Problems

Use the binomial expansion to obtain a polynomial of degree \(2\) which is a good approximation to \(\sqrt{1-x}\) when \(x\) is small.

1. By taking \(x=1/100\), show that \(\sqrt{11}\approx79599/24000\), and estimate, correct to 1 significant figure, the error in this approximation. (You may assume that the error is given approximately by the first neglected term in the binomial expansion.)
2. Find a rational number which approximates \(\sqrt{1111}\) with an error of about \(2 \times {10}^{-12}\).

Sketch the graph of the function \([x/N]\), for \(0 < x < 2N\), where the notation \([y]\) means the integer part of \(y\). (Thus \([2.9] = 2\), \ \([4]=4\).)

1. Prove that \[ \sum_{k=1}^{2N} (-1)^{[k/N]} k = 2N-N^2. \]
2. Let \[ S_N = \sum_{k=1}^{2N} (-1)^{[k/N]} 2^{-k}. \] Find \(S_N\) in terms of \(N\) and determine the limit of \(S_N\) as \(N\to\infty\).

The cuboid \(ABCDEFGH\) is such \(AE\), \(BF\), \(CG\), \(DH\) are perpendicular to the opposite faces \(ABCD\) and \(EFGH\), and \(AB =2, BC=1, AE={\lambda}\). Show that if \(\alpha\) is the acute angle between the diagonals \(AG\) and \(BH\) then $$\cos {\alpha} = |\frac {3-{\lambda}^2} {5+{\lambda}^2} |$$ Let \(R\) be the ratio of the volume of the cuboid to its surface area. Show that \(R<\frac{1}{3}\) for all possible values of \(\lambda\). Prove that, if \(R\ge \frac{1}{4}\), then \(\alpha \le \arccos \frac{1}{9}\).

Show Solution

---

Solution:

+ - ↺

Set \(A\) to be the origin, then \(B = \langle 2, 0, 0 \rangle, G = \langle 2, 1, \lambda \rangle, H = \langle 0, 1, \lambda \rangle\), in particular \begin{align*} && AG&= \langle 2, 1, \lambda \rangle \\ && BH &= \langle -2, 1, \lambda \rangle \\ \Rightarrow && \cos \alpha &= |\frac{-4+1+\lambda^2}{\sqrt{2^2+1^2+\lambda^2}\sqrt{(-2)^2+1^2+\lambda^2}}| \\ &&&= |\frac{-3+\lambda^2}{5+\lambda^2}| \end{align*} \begin{align*} && \text{Volume} &= 2\lambda \\ && \text{Surface area} &= 2\cdot2\lambda + 2\cdot\lambda + 2\cdot2 \\ \Rightarrow && R&= \frac{\lambda}{3\lambda + 2} < \frac{1}{3} \\ && \frac14 &\leq R \\ \Rightarrow && 3\lambda +2 &\leq 4\lambda \\ \Rightarrow &&2 & \leq \lambda \end{align*} Then \(\frac{\lambda^2-3}{5+\lambda^2}\) is increasing as \(\lambda\) increases, in particularly the smallest value is \(\frac{1}{9}\).

Let $$ \f(x) = P \, {\sin x} + Q\, {\sin 2x} + R\, {\sin 3x} \;. $$ Show that if \(Q^2 < 4R(P-R)\), then the only values of \(x\) for which \(\f(x) = 0\) are given by \(x=m\pi\), where \(m\) is an integer. \newline [You may assume that \(\sin 3x = \sin x(4\cos^2 x -1)\).] Now let $$ \g(x) = {\sin 2nx} + {\sin 4nx} - {\sin 6nx}, $$ where \(n\) is a positive integer and \(0 < x < \frac{1}{2}\pi \). Find an expression for the largest root of the equation \(\g(x)=0\), distinguishing between the cases where \(n\) is even and \(n\) is odd.

The curve \(C_1\) passes through the origin in the \(x\)--\(y\) plane and its gradient is given by $$ \frac{\d y}{\d x} =x(1-x^2)\e^{-x^2}. $$ Show that \(C_1\) has a minimum point at the origin and a maximum point at \(\left(1,{\frac12\, \e^{-1}} \right)\). Find the coordinates of the other stationary point. Give a rough sketch of \(C_1\). The curve \(C_2\) passes through the origin and its gradient is given by $$ \frac{\d y}{\d x}= x(1-x^2)\e^{-x^3}. $$ Show that \(C_2\) has a minimum point at the origin and a maximum point at \((1,k)\), where \phantom{} \(k > \frac12 \,\e^{-1}.\) (You need not find \(k\).)

Show that \[ \int_0^1 \frac{x^4}{1+x^2} \, \d x = \frac \pi {4} - \frac 23 \;. \] Determine the values of

1. \(\displaystyle \int_0^1 x^3 \; \tan ^{-1} \left(\frac {1-x} {1+x} \right) \,\d x \)
2. \(\displaystyle \int_0^1 \frac {(1-y)^3} {(1+y)^5} \; {{\tan}^{-1} y}\, \d y\)

In an Argand diagram, \(O\) is the origin and \(P\) is the point \(2+0\mathrm{i}\). The points \(Q\), \(R\) and \(S\) are such that the lengths \(OP\), \(PQ\), \(QR\) and \(RS\) are all equal, and the angles \(OPQ\), \(PQR\) and \(QRS\) are all equal to \({5{\pi}}/6\), so that the points \(O\), \(P\), \(Q\), \(R\) and \(S\) are five vertices of a regular 12-sided polygon lying in the upper half of the Argand diagram. Show that \(Q\) is the point \(2 + \sqrt 3 + \mathrm{i}\) and find \(S\). The point \(C\) is the centre of the circle that passes through the points \(O\), \(P\) and \(Q\). Show that, if the polygon is rotated anticlockwise about \(O\) until \(C\) first lies on the real axis, the new position of \(S\) is $$ - \tfrac{1}{2} (3\sqrt 2+ \sqrt6)(\sqrt3-\mathrm{i})\;. $$

The function \(\f\) satisfies \(\f(x+1)= \f(x)\) and \(\f(x)>0\) for all \(x\).

