Math · Calculus I · Worked example

Find a net change from a rate

Water flows into a tank at r(t) = 3t² − 2t liters per minute. How much water enters between t = 1 and t = 3 minutes?

r (t) = 3 t^{2} - 2 t

Recognize a net change

If V(t) is the volume of water in the tank, then V′(t) = r(t). The water added from t = 1 to t = 3 is V(3) − V(1), which Part 2 writes as an integral of the rate.

V (3) - V (1) = \int_{1}^{3} (3 t^{2} - 2 t) d t

\begin{array}{c} V (3) - V (1) \\ = \int_{1}^{3} (3 t^{2} - 2 t) d t \end{array}

Find an antiderivative

t³ − t² has derivative 3t² − 2t.

V (t) = t^{3} - t^{2}

Evaluate

Upper minus lower: (27 − 9) − (1 − 1) = 18.

\int_{1}^{3} (3 t^{2} - 2 t) d t = (27 - 9) - (1 - 1) = 18

\begin{array}{c} \int_{1}^{3} (3 t^{2} - 2 t) d t \\ = (27 - 9) - (1 - 1) = 18 \end{array}

Interpret

18 liters enter between minute 1 and minute 3. The rate is positive on this interval, so the net change is also the total inflow.

Result

18 liters.

Your turn

A particle moves along a line with velocity v(t) = t² − 4t + 3 m/s. Find its displacement and the total distance it travels from t = 0 to t = 3.

Show the answer and explanation

Displacement 0 m; distance 8/3 ≈ 2.67 m.

The displacement is ∫₀³ v(t) dt = 9 − 18 + 9 = 0. But v(t) = (t − 1)(t − 3) is positive on (0, 1) and negative on (1, 3): the particle moves 4/3 m forward, then 4/3 m back. The distance is ∫₀³ |v(t)| dt = 4/3 + 4/3 = 8/3 m.

\begin{array}{l} v (t) = t^{2} - 4 t + 3 \\ \int_{0}^{3} (t^{2} - 4 t + 3) d t = 9 - 18 + 9 = 0 \\ \int_{0}^{3} | t^{2} - 4 t + 3 | d t = \frac{4}{3} + \frac{4}{3} = \frac{8}{3} \end{array}

\begin{array}{c} v (t) = t^{2} - 4 t + 3 \\ \int_{0}^{3} (t^{2} - 4 t + 3) d t = 0 \\ \int_{0}^{3} | t^{2} - 4 t + 3 | d t = \frac{8}{3} \end{array}

Keep exploring

In Math, change the last row to V(2) − V(1) = 4. It checks: only 4 L arrive in the second minute and the other 14 L in the third, when the rate is highest.

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