AI HL · 13.5
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Accumulation: rate in, amount out
13.5.5
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Water flows into a pond at a rate of \(3t^2\) litres per minute, where t is the time in minutes. After 2 minutes the pond holds 15 litres. Find the volume of water in the pond after 5 minutes.
Have a go first. Then open the working one line at a time.
From \(t=2\) to \(t=5\)
We know the amount at t = 2. The water that flows in after that is the integral from 2 to 5.
From the calculator: \(\int_{2}^{5} 3t^2\,dt\) = 117
Integrating a rate gives the amount that has accumulated.
Volume = 15 + 117 = 132 litres
Add the amount that was there at the start.
The rate of growth of the mass m grams of a colony of lichen is modelled by \(\dfrac{dm}{dt}=kt+0.3\), where t is the time in weeks. When \(t=4\) the rate of growth is 1.1 grams per week. The colony has negligible mass at \(t=0\). Find the mass of the colony after 6 weeks.
Have a go first. Then open the working one line at a time.
\(4k+0.3=1.1\), so \(k=0.2\)
Put t = 4 and the rate 1.1 into the model.
Mass = \(\int_0^6 (0.2t+0.3)\,dt\)
The initial mass is negligible, so the mass is the accumulated amount from 0 to 6.
From the calculator: \(\int_{0}^{6} (0.2t+0.3)\,dt\) = 5.4
Give the answer with units: 5.4 grams.
A rain tank is filling at a rate of \(\dfrac{60}{t^2}\) litres per minute, where t is the time in minutes. The tank is empty when \(t=1\). At what time does the tank hold 50 litres?
Have a go first. Then open the working one line at a time.
Amount = \(\int_{1}^{T} 60t^{-2}\,dt\)
The tank is empty at t = 1, so the amount is the accumulation from 1 to T.
\(\int_{1}^{T} 60t^{-2}\,dt=50\)
This is an equation with T in the upper limit.
Solve for T: \(T=6\)
Use the calculator's solver on the equation. T = 6 minutes.
A small tank is filling at a rate of \(\dfrac{45}{t^2}\) litres per minute, where t is the time in minutes. The tank is empty when \(t=1\) and its capacity is 50 litres. Use technology to decide whether the tank will ever overflow if the filling goes on for ever. First find the amount in the tank when \(t=10\,000\).
Have a go first. Then open the working one line at a time.
Amount = \(\int_{1}^{T} 45t^{-2}\,dt\)
The tank is empty at t = 1.
T = 10: 40.5; T = 100: 44.6; T = 1 000: 45.0; T = 10 000: 45.0
From the calculator, with the upper limit set to each value of T.
The amount grows, but it stays below 45 litres.
The amount never reaches 50 litres, so the tank will never overflow.
Find \(\int (2x-3)(x^2+4)\,dx\). Give the coefficient of \(x^2\) in your answer.
Have a go first. Then open the working one line at a time.
\((2x-3)(x^2+4)=2 x^{3} - 3 x^{2} + 8 x - 12\)
You cannot integrate a product term by term. Multiply out first.
\(\int (2 x^{3} - 3 x^{2} + 8 x - 12)\,dx=\frac{x^{4}}{2} - x^{3} + 4 x^{2} - 12 x + c\)
Raise each power by 1 and divide by the new power. Do not forget the constant term, or the + c.
Find \(\int \dfrac{x^6+5x}{2x^3}\,dx\). Give the coefficient of \(\dfrac{1}{x}\) in your answer.
Have a go first. Then open the working one line at a time.
\(\dfrac{x^6+5x}{2x^3}=\frac{x^{3}}{2} + \frac{5}{2}x^{-2}\)
Divide every term on top by \(2x^3\). Subtract powers: \(x^6\div x^3=x^3\) and \(x\div x^3=x^{-2}\).
\(\int \left(\frac{1}{2}x^3+\frac{5}{2}x^{-2}\right)dx=\frac{x^4}{8}-\frac{5}{2x}+c\)
The integral of \(x^{-2}\) is \(-x^{-1}\). So the coefficient of \(\frac{1}{x}\) is −2.5.