AI HL · AHL 9
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Separable equations in context: growth, decay and harder forms
AHL 9.3
Screenshot and video to come
The mass \(m\) mg of a drug in the blood decreases at a rate proportional to \(m\). At first \(m=80\) and the mass is decreasing at 12 mg per hour. Find \(k\) in \(\dfrac{dm}{dt}=-km\), find \(m\) in terms of \(t\), and find how long it takes for the mass to halve.
Have a go first. Then open the working one line at a time.
\(12=k\times80\ \Rightarrow\ k=0.15\)
At the start dm/dt = −12 and m = 80.
\(\int\dfrac1m\,dm=\int-0.15\,dt\)
m > 0, so no modulus.
\(\ln m=-0.15t+c\ \Rightarrow\ m=Ae^{-0.15t}\)
At t = 0, m = 80, so A = 80.
\(m=80e^{-0.15t}\)
\(40=80e^{-0.15t}\ \Rightarrow\ t=\dfrac{\ln2}{0.15}=4.62\)
About 4.62 hours.
A culture of cells grows at a rate proportional to the number of cells \(N\). At first there are 1500 cells and the number is increasing at 90 cells per minute. Find \(k\) in \(\dfrac{dN}{dt}=kN\), then the number of cells after 20 minutes, to 3 significant figures.
Have a go first. Then open the working one line at a time.
\(90=k\times1500\ \Rightarrow\ k=0.06\)
Initial rate ÷ initial amount.
\(\int\dfrac1N\,dN=\int0.06\,dt\ \Rightarrow\ N=Ae^{0.06t}\)
N > 0.
\(N=1500e^{0.06t}\)
A = 1500.
\(N=1500e^{1.2}=4980\)
About 4980 cells.
A gas bubble is inflated at a rate inversely proportional to its volume \(V\) cm³. Initially \(V=250\) and the volume is increasing at 8 cm³ per second. Show that \(\dfrac{dV}{dt}=\dfrac{2000}{V}\) and find \(V\) in terms of \(t\). Give the value of \(k\) that you find.
Have a go first. Then open the working one line at a time.
\(\dfrac{dV}{dt}=\dfrac kV\)
Inversely proportional to V.
\(8=\dfrac{k}{250}\ \Rightarrow\ k=2000\)
So dV/dt = 2000/V, as required.
\(\int V\,dV=\int2000\,dt\)
\(\dfrac{V^2}{2}=2000t+c\)
At t = 0, V = 250: c = 31 250.
\(V^2=4000t+62\,500,\ \ V=\sqrt{4000t+62\,500}\)
V is positive.
A ball sinks through oil. Its velocity \(v\) m s⁻¹ after \(t\) seconds satisfies \(\dfrac{dv}{dt}=9-0.3v\), and \(v=0\) when \(t=0\). Find \(v\) in terms of \(t\), then the distance it falls in the first 4 seconds.
Have a go first. Then open the working one line at a time.
\(\int\dfrac{1}{9-0.3v}\,dv=\int1\,dt\)
\(-\dfrac{1}{0.3}\ln(9-0.3v)=t+c\)
9 − 0.3v > 0 at the start.
\(\ln(9-0.3v)=-0.3t+c_1\ \Rightarrow\ 9-0.3v=Ae^{-0.3t}\)
v = 0 at t = 0 gives A = 9.
\(v=30\left(1-e^{-0.3t}\right)\)
v approaches 30 m s⁻¹, since 9 ÷ 0.3 = 30.
From the calculator: \(\int_0^4 30\left(1-e^{-0.3t}\right)dt=50.1\)
Distance is the integral of velocity. About 50.1 m.
A cup of coffee at 90 °C is left in a room at 20 °C. Its temperature \(T\) °C after \(t\) minutes decreases at a rate proportional to \(T-20\). After 5 minutes the coffee is at 62 °C. Write down a differential equation for \(T\), find \(T\) in terms of \(t\), and find how long it takes to cool to 35 °C.
Have a go first. Then open the working one line at a time.
\(\dfrac{dT}{dt}=-k(T-20),\ k>0\)
The rate is proportional to T − 20. T falls towards 20, so the rate is negative.
\(\int\dfrac{1}{T-20}\,dT=\int-k\,dt\)
T > 20, so no modulus is needed.
\(\ln(T-20)=-kt+c\ \Rightarrow\ T=20+Ae^{-kt}\)
T = 90 when t = 0, so A = 70.
\(62=20+70e^{-5k}\ \Rightarrow\ k=\dfrac{\ln(70/42)}{5}=0.102\)
Keep the full value of k in later working.
\(T=20+70e^{-0.102t}\)
T approaches the room temperature, 20 °C.
\(35=20+70e^{-kt}\ \Rightarrow\ t=\dfrac{\ln(70/15)}{k}=15.1\)
About 15.1 minutes.
