Week 3 Exercises
3.3 Sheet 8 Successive Differentiation
Exercise 2
Show that \(y=4x\) is a solution to the differential equation \[ 2x^2\frac{d^2y}{dx^2}-x\frac{dy}{dx}+y=0. \]
Exercise 3
The displacement in metres for a particle moving along a line at time \(t\) seconds is given by the equation \[ s=2t^2-10t+4. \] Determine the velocity at times of \(1.5\) and \(10\) seconds.
Exercise 4
The velocity \(v\) of a particle moving along a line is given by \[ v=-3t^2+18t+3 \] where \(t\) is measured in seconds and \(v\) in metres per second. Determine the acceleration after \(2s\) and \(4s\).
Exercise 5
The displacement of a body, measured in meters, from a reference point after time \(t\) seconds is given by \[ s=4t^2-3t+2. \] Show that the acceleration of the body is constant at \(8ms^{-2}\) and that the velocity is zero after \(375ms\) (i.e. after \(375\) milliseconds have elapsed).
Exercise 6
The equation of motion of a particle is given by \[ s=3+5t-6t^2 \] where \(s\) is the displacement in metres and \(t\) is the time in seconds.
- Determine its position, velocity and acceleration after a time of \(4s\).
- Show that after \(415.6\overline{6}ms\) the body is momentarily at rest.
[Solutions: 1. \(dy/dx=3x^2-6x^{1/2}\), \(d^2y/dx^2=6x-3x^{-1/2}\), \(d^3y/dx^3=6+3x^{-3/2}/2\), \(d^4y/dx^4=-9x^{-5/2}/4\); 2. \(dy/dx=4\), \(d^2y/dx^2=0\); 3. \(v(1.5) = -4m/s\), \(v(10) = 30m/s\); 4. \(a(2) = 6m/s^2\), \(a(4) = -6\ m/s^2\); 5. \(v=8t-3\), \(a=8\), \(v(0.375)=0m/s\); 6. \(v = -12t+5\), \(a=-12\), \(s(4) = -73m\), \(v(4)=-43m/s\), \(a(4)=-12m/s^2\), \(v(0.4166\overline{6})=0m/s\);]
3.4 Sheet 8B More Examples of Successive Differentiation
Exercise 1
If \(y(t) = a + bt^2 + ct^4\); where \(a\) and \(b\) are constants, find \(dy/dt\) and \(d^2y/dt^2\).
Exercise 2
If \(y(t) = at -\frac 12gt^2\), where \(a\) and \(g\) are constants find \(dy/dt\) and \(d^2y/dt^2\).
Exercise 3
If a body moves according to the law \(s = 12 - 4.5t + 6.2t^2\), find its velocity in metres when \(t = 4\) seconds. What can you say about the variation of the acceleration with time?
Exercise 4
The angle \(\theta\) in radians turned through by a revolving wheel is connected by the time \(t\) (in seconds) that has elapsed since starting by the law \[ \theta = 2.1 - 3.2t + 4.8t^2. \] Find the angular velocity (in radians per second) of that wheel when \(1.5\) seconds have elapsed. Find also its angular acceleration.
Exercise 5
A slider moves so that, during the first part of its motion, its distance \(s\) (in metres) from its starting point is given by the expression \[ s = 6.8t^3 - 10.8t \] \(t\) being in seconds. Find the expression for the velocity and acceleration at any time; and hence find the velocity and acceleration after \(3\) seconds.
Exercise 6
The motion of a rising balloon is such that its height \(h\) in miles is given at any instant by the expression \[ h = 05+ \frac1{10}\sqrt[3]{t-125} \] \(t\) being in seconds. Find an expression for the velocity and acceleration at any time. Draw curves to show the variation of height, velocity and acceleration during the first ten minutes of the ascent.
Exercise 7
A body moves in such a way that the space described in the time \(t\) from starting is given by \(s = t^n\), where \(n\) is a constant. Find the value of \(n\) when the velocity is doubled from the 5th to the 10th second; find it also when the velocity is numerically equal to the acceleration at the end of the 10th second.
[Solutions: 1. \(dy/dt=2bt+4ct^3\), \(d^2y/dt^2=2b+12ct^2\); 2. \(dy/dt=a-gt\), \(d^2y/dt^2=-g\); 3. \(v=12.4t-4.5\), \(v(4)=45.1m/s\), \(a=12.4ms^{-2}\); 4. \(d\theta/dt(1.5)=11.2rad/s\), \(d^2\theta/dt^2 = 9.6rad/s^2\); 5. \(v=20.4t^2-10.8\), \(a=40.8t\), \(v(3)=172.8m/s\), \(a(3)=122.4m/s^2\); 6. \(v=(t-125)^{-2/3}/30\), \(a=-(t-125)^{-5/3}/45\); 7. \(n=2\), \(n=11\);]
3.5 Sheet 9 Maxima, minima and points of inflexion
For each of the functions given below determine the \(x\) and \(y\) coordinates of any maxima, minima or points of inflexion. Find the coordinates of the points where the curves cross the \(x\) and \(y\) axes. Using all this information, sketch the graph of the function.
- \(y=x^2-4x+4\),
- \(y=-2x^2+4x-5\),
- \(y=x^2-x-12\),
- \(y=2x^3-9x^2+12x+4\),
- \(y=x^3+2x^2-1\),
- \(y=x^3+6x^2-15x+3\),
- \(y=x^3-9x^2+24x-7\),
- \(y=x^4-8x^2\).
[Solutions: 1. minimum at \((2,0)\); 2. maximum at \((1,-3)\), passes through \((0,-5)\); 3. minimum at \((0.5,-12.25)\), passes through \((0,-12), (-3,0), (4,0)\); 4. maximum at \((1,9)\), minimum at \((2,8)\), point of inflection at \((1.5,8.5)\), passes through \((0,4), (-0.274,0)\); 5. minimum at \((0,-1)\), maximum at \((-4/3,5/27)\), point of inflection at \((-2/3,-11/27)\), passes through \((0,-1), (-1.618,0), (-1,0), (0.618,0)\); 6. local maximum at \((-5,103)\), minimum at \((1,-5)\), point of inflection at \((-2,49)\), passes through \((0,3), (-7.937,0), (0.22,0), (1.717,0)\); 7. maximum at \((2,13)\), minimum at \((4,9)\), point of inflection at \((3,11)\), passes through \((0,-7), (0.331,0)\); 8. maximum at \((0,0)\), minimum at \((-2,-16)\), minimum at \((2,-16)\), points of inflection at \((-\sqrt{4/3},-80/9)\) and \((\sqrt{4/3},-80/9)\), passes through \((0,0), (-2\sqrt{2},0), (2\sqrt{2},0)\);]