Problems

2017 Paper 3 Q5

D: 1700.0 B: 1484.0

The point with cartesian coordinates $(x,y)$ lies on a curve with polar equation $r=\f(\theta)\,$. Find an expression for $\dfrac{\d y}{\d x}$ in terms of $\f(\theta)$, $\f'(\theta)$ and $\tan\theta\,$. Two curves, with polar equations $r=\f(\theta)$ and $r=\g(\theta)$, meet at right angles. Show that where they meet \[ \f'(\theta) \g'(\theta) +\f(\theta)\g(\theta) = 0 \,. \] The curve $C$ has polar equation $r=\f(\theta)$ and passes through the point given by $r=4$, $\theta = - \frac12\pi$. For each positive value of $a$, the curve with polar equation $r= a(1+\sin\theta)$ meets~$C$ at right angles. Find $\f(\theta)\,$. Sketch on a single diagram the three curves with polar equations $r= 1+\sin\theta\,$, \ $r= 4(1+\sin\theta)$ and $r=\f(\theta)\,$.

Solution: $(x, y) = (f(\theta)\cos(\theta), f(\theta)\sin(\theta))$ so \begin{align*} \frac{dy}{d\theta} &= -f(\theta)\sin(\theta) + f'(\theta)\cos(\theta) \\ \frac{dx}{d\theta} &= f(\theta)\cos(\theta) + f'(\theta)\sin(\theta) \\ \frac{dy}{dx} &= \frac{-f(\theta)\sin(\theta) + f'(\theta)\cos(\theta)}{f(\theta)\cos(\theta) + f'(\theta)\sin(\theta) } \\ &= \frac{-f(\theta)\tan(\theta) + f'(\theta)}{f(\theta) + f'(\theta)\tan(\theta) } \end{align*} If the curves meet at right angles then the product of their gradients is $-1$, ie \begin{align*} \frac{-f(\theta)\tan(\theta) + f'(\theta)}{f(\theta) + f'(\theta)\tan(\theta) } \cdot \frac{-g(\theta)\tan(\theta) + g'(\theta)}{g(\theta) + g'(\theta)\tan(\theta) } &= -1 \\ f(\theta)g(\theta)\tan^2 \theta - f(\theta)g'(\theta)\tan \theta - f'(\theta)g(\theta)\tan \theta + f'(\theta)g'(\theta) &= \\ \quad - \l f(\theta)g(\theta) + f(\theta)g'(\theta)\tan(\theta) + f'(\theta)g(\theta)\tan(\theta) + f'(\theta)g'(\theta)\tan^2 \theta \r \\ \tan^2\theta \l f(\theta)g(\theta) + f'(\theta)g'(\theta) \r + f'(\theta)g'(\theta) + f(\theta)g(\theta) &= 0 \\ (\tan^2\theta + 1) \l f(\theta)g(\theta) + f'(\theta)g'(\theta) \r &= 0 \\ f(\theta)g(\theta) + f'(\theta)g'(\theta) &= 0 \end{align*} $g(\theta) = a(1+\sin\theta), g'(\theta) = a\cos\theta$ Therefore $f'(\theta)a\cos \theta+f(\theta)a(1+\sin(\theta)) = 0$ \begin{align*} && \frac{f'(\theta)}{f(\theta)} &= -\sec(\theta) - \tan(\theta) \\ \Rightarrow && \ln(f(\theta)) &= -\ln |\tan(\theta) + \sec(\theta)| + \ln |\cos(\theta)| + C \\ \Rightarrow && f(\theta) &= A \frac{\cos \theta}{\tan \theta + \sec \theta} \\ &&&= A \frac{\cos^2 \theta}{\sin \theta + 1} \\ &&&= A \frac{1-\sin^2 \theta}{\sin \theta + 1} \\ &&&= A (1-\sin \theta) \end{align*} When $\theta = -\frac12 \pi, r = 4$, so $A = 2$.

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2017 Paper 3 Q6

D: 1700.0 B: 1500.0

In this question, you are not permitted to use any properties of trigonometric functions or inverse trigonometric functions. The function $\T$ is defined for $x>0$ by \[ \T(x) = \int_0^x \! \frac 1 {1+u^2} \, \d u\,, \] and $\displaystyle T_\infty = \int_0^\infty \!\! \frac 1 {1+u^2} \, \d u\,$ (which has a finite value).