A tank holds 200 litres of pure water. Salt solution of concentration 0.2 kg per litre flows in at 10 litres per minute. The mixture is stirred and flows out at 10 litres per minute, so the volume stays at 200 litres. After \(t\) minutes the tank holds \(m\) kg of salt. Write down a differential equation for \(m\), solve it, and find the mass of salt after 20 minutes.
Have a go first. Then open the working one line at a time.
\(0.2\times10=2\) kg per minute
Concentration times flow.
\(\dfrac{m}{200}\times10=0.05m\) kg per minute
The tank is stirred, so the concentration is m ÷ 200.
\(\dfrac{dm}{dt}=2-0.05m\)
Rate of change = rate in − rate out.
\(\int\dfrac{1}{2-0.05m}\,dm=\int dt\ \Rightarrow\ 2-0.05m=Ae^{-0.05t}\)
m = 0 at t = 0, so A = 2.
\(m=40\left(1-e^{-0.05t}\right)\), so \(m(20)=25.3\)
The salt approaches 40 kg, since 2 ÷ 0.05 = 40.
A spherical sweet dissolves. Its volume decreases at a rate proportional to its surface area. When the volume is 18 cm³ the volume is decreasing at 2.4 cm³ per minute. Find \(\dfrac{dr}{dt}\), where \(r\) cm is the radius, and the time for the sweet to dissolve completely, in minutes.
Have a go first. Then open the working one line at a time.
\(\dfrac{dV}{dt}=-kS=-k\times4\pi r^2\)
A decrease proportional to surface area.
\(\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}\ \Rightarrow\ \dfrac{dr}{dt}=-k\)
The 4πr² cancels: the radius shrinks at a constant rate.
\(18=\tfrac43\pi r^3\ \Rightarrow\ r=1.626\)
Surface area S = 4π × 1.626² = 33.21 cm².
\(2.4=k\times33.21\ \Rightarrow\ k=0.0723\)
So dr/dt = −0.0723 cm per minute.
\(t=\dfrac{1.626}{0.0723}=22.5\)
The radius falls from 1.63 cm to 0 at a constant rate.
Given that \(\dfrac{dy}{dx}=e^{2x+y}\) and \(y=0\) when \(x=0\), find \(y\) in terms of \(x\). The general solution is \(e^{-y}=A-\tfrac12e^{2x}\). Find \(A\).
Have a go first. Then open the working one line at a time.
\(\dfrac{dy}{dx}=e^{2x}e^{y}\)
This is what makes it separable.
\(\int e^{-y}\,dy=\int e^{2x}\,dx\)
Divide by e^y.
\(-e^{-y}=\tfrac12e^{2x}+c\)
\(-1=\tfrac12+c\ \Rightarrow\ c=-1.5,\ A=1.5\)
e⁰ = 1.
\(e^{-y}=1.5-\tfrac12e^{2x}\ \Rightarrow\ y=-\ln\left(1.5-\tfrac12e^{2x}\right)\)
Given that \(y\dfrac{dy}{dx}=6x^2\) and that \(y=-3\) when \(x=1\), find \(y\) in terms of \(x\). Then find \(y\) when \(x=2\), to 3 significant figures.
Have a go first. Then open the working one line at a time.
\(\int y\,dy=\int6x^2\,dx\ \Rightarrow\ \dfrac{y^2}{2}=2x^3+c\)
\(\dfrac92=2+c\ \Rightarrow\ c=2.5\)
Square the −3 first.
\(y^2=4x^3+5\)
\(y=-\sqrt{4x^3+5}\)
y is negative at x = 1, so take the negative root.
\(y=-\sqrt{37}=-6.08\)
Given \(\dfrac{dy}{dx}=\dfrac{2\sin x}{\sin y}\), show that the general solution is \(\cos y-2\cos x=c\). A solution has \(y=\dfrac\pi3\) when \(x=\dfrac\pi2\). For \(0\le x\le\pi\) and \(-\pi\le y\le\pi\), find the positive value of \(y\) when \(x=\dfrac{2\pi}{3}\), to 3 significant figures.
Have a go first. Then open the working one line at a time.
\(\int\sin y\,dy=\int2\sin x\,dx\)
Multiply by sin y.
\(-\cos y=-2\cos x+c_1\)
\(\cos y-2\cos x=c\)
c = −c₁. This is the result to be shown.
\(\cos\tfrac\pi3-2\cos\tfrac\pi2=\tfrac12\ \Rightarrow\ c=\tfrac12\)
\(\cos y=\tfrac12+2\cos\tfrac{2\pi}{3}=-\tfrac12\)
\(y=\pm\dfrac{2\pi}{3}=\pm2.09\)
Both lie in −π ≤ y ≤ π. The positive one is 2.09.