Integrals: Real-World Scenarios (10)

easy • 3 min read
Tags Integrals

Question

A water tank is being filled at a variable rate of r(t)=6t−t2r(t) = 6t - t^2 litres per minute, where tt is the time in minutes after the tap was opened. (a) How much water flows in during the first 4 minutes? (b) When is the rate of flow maximum? (c) What is the total water filled by the time the rate becomes zero again?

Solution — Step by Step

Total water = integral of rate over the time interval:

V=∫04(6t−t2) dt=[3t2−t33]04=48−643=144−643=803V = \int_0^4 (6t - t^2) \, dt = \left[ 3t^2 - \frac{t^3}{3} \right]_0^4 = 48 - \frac{64}{3} = \frac{144 - 64}{3} = \frac{80}{3}

So approximately 26.67 litres in the first 4 minutes.

Differentiate the rate:

drdt=6−2t=0  ⟹  t=3 min\frac{dr}{dt} = 6 - 2t = 0 \implies t = 3 \text{ min}

Second derivative is −2<0-2 < 0, so this is a maximum. Maximum rate: r(3)=18−9=9r(3) = 18 - 9 = 9 L/min.

r(t)=6t−t2=t(6−t)=0  ⟹  t=0r(t) = 6t - t^2 = t(6 - t) = 0 \implies t = 0 or t=6t = 6 min.

Total volume by t=6t = 6 min:

V=∫06(6t−t2) dt=[3t2−t33]06=108−72=36 LV = \int_0^6 (6t - t^2) \, dt = \left[ 3t^2 - \frac{t^3}{3} \right]_0^6 = 108 - 72 = 36 \text{ L}

(a) 80/3≈26.6780/3 \approx 26.67 L, (b) at t=3t = 3 min, max rate = 9 L/min, (c) total = 36 L by t=6t = 6 min.

Why This Works

When a rate r(t)r(t) is given, the cumulative quantity over time is ∫r(t) dt\int r(t) \, dt. This is just the fundamental theorem — accumulation reverses differentiation.

The rate r(t)=6t−t2r(t) = 6t - t^2 peaks at t=3t = 3 and returns to zero at t=6t = 6, modelling a tap that opens, peaks, then closes itself (perhaps due to back-pressure). The integral ∫06r dt\int_0^6 r \, dt is the area under this rate curve and gives the total accumulated volume.

Whenever a problem gives “rate of flow” or “rate of change”, expect to integrate to get totals. Whenever it gives “total” and asks “rate at t=t = …”, expect to differentiate.

Alternative Method

Average rate × time: average rate over [0,6][0, 6] is (1/6)∫06r dt=6(1/6) \int_0^6 r \, dt = 6 L/min. Multiply by interval length: 6⋅6=366 \cdot 6 = 36 L. Same answer for total volume.

Common Mistake

Students sometimes integrate the rate over the wrong interval, e.g. from −1-1 to 44 instead of 00 to 44. Always check the lower limit — usually it’s the moment the tap opens or the process begins, often t=0t = 0.

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