By making an appropriate substitution in the integral for $\T(x)$, show that \[\T(x) = \T_\infty - \T(x^{-1})\,.\]
Let $v= \dfrac{u+a}{1-au}$, where $a$ is a constant. Verify that, for $u\ne a^{-1}$, \[ \frac{\d v}{\d u} = \frac{1+v^2}{1+u^2} \,. \] Hence show that, for $a>0$ and $x< \dfrac1a\,$, \[ \T(x) = \T\left(\frac{x+a}{1-ax}\right) -\T(a) \,. \] Deduce that \[ \T(x^{-1}) = 2\T_\infty -\T\left(\frac{x+a}{1-ax}\right) -\T(a^{-1}) \] and hence that, for $b>0$ and $y>\dfrac1b\,$, \[ \T(y) =2\T_\infty - \T\left(\frac{y+b}{by-1}\right) - \T(b) \,. \]
Use the above results to show that $\T(\sqrt3)= \tfrac23 \T_\infty \,$ and $\T(\sqrt2 -1)= \frac14 \T_\infty\,$.

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2017 Paper 3 Q7

D: 1700.0 B: 1500.0

Show that the point $T$ with coordinates \[ \left( \frac{a(1-t^2)}{1+t^2} \; , \; \frac{2bt}{1+t^2}\right) \tag{$*$} \] (where $a$ and $b$ are non-zero) lies on the ellipse \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} =1 \,. \]

The line $L$ is the tangent to the ellipse at $T$. The point $(X,Y)$ lies on $L$, and $X^2\ne a^2$. Show that \[ (a+X)bt^2 -2aYt +b(a-X) =0 \,.\] Deduce that if $a^2Y^2>(a^2-X^2)b^2$, then there are two distinct lines through $(X,Y)$ that are tangents to the ellipse. Interpret this result geometrically. Show, by means of a sketch, that the result holds also if $X^2=a^2\,$.
The distinct points $P$ and $Q$ are given by $(*)$, with $t=p$ and $t=q$, respectively. The tangents to the ellipse at $P$ and $Q$ meet at the point with coordinates $(X,Y)$, where $X^2\ne a^2\,$. Show that \[ (a+X)pq = a-X\] and find an expression for $p+q$ in terms of $a$, $b$, $X$ and $Y$. Given that the tangents meet the $y$-axis at points $(0,y_1)$ and $(0,y_2)$, where $y_1+y_2 = 2b\,$, show that \[ \frac{X^2}{a^2} +\frac{Y}{b}= 1 \,. \]

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2017 Paper 3 Q8

D: 1700.0 B: 1500.0

Prove that, for any numbers $a_1, a_2, \ldots\,,$ and $b_1, b_2, \ldots\,,$ and for $n\ge1$, \[ \sum_{m=1}^n a_m(b_{m+1} -b_m) = a_{n+1}b_{n+1} -a_1b_1 -\sum_{m=1}^n b_{m+1}(a_{m+1} -a_m) \,. \]

By setting $b_m = \sin mx$, show that \[ \sum_{m=1}^n \cos (m+\tfrac12)x = \tfrac12 \big(\sin (n+1)x - \sin x \big) \cosec \tfrac12 x \,. \] Note: $\sin A - \sin B = \displaystyle 2 \cos \big( \tfrac{{\displaystyle A+B\vphantom{_1}}} {\displaystyle 2\vphantom{^1}} \big)\, \sin\big( \tfrac{{\displaystyle A-B\vphantom{_1}}}{\displaystyle 2\vphantom{^1}} \big)\, $.
Show that \[ \sum_{m=1}^n m\sin mx = \big (p \sin(n+1)x +q \sin nx\big) \cosec^2 \tfrac12 x \,, \] where $p$ and $q$ are to be determined in terms of $n$. Note: $2\sin A \sin B = \cos (A-B) - \cos (A+B)\,$; Note: $2\cos A \sin B = \sin (A+B) - \sin (A-B)\,$.

