Setting up and integrating Problem set B One question at a time

Write a differential equation for each situation. Define your variables first.

1. (a) The number of subscribers to a channel grows at a rate proportional to that number.

   • \(C\) = number of subscribers, \(t\) = time in weeks
   • Proportional to C: \(\dfrac{dC}{dt}=kC\)

2. (b) The area of a spreading stain increases at a rate proportional to the fourth root of its area.

   • \(A\) = area in cm², \(t\) = time in minutes
   • Fourth root is the power 1/4: \(\dfrac{dA}{dt}=kA^{1/4}\)

Write a differential equation for each situation. Define your variables first.

1. (a) The depth of silt in a harbour increases at a rate inversely proportional to the depth.

   • \(d\) = depth of silt in cm, \(t\) = time in years
   • Inverse proportion: \(\dfrac{dd}{dt}=\dfrac kd\)

2. (b) The length of a crack in a window increases at a rate inversely proportional to the square of its length.

   • \(l\) = length in mm, \(t\) = time in hours
   • Inverse to the square: \(\dfrac{dl}{dt}=\dfrac{k}{l^2}\)

Write a differential equation for each situation. Define your variables first.

1. (a) The volume of fuel in a lamp decreases at a rate proportional to the cube root of the volume.

   • \(V\) = volume of fuel in ml, \(t\) = time in hours
   • A decrease: \(\dfrac{dV}{dt}=-kV^{1/3}\) with \(k>0\)

2. (b) The current in a coil decreases at a rate inversely proportional to the current.

   • \(I\) = current in A, \(t\) = time in ms
   • Inverse proportion and a decrease: \(\dfrac{dI}{dt}=-\dfrac kI\)

Find the general solution of each differential equation.

1. (a) \(\dfrac{dy}{dx}=4x^{3} -\frac{10}{\sqrt{x}}\)

   • Integrate both sides with respect to \(x\)
   • \(y=-20 \sqrt{x} + x^{4}+c\)

2. (b) \(\dfrac{dy}{dt}=6e^{-\frac{t}{3}}\)

   • Integrate both sides with respect to \(t\)
   • \(y=-18e^{-\frac{t}{3}}+c\)

3. (c) \(\dfrac{dy}{dt}=-3 + \frac{12}{t}\), \(t>0\)

   • Integrate both sides with respect to \(t\)
   • \(y=-3t + 12 \ln|t|+c\)

4. (d) \(\dfrac{ds}{dt}=3 \sin(4t)\)

   • Integrate both sides with respect to \(t\)
   • \(s=-\frac{3 \cos(4t)}{4}+c\)

A gym has 800 members. After \(t\) days, \(J\) of them have joined a step challenge. The rate at which \(J\) increases is proportional to the number who have joined and to the number who have not. Write a differential equation.

• \(J\) = number who have joined, \(t\) = days
• The number who have not joined is \(800-J\)
• \(\dfrac{dJ}{dt}=kJ(800-J)\)

A joke goes round a school of 900 students. After \(t\) minutes, \(J\) of them have heard it. The rate at which \(J\) increases is proportional to the number who have heard it and to the number who have not, and is inversely proportional to \(t\). Write a differential equation.

• \(J\) = number who have heard the joke, \(t\) = minutes
• Not heard it: \(900-J\)
• Multiply, then divide by \(t\): \(\dfrac{dJ}{dt}=\dfrac{kJ(900-J)}{t}\)

A drone climbs with velocity \(v=\dfrac{dh}{dt}=5e^{-0.25t}\) m s⁻¹, where \(h\) is its height above the ground in metres and \(t\) is the time in seconds. It starts on the ground.

1. (a) Find the general solution for \(h\) in terms of \(t\).

   • Integrate with respect to \(t\)
   • \(h=-20e^{-\frac{t}{4}}+c\)

2. (b) Given that \(h=0\) when \(t=0\), find the height after 8 seconds.

   • At \(t=0\), \(h=0\): \(c=20\)
   • At \(t=8\): \(h=20 -\frac{20}{e^{2}}=17.3\)
   • 17.3 m

Rearrange each equation to make the derivative the subject, then find the general solution.

1. (a) \(\dfrac{dF}{dm}+\frac{3}{m^{2}}=4\), \(m>0\)

   • \(\dfrac{dF}{dm}=4-\frac{3}{m^{2}}\)
   • Integrate both sides: \(F=4m + \frac{3}{m}+c\)

2. (b) \(\dfrac{dG}{dt}+6e^{-3t}=2\)

   • \(\dfrac{dG}{dt}=2-6e^{-3t}\)
   • Integrate both sides: \(G=2t + 2e^{-3t}+c\)