Problems

In this question, the \(x\)- and \(y\)-axes are horizontal and the \(z\)-axis is vertically upwards.

1. A particle \(P_\alpha\) is projected from the origin with speed \(u\) at an acute angle \(\alpha\) above the positive \(x\)-axis. The curve \(E\) is given by \(z = A - Bx^2\) and \(y = 0\). If \(E\) and the trajectory of \(P_\alpha\) touch exactly once, show that \[u^2 - 2gA = u^2(1 - 4AB)\cos^2\alpha\,.\] \(E\) and the trajectory of \(P_\alpha\) touch exactly once for all \(\alpha\) with \(0 < \alpha < \frac{1}{2}\pi\). Write down the values of \(A\) and \(B\) in terms of \(u\) and \(g\).

2. Describe the set of points which can be hit by particles from the explosion, explaining your answer.
3. Show that, at a time \(t\) after the explosion, the particles lie on a sphere whose centre and radius you should find.
4. Another particle \(Q\) is projected horizontally from the point \((0, 0, A)\) with speed \(u\) in the positive \(x\) direction. Show that, at all times, \(Q\) lies on the curve \(E\).
5. Show that for particles \(Q\) and \(P_\alpha\) to collide, \(Q\) must be projected a time \(\dfrac{u(1-\cos\alpha)}{g\sin\alpha}\) after the explosion.

Two particles, \(A\) of mass \(m\) and \(B\) of mass \(M\), are fixed to the ends of a light inextensible string \(AB\) of length \(r\) and lie on a smooth horizontal plane. The origin of coordinates and the \(x\)- and \(y\)-axes are in the plane. Initially, \(A\) is at \((0,\,0)\) and \(B\) is at \((r,\,0)\). \(B\) is at rest and \(A\) is given an instantaneous velocity of magnitude \(u\) in the positive \(y\) direction. At a time \(t\) after this, \(A\) has position \((x,\,y)\) and \(B\) has position \((X,\,Y)\). You may assume that, in the subsequent motion, the string remains taut.

1. Explain by means of a diagram why \[X = x + r\cos\theta\] \[Y = y - r\sin\theta\] where \(\theta\) is the angle clockwise from the positive \(x\)-axis of the vector \(\overrightarrow{AB}\).
2. Find expressions for \(\dot{X}\), \(\dot{Y}\), \(\ddot{X}\) and \(\ddot{Y}\) in terms of \(\ddot{x}\), \(\ddot{y}\), \(\dot{x}\), \(\dot{y}\), \(r\), \(\ddot{\theta}\), \(\dot{\theta}\) and \(\theta\), as appropriate. Assume that the tension \(T\) in the string is the only force acting on either particle.
3. Show that \[\ddot{x}\sin\theta + \ddot{y}\cos\theta = 0\] \[\ddot{X}\sin\theta + \ddot{Y}\cos\theta = 0\] and hence that \(\theta = \dfrac{ut}{r}\).
4. Show that \[m\ddot{x} + M\ddot{X} = 0\] \[m\ddot{y} + M\ddot{Y} = 0\] and find \(my + MY\) in terms of \(t\) and \(m, M, u, r\) as appropriate.
5. Show that \[y = \frac{1}{m+M}\left(mut + Mr\sin\!\left(\frac{ut}{r}\right)\right).\]
6. Show that, if \(M > m\), then the \(y\) component of the velocity of particle \(A\) will be negative at some time in the subsequent motion.

Let \(C_1\) be the curve given by the parametric equations \[ x = ct\,, \quad y = \frac{c}{t}\,, \] where \(c > 0\) and \(t \neq 0\), and let \(C_2\) be the circle \[ (x-a)^2 + (y-b)^2 = r^2\,. \] \(C_1\) and \(C_2\) intersect at the four points \(P_i\) (\(i = 1,2,3,4\)), and the corresponding values of the parameter \(t\) at these points are \(t_i\).