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2017 Paper 3 Q9

D: 1700.0 B: 1500.9

Two particles $A$ and $B$ of masses $m$ and $2 m$, respectively, are connected by a light spring of natural length $a$ and modulus of elasticity $\lambda$. They are placed on a smooth horizontal table with $AB$ perpendicular to the edge of the table, and $A$ is held on the edge of the table. Initially the spring is at its natural length. Particle $A$ is released. At a time $t$ later, particle $A$ has dropped a distance $y$ and particle $ B$ has moved a distance $x$ from its initial position (where $x < a$). Show that $ y + 2x= \frac12 gt^2$. The value of $\lambda$ is such that particle $B$ reaches the edge of the table at a time $T$ given by $T= \sqrt{6a/g\,}\,$. By considering the total energy of the system (without solving any differential equations), show that the speed of particle $B$ at this time is $\sqrt{2ag/3\,}\,$.

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2017 Paper 3 Q10

D: 1700.0 B: 1484.0

A uniform rod $PQ$ of mass $m$ and length $3a$ is freely hinged at $P$. The rod is held horizontally and a particle of mass $m$ is placed on top of the rod at a distance~$\ell$ from $P$, where $\ell <2a$. The coefficient of friction between the rod and the particle is $\mu$. The rod is then released. Show that, while the particle does not slip along the rod, \[ (3a^2+\ell^2)\dot \theta^2 = g(3a+2\ell)\sin\theta \,, \] where $\theta$ is the angle through which the rod has turned, and the dot denotes the time derivative. Hence, or otherwise, find an expression for $\ddot \theta$ and show that the normal reaction of the rod on the particle is non-zero when~$\theta$ is acute. Show further that, when the particle is on the point of slipping, \[ \tan\theta = \frac{\mu a (2a-\ell)}{2(\ell^2 + a\ell +a^2)} \,. \] What happens at the moment the rod is released if, instead, $\ell>2a$?

Solution:

By energy considerations, the initial energy is $0$.

\begin{tabular}{l|c|c} & Inital & \@ $\theta$ \\ \hline Rotational KE of rod & $0$ & $\frac{1}{2}I\dot{\theta}^2 = \frac{1}{2} \frac{1}{3} m (3a)^2 \dot{\theta}^2 = \frac32 m a^2 \dot{\theta}^2$\\ KE of particle & $0$ & $\frac12 m \ell^2\dot{\theta}^2$\\ GPE of rod & $0$ & $-\frac{3}{2}mga \sin \theta$\\ GPE of particle & $0$ & $-mg \ell \sin \theta$ \\ \hline Total & $0$ & $\frac12m \l \l 3a^2 + \ell^2\r \dot{\theta}^2 - \l 3a + 2\ell \r g \sin \theta \r$ \end{tabular}

Therefore: \begin{align*} && \l 3a^2 + \ell^2\r \dot{\theta}^2 &= \l 3a + 2\ell \r g \sin \theta \\ \Rightarrow && \l 3a^2 + \ell^2\r 2\dot{\theta} \ddot{\theta} &= \l 3a + 2\ell \r g \cos\theta \dot{\theta} \tag{$\frac{\d}{\d t}$} \\ \Rightarrow && 2\l 3a^2 + \ell^2\r \ddot{\theta} &= \l 3a + 2\ell \r g \cos\theta \\ \Rightarrow && \ddot{\theta} &= \boxed{\frac{3a + 2\ell }{2(3a^2 + \ell^2)}g \cos\theta} \\ \end{align*} \begin{align*} \text{N}2(\perp PQ): && mg \cos \theta - R &= m \ell \ddot{\theta} \\ && R &= mg \cos \theta - m \ell \l \frac{3a + 2\ell }{2(3a^2 + \ell^2)}g \cos\theta \r \\ && &= mg\cos \theta \l 1 - \ell \frac{3a + 2\ell }{2(3a^2 + \ell^2)} \r \\ && &= mg \cos \theta \l \frac{6a^2 + 2\ell^2 - 3a\ell - 2\ell^2}{2(3a^2 + \ell^2)} \r \\ && &= mg \cos \theta \l \frac{3a(2a - \ell)}{2(3a^2 + \ell^2)} \r > 0 \tag{since $2a > \ell$} \end{align*} At limiting equilibrium, $F = \mu R$. \begin{align*} \text{N}2(\parallel PQ): && \mu R - mg \sin \theta &= m \ell \dot{\theta}^2 \\ \Rightarrow && \mu mg \cos \theta \l \frac{3a(2a - \ell)}{2(3a^2 + \ell^2)} \r - mg \sin \theta &= m \ell \frac{(3a+2\ell)}{(3a^2+\ell^2)} g \sin \theta \\ \Rightarrow && \mu \l 3a(2a - \ell) \r - \l 2(3a^2 + \ell^2) \r \tan \theta &= 2\ell (3a+2\ell) \tan \theta \\ \Rightarrow && \mu \l 3a(2a - \ell) \r &= \l 6a\ell + 6a^2 + 6\ell^2 \r \tan \theta \\ \Rightarrow && \tan \theta &= \boxed{\frac{\mu a(2a-\ell)}{2(a^2 + a\ell + \ell^2)}} \end{align*} If $\ell > 2a$, then the initial reaction force will be $0$, ie the particle will have no contact with the rod. In other words, the rod will rotate faster than the particle will free-fall and the particle immediately loses contact with the rod.

