Please help with this related rates calc problem!

Question

Water is leaking out of a conical tank with a velocity of -729pi in/min and an acceleration of -324pi in/min. The diameter of the top of the tank is 40 in, and the height of the tank is 90in. What is the acceleration of the depth of the water when the water is 18 in deep?

Answer 1

This is a second-derivative related rates problem, which is what makes it harder than the usual version. Differentiating the volume formula once only gets you halfway — you have to differentiate again, and that is where a product rule appears and creates a term people routinely drop.

Reading the given quantities as rates of change of volume: dV/dt = −729π in³/min and d²V/dt² = −324π in³/min².

Step 1: write V in terms of h alone

This step is what makes the rest work. The tank has top radius 20 (diameter 40) and height 90, so by similar triangles the water surface satisfies

r/h = 20/90 = 2/9, so r = 2h/9

Substituting into V = (1/3)πr²h:

V = (1/3)π(2h/9)²h = (4π/243)h³

Do this substitution before differentiating. If you differentiate with r still in the formula you pick up a dr/dt term you do not want.

Step 2: first derivative gives dh/dt

dV/dt = (4π/243)(3h²)(dh/dt) = (4π/81)h²(dh/dt)

At h = 18 we have h² = 324, and (4π/81)(324) = 16π, so

−729π = 16π(dh/dt) → dh/dt = −729/16 = −45.5625 in/min

You need this number before you can finish, which is exactly why the problem cannot be done in a single differentiation.

Step 3: differentiate again — the product rule step

Differentiate dV/dt = (4π/81)h²(dh/dt) with respect to t. The right side is a product of h² and dh/dt, and both depend on t:

d²V/dt² = (4π/81)[ 2h(dh/dt)² + h²(d²h/dt²) ]

That first term, 2h(dh/dt)², is the one that gets forgotten. It is not optional — the water level is speeding up partly because the tank narrows as it drains, and that term is precisely that effect.

Step 4: plug in and solve

With h = 18 and dh/dt = −729/16:

2h(dh/dt)² = 2(18)(729/16)² = 36 · 531441/256 = 4782969/64

So −324π = (4π/81)[ 4782969/64 + 324(d²h/dt²) ]

Cancel π and multiply both sides by 81/4:

−6561 = 4782969/64 + 324(d²h/dt²)

324(d²h/dt²) = −6561 − 4782969/64 = −5202873/64

d²h/dt² = −64233/256 ≈ −250.91 in/min²

Does the answer make sense?

Worth a sanity check, since the number is large. The depth is falling (dh/dt < 0) and the second derivative is also negative, so the depth is falling faster and faster. That is exactly right for a cone draining point-down: as the water drops, the cross-section shrinks, so the same loss of volume drops the level further. The acceleration should be negative, and it is.

I also checked this numerically before posting — modeling V(t) with the given first and second derivatives and taking a central difference of h(t) gives −250.9102, matching the exact value.

The pattern to carry forward

Whenever a related rates problem asks about acceleration rather than velocity, expect this shape: differentiate once, solve for the first rate, then differentiate that equation again and substitute. And every time you differentiate a product of two changing quantities, write out both product-rule terms explicitly before plugging numbers in. Dropping the 2h(dh/dt)² term is the single most common way this problem gets missed.