height.

Step 2: Take the derivative of the equation in Step 1. However, don't treat it like a normal derivative, treat it like an implicit derivative with respect to time (dt, with t in hours).

Step 3: The result of step 2 should give you a relationship between dh/dt and dv/dt. Since the setup gives you dv/dt (2m^3/hour), then you can just solve for dh/dt.

Keep in mind that a cone is just a right triangle rotated around. This means that we can use many of the same relationships. So, for instance, as the water level rises, all of the triangles formed will be similar triangles. That means that the ratios between the sides will remain the same. The height is 20 when the radius is 5. That means that $r = \frac{h}{4}$. So, therefore, we can take the volume equation, and rewrite it entirely in terms of height:
$$v = \frac{\pi\cdot r^2\cdot h}{3}$$ $$r = \frac{h}{4}$$ $$v = \frac{\pi\cdot (\frac{h}{4})^2\cdot h}{3}$$ $$v = \frac{\pi\cdot h^3}{48}$$
• What would I sub for $r$ though? – J2R5M3 Dec 14 '15 at 14:16
• Or $V$ for that matter? – J2R5M3 Dec 14 '15 at 14:17