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2017 Paper 3 Q11

D: 1700.0 B: 1484.0

A railway truck, initially at rest, can move forwards without friction on a long straight \mbox{horizontal} track. On the truck, $n$ guns are mounted parallel to the track and facing backwards, where $n>1$. Each of the guns is loaded with a single projectile of mass $m$. The mass of the truck and guns (but not including the projectiles) is $M$. When a gun is fired, the projectile leaves its muzzle horizontally with a speed $v-V$ relative to the ground, where~$V$ is the speed of the truck immediately before the gun is fired.

All $n$ guns are fired simultaneously. Find the speed, $u$, with which the truck moves, and show that the kinetic energy, $K$, which is gained by the system (truck, guns and projectiles) is given by \[ K= \tfrac{1}{2}nmv^2\left(1 +\frac{nm}{M} \right) . \]
Instead, the guns are fired one at a time. Let $u_r$ be the speed of the truck when $r$ guns have been fired, so that $u_0= 0$. Show that, for $1\le r \le n\,$, \[ u_r - u_{r-1} = \frac{mv}{M+(n-r)m} \tag{$*$} \] and hence that $u_n < u\,$.
Let $K_r$ be the total kinetic energy of the system when $r$ guns have been fired (one at a time), so that $K_0 = 0$. Using $(*)$, show that, for $1\le r\le n\,$, \[ K_r -K_{r-1} = \tfrac 12 mv^2 + \tfrac12 mv (u_r-u_{r-1}) \] and hence show that \[ K_n = \tfrac{1}{2}nmv^2 +\tfrac{1}{2}mvu_n \,. \] Deduce that \(K_n

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2017 Paper 3 Q12

D: 1700.0 B: 1500.2

The discrete random variables $X$ and $Y$ can each take the values $1$, $\ldots\,$, $n$ (where $n\ge2$). Their joint probability distribution is given by \[ \P(X=x, \ Y=y) = k(x+y) \,, \] where $k$ is a constant.

Show that \[ \P(X=x) = \dfrac{n+1+2x}{2n(n+1)}\,. \] Hence determine whether $X$ and $Y$ are independent.
Show that the covariance of $X$ and $Y$ is negative.

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2017 Paper 3 Q13

D: 1700.0 B: 1500.0

The random variable $X$ has mean $\mu$ and variance $\sigma^2$, and the function ${\rm V}$ is defined, for $-\infty < x < \infty$, by \[ {\rm V}(x) = \E \big( (X-x)^2\big) . \] Express ${\rm V}(x)$ in terms of $x$, $ \mu$ and $\sigma$. The random variable $Y$ is defined by $Y={\rm V}(X)$. Show that \[ \E(Y) = 2 \sigma^2 %\text{ \ \ and \ \ } %\Var(Y) = \E(X-\mu)^4 -\sigma^4 . \tag{$*$} \] Now suppose that $X$ is uniformly distributed on the interval $0\le x \le1\,$. Find ${\rm V}(x)\,$. Find also the probability density function of $Y\!$ and use it to verify that $(*)$ holds in this case.

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