If \(Q=ax^2+2bx+c\), and \[ I_n = \int \frac{dx}{Q^{n+1}}, \] show by differentiating \((Ax+B)/Q^n\) (where \(A, B\) are adjustable constants), or otherwise, that \[ 2n(ac-b^2)I_n = \frac{ax+b}{Q^n} + (2n-1)aI_{n-1}. \] Obtain a similar formula of reduction for \[ J_n = \int \frac{x\,dx}{Q^{n+1}}. \] Evaluate \[ \int_0^1 \frac{dx}{(x^2-x+1)^3}. \]
From the parallelogram of forces show that, if two couples acting in a plane are in equilibrium, their moments are equal and opposite. Show, conversely, that two co-planar couples of equal and opposite moment are in equilibrium. A force acting in a plane has moments \(M_1, M_2, M_3\) about points whose coordinates referred to axes \(Ox, Oy\) in the plane are \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) respectively. Show that the equation of the line of action of the force is \[ x \begin{vmatrix} 1 & y_1 & M_1 \\ 1 & y_2 & M_2 \\ 1 & y_3 & M_3 \end{vmatrix} + y \begin{vmatrix} x_1 & 1 & M_1 \\ x_2 & 1 & M_2 \\ x_3 & 1 & M_3 \end{vmatrix} = \begin{vmatrix} x_1 & y_1 & M_1 \\ x_2 & y_2 & M_2 \\ x_3 & y_3 & M_3 \end{vmatrix}. \]
A light inelastic string, of length \(2l\), is fixed at its upper end; it carries a particle of mass \(m\) at its mid-point and a particle of mass \(M\) at its lower end. The particles move in a vertical plane so that the upper and lower portions of the string make angles \(\theta\) and \(\phi\) respectively with the vertical, and on the same side of it. If the angular displacements \(\theta, \phi\) are small, write down the equations of motion of \(m\) and \(M\), neglecting quantities of order higher than the first. Show that solutions of these equations can be obtained by assuming that \(\phi=k\theta\), where \(k\) is a constant. In particular, describe the corresponding motions when \(m=3M\).
Justify the rule for writing down the equations of motion of a rigid lamina in a plane (sometimes referred to as ``the principle of the independence of the motions of translation and rotation''). A uniform straight rod, of mass \(M\) and length \(2l\), moves in a vertical plane, with the lower end of the rod sliding on a horizontal table. At any instant the rod makes an acute angle \(\theta\) with the vertical. Find the vertical acceleration of the mass-centre in terms of \(\theta\) and its time-derivatives. If the table is smooth and the rod is released from rest when \(\theta=\beta\), find the initial angular acceleration of the rod and the initial value of the reaction between the rod and the table.
If \(\bar{a}, \bar{b}, \bar{c}\) are the complex conjugates of \(a, b, c\), respectively, and if \(p, q, r\) are real, show that the equation \[ \begin{vmatrix} a-z & p & \bar{b} \\ c & q-z & b \\ \bar{c} & r & \bar{a}-z \end{vmatrix} = 0 \] has either three real roots or one real root and a pair of conjugate complex roots. It is given that \(q\) is a root when \(a=i, b=c=p=r=1\). Find \(q\) and solve the equation completely.
If the polynomial \[ ax^3+x^2-3bx+3b^2 \] has two coincident zeros show that, in general, it is a perfect cube. Hence, or otherwise, show that if \[ x^4+4ax^3+2x^2-4bx+3b^2 \] has three equal zeros then the fourth is identical with them, and find for what values of \(a\) and \(b\) this is the case.
Solution: Suppose \(ax^3+x^2-3bx+3b^2 = a(x-\alpha)^2(x-\beta)\), then if \(F(x) = ax^3+x^2-3bx+3b^2\) we must have \(F(\alpha) = F'(\alpha) = 0\), therefore: \begin{align*} && 0 &= a \alpha^3 + \alpha^2 - 3b\alpha + 3b^2 \\ &&0 &= 3a\alpha^2+2\alpha-3b \\ \Rightarrow && 3b &= \alpha(3a\alpha + 2) \\ \Rightarrow && 0 &= a \alpha^3 + \alpha^2 -\alpha^2(3a\alpha + 2) + \alpha^2(3a\alpha+2)^2/3 \\ &&&= \alpha^2 ((a-3a)\alpha-1 + \frac13 (9a^2\alpha^2+12a \alpha+4)) \\ &&&= \frac13 \alpha^2 (9a\alpha^2+6a\alpha+1) \\ &&&= \frac13 \alpha^2 (3a\alpha+1)^2 \end{align*} Therefore \(\alpha = 0\) or \(\alpha = - \frac{1}{3a}\). If \(\alpha = 0\), then \(b = 0\) and we don't (necessarily) have cubed roots. But in the general form (say when \(b \neq 0\)) we must have \(2\alpha + \beta = -\frac{1}{a} \Rightarrow \beta = -\frac{1}{3a} \Rightarrow \beta = \alpha\), therefore we have repeated roots. Similarly, if \(F(x) = x^4+4ax^3+2x^2-4bx+3b^2\) \(F'(x) = 4x^3+12ax^2+2x-4b = 12a(\frac1{3a} x^3+x^2+\frac{1}{6a} - \frac{b}{3a})\) then we must have \(\alpha\) is a root of \(F''(x) = 12x^2+24ax+4 = 4(3x^2+6ax+1)\)
Find the area common to a circle of radius \(a\) and an ellipse with semi-axes \(b\) and \(c\), where the circle and ellipse have the same centre and \(b>a>c\). If \(b=a(1+\epsilon), c=a(1-\epsilon)\), where \(\epsilon\) is a positive number small compared with unity, show that this area is approximately \[ \pi a^2 - 2\epsilon a^2. \]
Find the stationary values of
The sides of a triangle are \(a, b, c\) and the corresponding angles \(A, B, C\). Prove that
State and prove De Moivre's theorem for a (positive or negative) rational index. Evaluate \[ 32\int_0^\infty e^{-x}\cos^6 x \,dx, \] giving the answer to two decimal places.
Solution: Let \(z = e^{ix}\) \begin{align*} && 2\cos x &= z + \frac{1}{z} \\ \Rightarrow && 64 \cos^6x &= z^6 + 6z^4 + 15z^2 + 20 + 15z^{-2} + 6z^{-4} + z^{-6} \\ &&&= 2\cos 6x + 12\cos 4x + 30 \cos 2x + 20 \end{align*} \begin{align*} &&I &= \int_0^{\infty} e^{-x} \cos k x \d x \\ &&&= \left [ -e^{-x} \cos k x \right]_0^{\infty} -k\int_0^{\infty} e^{-x} \sin k x \d x \\ &&&= 1 - k \left ( [-e^{-x} \sin k x]_0^{\infty} + k\int_0^{\infty} e^{-x} \cos k x \d x\right) \\ \Rightarrow && (1+k^2)I &= 1 \\ \Rightarrow && I &= \frac{1}{1+k^2} \end{align*} Therefore \begin{align*} && I &= 32\int_0^\infty e^{-x}\cos^6 x \,dx \\ &&&= \frac12\int_0^\infty e^{-x}\left ( 2\cos 6x + 12\cos 4x + 30 \cos 2x + 20\right) \,dx \\ &&&= \frac{1}{6^2+1} + \frac{6}{4^2+1} + \frac{15}{2^2+1} + \frac{10}{0^2+1} \\ &&&= \frac{1}{37} + \frac{6}{17} + \frac{15}{5} + 10 \\ &&&= 13 + \frac{6}{17} + \frac{1}{37} \end{align*}