1. Show that \(t_i\) are the roots of the equation \[ c^2 t^4 - 2act^3 + (a^2 + b^2 - r^2)t^2 - 2bct + c^2 = 0\,. \qquad (*) \]
2. Show that \[ \sum_{i=1}^{4} t_i^2 = \frac{2}{c^2}(a^2 - b^2 + r^2) \] and find a similar expression for \(\displaystyle\sum_{i=1}^{4} \frac{1}{t_i^2}\).
3. Hence show that \(\displaystyle\sum_{i=1}^{4} OP_i^2 = 4r^2\), where \(OP_i\) denotes the distance of the point \(P_i\) from the origin.
4. Suppose that the curves \(C_1\) and \(C_2\) touch at two distinct points. By considering the product of the roots of \((*)\), or otherwise, show that the centre of circle \(C_2\) must lie on either the line \(y = x\) or \(y = -x\).

1. Show that when \(\alpha\) is small, \(\cos(\theta + \alpha) - \cos\theta \approx -\alpha\sin\theta - \frac{1}{2}\alpha^2\cos\theta\). Find the limit as \(\alpha \to 0\) of \[ \frac{\sin(\theta+\alpha) - \sin\theta}{\cos(\theta+\alpha) - \cos\theta} \qquad (*) \] in the case \(\sin\theta \neq 0\). In the case \(\sin\theta = 0\), what happens to the value of expression \((*)\) when \(\alpha \to 0\)?
2. A circle \(C_1\) of radius \(a\) rolls without slipping in an anti-clockwise direction on a fixed circle \(C_2\) with centre at the origin \(O\) and radius \((n-1)a\), where \(n\) is an integer greater than \(2\). The point \(P\) is fixed on \(C_1\). Initially the centre of \(C_1\) is at \((na, 0)\) and \(P\) is at \(\big((n+1)a, 0\big)\).
   1. Let \(Q\) be the point of contact of \(C_1\) and \(C_2\) at any time in the rolling motion. Show that when \(OQ\) makes an angle \(\theta\), measured anticlockwise, with the positive \(x\)-axis, the \(x\)-coordinate of \(P\) is \(x(\theta) = a(n\cos\theta + \cos n\theta)\), and find the corresponding expression for the \(y\)-coordinate, \(y(\theta)\), of \(P\).
   2. Find the values of \(\theta\) for which the distance \(OP\) is \((n-1)a\).
   3. Let \(\theta_0 = \dfrac{1}{n-1}\pi\). Find the limit as \(\alpha \to 0\) of \[ \frac{y(\theta_0 + \alpha) - y(\theta_0)}{x(\theta_0 + \alpha) - x(\theta_0)}\,. \] Hence show that, at the point \(\big(x(\theta_0),\, y(\theta_0)\big)\), the tangent to the curve traced out by \(P\) is parallel to \(OP\).

1. A curve has parametric equations \[ x = -4\cos^3 t, \qquad y = 12\sin t - 4\sin^3 t. \] Find the equation of the normal to this curve at the point \[ \bigl(-4\cos^3\phi,\; 12\sin\phi - 4\sin^3\phi\bigr), \] where \(0 < \phi < \tfrac{1}{2}\pi\). Verify that this normal is a tangent to the curve \[ x^{2/3} + y^{2/3} = 4 \] at the point \((8\cos^3\phi,\; 8\sin^3\phi)\).
2. A curve has parametric equations \[ x = \cos t + t\sin t, \qquad y = \sin t - t\cos t. \] Find the equation of the normal to this curve at the point \[ \bigl(\cos\phi + \phi\sin\phi,\; \sin\phi - \phi\cos\phi\bigr), \] where \(0 < \phi < \tfrac{1}{2}\pi\). Determine the perpendicular distance from the origin to this normal, and hence find the equation of a curve, independent of \(\phi\), to which this normal is a tangent.

A curve \(C\) is determined by the parametric equations \[ x=at^2 \, , \; y = 2at\,, \] where \(a > 0\).

1. Show that the normal to \(C\) at a point \(P\), with non-zero parameter \(p\), meets \(C\) again at a point \(N\), with parameter \(n\), where \[ n= - \left( p + \frac{2}{p} \right). \]
2. Show that the distance \(\left| PN \right|\) is given by \[ \vert PN\vert^2 = 16a^2\frac{(p^2+1)^3}{p^4} \] and that this is minimised when \(p^2=2\,\).
3. The point \(Q\), with parameter \(q\), is the point at which the circle with diameter \(PN\) cuts \(C\) again. By considering the gradients of \(QP\) and \(QN\), show that \[ 2 = p^2-q^2 + \frac{2q}p. \] Deduce that \(\left| PN \right|\) is at its minimum when \(Q\) is at the origin.