1. Give an example of such a function.
2. The function \(\F\) satisfies \[ \frac{\d \F}{\d x} =\f(x) \] and \(\F(0)=0\). Show that \(\F(n) = n\F(1)\), for any positive integer \(n\).
3. Let \(y\) be the solution of the differential equation \[ \frac{\d y}{\d x} +\f(x) y=0 \] that satisfies \(y=1\) when \(x=0\). Show that \(y(n) \to 0\) as \(n\to\infty\), where \(n= 1,\,2,\, 3,\, \ldots\)

A particle of unit mass is projected vertically upwards with speed \(u\). At height \(x\), while the particle is moving upwards, it is found to experience a total force \(F\), due to gravity and air resistance, given by \(F=\alpha \e^{-\beta x}\), where \(\alpha\) and \(\beta\) are positive constants. Calculate the energy expended in reaching this height. Show that \[ F= {\textstyle \frac12} \beta v^2+ \alpha - {\textstyle \frac12} \beta u^2 \;, \] where \(v\) is the speed of the particle, and explain why \( \alpha = \frac12 \beta u^2 +g\), where \(g\) is the acceleration due to gravity. Determine an expression, in terms of \(y\), \(g\) and \(\beta\), for the air resistance experienced by the particle on its downward journey when it is at a distance \(y\) below its highest point.

Two particles \(A\) and \(B\) of masses \(m\) and \(km\), respectively, are at rest on a smooth horizontal surface. The direction of the line passing through \(A\) and \(B\) is perpendicular to a vertical wall which is on the other side of \(B\) from \(A\). The particle \(A\) is now set in motion towards \(B\) with speed \(u\). The coefficient of restitution between \(A\) and \(B\) is \(e_1\) and between \(B\) and the wall is \(e_2\). Show that there will be a second collision between \(A\) and \(B\) provided $$ k< \frac {1+e_2(1+e_1)} {e_1}\;. $$ Show that, if \(e_1=\frac13\), \(e_2=\frac12\) and \(k<5\), then the kinetic energy of \(A\) and \(B\) immediately after \(B\) rebounds from the wall is greater than \(mu^2/27\).

Show Solution

---

Solution: First collision:

+ - ↺

Since the \(e = e_1\), the speed of approach is \(u\) the speed of separation will be \(e_1u\) and so \(v_B = v_A + e_1u\). \begin{align*} \text{COM}: && mu &= mv_A + km(v_A + e_1u) \\ \Rightarrow && v_A(1+k) &= u(1-ke_1) \\ \Rightarrow && v_A &= \frac{1-ke_1}{1+k} u \\ && v_B &= \frac{1-ke_1}{1+k} u + e_1 u \\ &&&= \frac{1-ke_1 + e_1+ke_1}{1+k}u \\ &&&= \frac{1+e_1}{1+k}u \end{align*} Once the ball rebounds from the wall it will have velocity (still taking towards the wall as +ve) of \(-\frac{1+e_1}{1+k}e_2u\). There will be another collision if it is travelling faster than \(A\), ie if: \begin{align*} -\frac{1+e_1}{1+k}e_2u &< \frac{1-ke_1}{1+k} u \\ \Leftrightarrow && 0 &< (1-ke_1) + (1+e_1)e_2 \\ \Leftrightarrow && ke_1 &< 1 +e_2 (1+e_1) \\ \Leftrightarrow && k &< \frac{1 +e_2 (1+e_1)}{e_1} \\ \end{align*} If \(e_1 = \frac13, e_2 = \frac12\), then \(v_A = \frac{1-\frac13k}{1+k}u = \frac{3-k}{3(1+k)}u\) and \(v_B = \frac{4}{3(1+k)}u\). Therefore \begin{align*} && \text{total k.e.} &= \underbrace{\frac12 m v_A^2}_{\text{k.e. of }A} + \underbrace{\frac12 (km) (e_2 v_B)^2}_{\text{k.e. of }B} \\ &&&= \frac12 m \frac{(3-k)^2}{9(1+k)^2}u^2 + \frac12 km \frac14 \frac{16}{9(1+k)^2}u^2 \\ &&&= \frac12mu^2 \frac{1}{9(1+k)^2}\left ( (3-k)^2+4k \right) \\ &&&= \frac12mu^2 \frac{1}{9(1+k)^2}\left ( 9-2k+k^2 \right) \\ &&&= \frac{mu^2}{18} \frac{9-2k+k^2}{1+2k+k^2} \end{align*} We wish to minimize this as a function of \(k\). \begin{align*} \frac{\d}{\d k} \left ( \frac{9-2k+k^2}{1+2k+k^2}\right) &= \frac{(1+k)^2(2k-2)-2(1+k)(k^2-2k+9)}{(1+k)^4} \\ &= \frac{2(k^2-1) - 2(k^2-2k+9)}{(1+k)^3} \\ &= \frac{2(2k-10)}{(1+k)^3} \end{align*} Therefore the minimum will be when \(k = 5\) can't be a maximum by considering \(k \to 0\). This value is \(\frac{2}{3}\) and therefore \(\frac{mu^2}{18} \frac{2}{3} = \frac{mu^2}{27}\) is the smallest energy (which isn't quite achievable since \(k < 5\).