Related Rates / Conical Tank

Question

A right conical tank with the point oriented down, a height of 18 feet, and a radius of 4 feet has sprung a leak. How fast is the volume of water in the tank changing when the water is 13 feet high and the water level changes at a rate of 15 inches per minute? Answer must be negative

Answer 1

V = (1/3)πr²h. Given dh/dt = -1.25 ft/min = -5/4 ft/min. Find dV/dt when h = 13.

Using similar triangles, r/h = 4/18 = 2/9. So, 9r = 2h. r = (2/9)h

V = (1/3)π(4h²/81)h = (4/243)πh³

dV/dt = (4/81)πh²(dh/dt) = (4/81)π(13)²(-1.25) = (4/81)π(169)(-5/4) = -845π/81 ft³/min

Answer 2

To begin, let's establish our knowns and what we want to find. This will be always a good first step in

performing a related rates problem.

We want to find what dV/dt is when h = 13 ft, and we have that dh/dt = -15 in /min = -1.25 ft/in.

Note, V and h are functions of t (time); as water flows out continually, and thus, the volume

and height of the water currently in the cone is changing.

As well, dh/dt is negative, as water is flowing out, which causes our height of the remaining water to

decrease.

Now, the volume of a cone is found using the following formula:

V = π/3 r^2 h.

We do not have much information about the radius, namely what dr/dt is.

This indicates that we probably don't want to work with any r terms.

Thus, we can use similar triangles (see the video above for details), to make V solely a function

of h:

r/h = 4/18,

r = (2/9)h.

We can plug in h = 13 in the above equation to get

r = (2/9)13

r = 26/9

We can plug r into our volume formula, giving us,

V = π/3 r^2 h

V = π/3 (26/9)^2 h

V= π/3 (676/81) h

V = 676π/243 h.

Then, we take the derivative of V (with respect to t, time), to get

dV/dt = 676π/81 dh/dt.

Therefore, we plug in dh/dt = -1.25 = -5/4 to get

dV/dt = 676π/81 (-5/4).

dV/dt = -845π/81.

Thus, dV/dt is approximately -32.7734 ft^3, i.e., when h = 13 ft, we are losing approximately

32.7734 ft^3 of water volume.