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 \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} =1 \,. \]

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 \(X^2=a^2\,\).
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 \[ \frac{X^2}{a^2} +\frac{Y}{b}= 1 \,. \]

The curve \(C_1\) has parametric equations \(x=t^2\), \(y= t^3\), where \(-\infty < t < \infty\,\). Let \(O\) denote the point \((0,0)\). The points \(P\) and \(Q\) on \(C_1\) are such that \(\angle POQ\) is a right angle. Show that the tangents to \(C_1\) at \(P\) and \(Q\) intersect on the curve \(C_2\) with equation \(4y^2=3x-1\). Determine whether \(C_1\) and \(C_2\) meet, and sketch the two curves on the same axes.

Show Solution

---

Solution: \(\angle POQ = 90^\circ\) means that if \(P(p^2,p^3)\) and \(Q(q^2,q^3)\) are our points then \(OP^2+OQ^2 = PQ^2\), so \begin{align*} && p^4+p^6+q^4+q^6 &= (p^2-q^2)^2+(p^3-q^3)^2 \\ &&&= p^4+q^4-2p^2q^2+p^6+q^6-2p^3q^3 \\ \Rightarrow && 0 &= 2p^2q^2(1+pq) \\ \Rightarrow && pq &= -1 \\ \\ && \frac{\d y}{ \d x} &= \frac{\frac{\d y }{\d t}}{\frac{\d x}{\d t}} \\ &&&= \frac{3t^2}{2t} = \tfrac32t \\ \Rightarrow && \frac{y-p^3}{x-p^2} &= \tfrac32p \\ \Rightarrow && 2(y-p^3) &=3p(x-p^2) \\ && 2(y-q^3) &=3q(x-q^2) \\ \Rightarrow && 2(q^3-p^3) &= (3p-3q)x+3(q^3-p^3) \\ && p^3-q^3 &= 3(p-q)x \\ \Rightarrow && x &= \tfrac13(p^2+q^2+pq) \\ && 2y &= 3p(\tfrac13(p^2+q^2+pq)-p^2)+2p^3 \\ &&&= p(p^2+q^2+pq)-p^3 \\ &&&= pq^2+p^2q \\ &&&= -p-q \\ &&y&= -\frac{p+q}{2} \\ \\ && 4y^2 &= p^2+q^2 \\ && 3x-1 &= p^2+q^2 \\ \end{align*} To check if they meet, try \(4t^6=3t^2 - 1\). Consider \(y = 4x^3-3x+1\) \(y(0) = 1\) and \(y' = 12x^2-3 = 3(4x^2-1)\) which has roots at \(\pm \tfrac12\), therefore we need to test \(y(\tfrac12) = \tfrac12-\tfrac32 + 1 = 0\), so there is a one intersection at \(x = \tfrac1{2}, y = \tfrac1{2\sqrt{2}}\)

+ - ↺

The midpoint of a rod of length \(2b\) slides on the curve \(y =\frac14 x^2\), \(x\ge0\), in such a way that the rod is always tangent, at its midpoint, to the curve. Show that the curve traced out by one end of the rod can be written in the form \begin{align*} x& = 2 \tan\theta - b \cos\theta \\ y& = \tan^2\theta - b \sin\theta \end{align*} for some suitably chosen angle \(\theta\) which satisfies \(0\le \theta < \frac12\pi\,\). When one end of the rod is at a point \(A\) on the \(y\)-axis, the midpoint is at point \(P\) and \(\theta = \alpha\). Let \(R\) be the region bounded by the following:

• the curve \(y=\frac14x^2\) between the origin and \(P\);
• the \(y\)-axis between \(A\) and the origin;
• the half-rod \(AP\).

Show Solution

---

Solution: At the point \((2t, t^2)\) the gradient is \(t\). Suppose \(\tan \theta = t\), then the point \(b\) away in each direction is \(\binom{2t}{t^2} \pm b \binom{\cos \theta}{\sin \theta}\), ie one end can be written in the form \((x,y) = (2\tan \theta - b \cos \theta, \tan^2 \theta - b \sin \theta)\). Notice we must have \(2\tan \alpha- b \cos \alpha= 0 \Rightarrow b = 2 \frac{\sin \alpha}{\cos ^2 \alpha}\), therefore the coordinates are \((2 \tan \alpha - 2 \tan \alpha, \tan^2 \alpha - 2\tan^2 \alpha) = (0, -\tan^2 \alpha)\) and \((4 \tan \alpha, 3\tan^2 \alpha)\)

+ - ↺

The area we can find by calculating the integrate of \(\tan^2 \alpha + \frac14x^2\) between \(0\) and \(2 \tan \alpha\) and then subtracting the triangle, ie \begin{align*} &&A &= 2\tan^3 \alpha + \frac1{12} (2 \tan \alpha)^3 - \frac12 \cdot 2 \tan \alpha \cdot (2 \tan^2 \alpha) \\ &&&= \left (2 + \frac23 -2\right) \tan^3 \alpha \\ &&&= \frac23 \tan^3 \alpha \end{align*}

A curve is given parametrically by \begin{align*} x&= a\big( \cos t +\ln \tan \tfrac12 t\big)\,,\\ y&= a\sin t\,, \end{align*} where \(0 < t < \frac12 \pi\) and \(a\) is a positive constant. Show that \(\ds \frac{\d y}{\d x} = \tan t\) and sketch the curve. Let \(P\) be the point with parameter \(t\) and let \(Q\) be the point where the tangent to the curve at \(P\) meets the \(x\)-axis. Show that \(PQ=a\). The {\sl radius of curvature}, \(\rho\), at \(P\) is defined by \[ \rho= \frac {\big(\dot x ^2+\dot y^2\big)^{\frac32}} {\vert \dot x \ddot y - \dot y \ddot x\vert \ \ } \,, \] where the dots denote differentiation with respect to \(t\). Show that \(\rho =a\cot t\). The point \(C\) lies on the normal to the curve at \(P\), a distance \(\rho\) from \(P\) and above the curve. Show that \(CQ\) is parallel to the \(y\)-axis.

A curve is defined parametrically by \[ x=t^2 \;, \ \ \ y=t (1 + t^2 ) \;. \] The tangent at the point with parameter \(t\), where \(t\ne0\,\), meets the curve again at the point with parameter \(T\), where \(T\ne t\,\). Show that \[ T = \frac{1 - t^2 }{2t} \mbox { \ \ \ and \ \ \ } 3t^2\ne 1\;. \] Given a point \(P_0\,\) on the curve, with parameter \(t_0\,\), a sequence of points \(P_0 \, , \; P_1 \, , \; P_2 \, , \ldots\) on the curve is constructed such that the tangent at \(P_i\) meets the curve again at \(P_{i+1}\). If \(t_0 = \tan \frac{ 7 } {18}\pi\,\), show that \(P_3 = P_0\) but \(P_1\ne P_0\,\). Find a second value of \(t_0\,\), with \(t_0>0\,\), for which \(P_3 = P_0\) but \(P_1\ne P_0\,\).

Show that the equation \(x^3 + px + q=0\) has exactly one real solution if \(p \ge 0\,\). A parabola \(C\) is given parametrically by \[ x = at^2, \: \ \ y = 2at \: \: \: \ \ \ \ \ \ \l a > 0 \r \;. \] Find an equation which must be satisfied by \(t\) at points on \(C\) at which the normal passes through the point \(\l h , \; k \r\,\). Hence show that, if \(h \le 2a \,\), exactly one normal to \(C\) will pass through \(\l h , \; k \r \, \). Find, in Cartesian form, the equation of the locus of the points from which exactly two normals can be drawn to \(C\,\). Sketch the locus.

Show Solution

---

Solution: If \(p \geq 0\) then the derivative is \(x^2+p \geq 0\) and in particular the function is increasing. Therefore it will have exactly \(1\) real root (as for very large negative \(x\) it is negative, and vice-versa fo positive \(x\)). \begin{align*} && \frac{\d y}{\d x} &= \frac{\dot{y}}{\dot{x}} \\ &&&= \frac{2a}{2at} \\ &&&= \frac{1}{t} \\ \text{eq of normal} && \frac{k-2at}{h-at^2} &= -t \\ \Rightarrow && k-2at &= at^3-th \\ && 0 &= at^3+(2a-h)t-k \end{align*} Since \(a > 0\) this is the same constraint as the first part, in particular \(2a-h \geq 0 \Leftrightarrow 2a \geq h\). If exactly two normals can be drawn to \(C\) we must have that our equation has a repeated root, ie \begin{align*} && 0 &= at^3+(2a-h)t-k\\ && 0 &= 3at^2+2a-h\\ \Rightarrow && 0 &= 3at^3+ 3(2a-h)t-3k \\ && 0 &= 3at^3+(2a-h)t \\ \Rightarrow && 0 &= 2(2a-h)t-3k \\ \Rightarrow && t &= \frac{3k}{2(2a-h)} \\ \Rightarrow && 0 &= 3a \left (\frac{3k}{2(2a-h)} \right)^2+2a-h \\ && 0 &= 27ak^2+4(2a-h)^3 \end{align*}

+ - ↺

A closed curve is given by the equation $$ x^{2/n} + y^{2/n} = a^{2/n} \eqno(*) $$ where \(n\) is an odd integer and \(a\) is a positive constant. Find a parametrization \(x=x(t)\), \(y=y(t)\) which describes the curve anticlockwise as \(t\) ranges from \(0\) to \(2\pi\). Sketch the curve in the case \(n=3\), justifying the main features of your sketch. The area \(A\) enclosed by such a curve is given by the formula $$ A= {1\over 2} \int_0^{2\pi} \left[ x(t) {\d y(t)\over \d t} - y(t) {\d x(t)\over \d t} \right] \,\d t \,. $$ Use this result to find the area enclosed by (\(*\)) for \(n=3\).

Two curves are given parametrically by \[ x_{1}=(\theta+\sin\theta),\qquad y_{1}=(1+\cos\theta),\tag{1} \]and \[ x_{2}=(\theta-\sin\theta),\qquad y_{1}=-(1+\cos\theta),\tag{2} \] Find the gradients of the tangents to the curves at the points where \(\theta= \pi/2\) and \(\theta=3\pi/2\). Sketch, using the same axes, the curves for \(0\le\theta \le 2\pi\). Find the equation of the normal to the curve (1) at the point with parameter \(\theta\). Show that this normal is a tangent to the curve (2).

Sketch the curve \(C_{1}\) whose parametric equations are \(x=t^{2},\) \(y=t^{3}.\) The circle \(C_{2}\) passes through the origin \(O\). The points \(R\) and \(S\) with real non-zero parameters \(r\) and \(s\) respectively are other intersections of \(C_{1}\) and \(C_{2}.\) Show that \(r\) and \(s\) are roots of an equation of the form \[ t^{4}+t^{2}+at+b=0, \] where \(a\) and \(b\) are real constants. By obtaining a quadratic equation, with coefficients expressed in terms of \(r\) and \(s\), whose roots would be the parameters of any further intersections of \(C_{1}\) and \(C_{2},\) or otherwise, show that \(O\), \(R\) and \(S\) are the only real intersections of \(C_{1}\) and \(C_{2}.\)

Show Solution

---

Solution:

+ - ↺

Suppose the circle has centre \((c,d)\), then \begin{align*} && c^2+d^2 &= (t^2-c)^2+(t^3-d)^2 \\ \Rightarrow && 0 &= t^4-2ct^2+t^6-2t^3d \\ \Rightarrow && 0 &= t^4+t^2-2td-2c \end{align*} So by setting \(a = -2d\) and \(b = -2c\) we have the desired equation. By matching the coefficients of \(t^4, t^3, t^2\) we must have: \begin{align*} && 0 &= (t^2-(r+s)t+rs)(t^2+t(r+s)-rs+(r+s)^2+1) \\ \Rightarrow && 0 &= t^2+(r+s)t-rs+(r+s)^2+1 \\ && \Delta &= (r+s)^2 -4(1-rs+(r+s)^2) \\ &&&= -4+4rs-3(r+s)^2 \\ &&&=-4-2(r+s)^2-(r-s)^2 < 0 \end{align*} Therefore there are no further (real) solutions. Hence \(O, R, S\) are the only solutions.