Problems

Filters
Clear Filters

7 problems found

2018 Paper 1 Q9
D: 1516.0 B: 1500.0
energy conservation kinematics inclined plane optimisation inequality SUVAT go-kart minimisation

A straight road leading to my house consists of two sections. The first section is inclined downwards at a constant angle \(\alpha\) to the horizontal and ends in traffic lights; the second section is inclined upwards at an angle \(\beta\) to the horizontal and ends at my house. The distance between the traffic lights and my house is \(d\). I have a go-kart which I start from rest, pointing downhill, a distance \(x\) from the traffic lights on the downward-sloping section. The go-kart is not powered in any way, all resistance forces are negligible, and there is no sudden change of speed as I pass the traffic lights. Given that I reach my house, show that \(x \sin \alpha\ge d \sin\beta\,\). Let \(T\) be the total time taken to reach my house. Show that \[ \left(\frac{g\sin\alpha}2 \right)^{\!\frac12} T = (1+k) \sqrt{x} - \sqrt{k^2 x -kd\;} \,, \] where \(k = \dfrac{\sin\alpha}{\sin\beta}\,\). Hence determine, in terms of \(d\) and \(k\), the value of \(x\) which minimises \(T\). [You need not justify the fact that the stationary value is a minimum.]

Solution: Applying conservation of energy, since there are no external forces (other than gravity) the condition to reach the house (with any speed) is the initial GPE is larger than the final GPE, ie: \begin{align*} && m g x \sin \alpha &\geq m g d \sin \beta \\ \Rightarrow && x \sin \alpha &\geq d \sin \beta \end{align*} Let \(T_1\) be the time taken on the downward section, and \(T_2\) the time taken on the upward section, then: \begin{align*} && s &= ut + \frac12 a t^2 \\ \Rightarrow && x &= \frac12 g \sin \alpha T_1^2 \\ \Rightarrow && T_1^2 &= \frac{2x}{g \sin \alpha} \\ && v &= u + at \\ \Rightarrow && v &= T_1 g \sin \alpha \\ && mg x \sin \alpha &= mg d \sin \beta + \frac12 m w^2 \\ \Rightarrow && w &= \sqrt{2(x \sin \alpha - d \sin \beta)} \\ && w &= v - g \sin \beta T_2 \\ \Rightarrow && T_2 &= \frac{v - w}{g \sin \beta} \\ \Rightarrow && T &= T_1 + T_2 \\ &&&= \sqrt{\frac{2x}{g \sin \alpha}} + \frac{\sqrt{\frac{2x}{g \sin \alpha}} g \sin \alpha- \sqrt{2(x \sin \alpha - d \sin \beta)}}{g \sin \beta} \\ &&&= \left ( \frac{2}{g \sin \alpha} \right)^{\tfrac12} \left ( \sqrt{x} + \sqrt{x}k - \sqrt{k^2x-kd}\right) \end{align*} Differentiating wrt to \(x\), we obtain: \begin{align*} && \frac{\d T}{\d x} &= C(-(1+k)x^{-1/2}+k^2(k^2 x - kd)^{-1/2}) \\ \text{set to }0: && 0 &= k^2(k^2 x - kd)^{-1/2} - (1+k)x^{-1/2} \\ \Rightarrow && \sqrt{x} k^2 &= \sqrt{k^2x - kd} (1+k) \\ \Rightarrow && x k^4 &= (k^2x-kd)(1+k)^2 \\ \Rightarrow && x(k^4-k^2(1+k)^2) &= -kd(1+k)^2 \\ \Rightarrow && x(2k^2+k) &= d \\ \Rightarrow && x &= \frac{d}{(2k^2+k)} \end{align*}

View
2017 Paper 2 Q5
D: 1600.0 B: 1484.0
parametric equations parabola normal to curve conics optimisation circle gradient minimisation

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.

Solution:

  1. \(\,\) \begin{align*} && \frac{\d x}{\d t} &= 2at \\ && \frac{\d y}{\d t} &= 2a \\ \Rightarrow && \frac{\d y}{\d x} &= \frac1t \\ && -p &= \text{grad of normal} \\ &&&= \frac{y-2ap}{x-ap^2} \\ \Rightarrow && y &= -px + ap^3+2ap \\ && 2an &= -pan^2 + ap^3 + 2ap \\ \Rightarrow && 0 &= pan^2+2an-ap(2+p^2) \\ \Rightarrow && n &= p, -\left ( p + \frac2{p}\right) \\ \Rightarrow && n &= -\left ( p + \frac2{p}\right) \end{align*}
  2. \(\,\) \begin{align*} && |PN|^2 &= (ap^2-an^2)^2 +(2ap-2an)^2 \\ &&&= a^2(p-n)^2(p+n)^2+4a^2(p-n)^2 \\ &&&= a^2(p-n)^2((p+n)^2+4) \\ &&&= a^2\left(p+p+\frac2p \right)^2 \left ( \left ( -\frac2p\right)^2+4\right)\\ &&&= a^2\left(\frac{2p^2+2}p \right)^2 \left ( \frac{4}{p^2}+4\right)\\ &&&= 16a^2 \frac{(p^2+1)^3}{p^4} \\ \\ && \frac{\d |PN|^2}{\d p^2} &= 16a^2\frac{3(p^2+1)^2p^4-2(p^2+1)^3p^2}{p^8} \\ &&&= 16a^2(p^2+1)^2 \frac{3p^2-2(p^2+1)}{p^6} \\ &&&= 16a^2(p^2+1)^2 \frac{p^2-2}{p^6} \end{align*} Therefore minimized when \(p^2=2\) (clearly a minimum by considering behaviour as \(p^2 \to 0, \infty\))
  3. If \(PN\) is the diameter of \(PNQ\) then \(QP\) and \(QN\) are perpendicular, ie \begin{align*} && -1 &= \frac{2ap-2aq}{ap^2-aq^2} \cdot \frac{2aq-2an}{aq^2-an^2} \\ &&&= \frac{2}{p+q} \cdot \frac{2}{q+n} \\ &&&= \frac{2}{p+q} \cdot \frac{2}{q - p -\frac{2}{p}} \\ \Rightarrow && 4 &= (p+q)(p+\frac2{p}-q) \\ &&&= p^2-q^2 + \frac{2q}{p} + 2 \\ \Rightarrow && 2 &= p^2 - q^2 + \frac{2q}{p} \end{align*} Therefore \(q = 0 \Rightarrow p^2 = 2 \Rightarrow |PN|\) is at it's minimum.

View
2003 Paper 2 Q1
D: 1600.0 B: 1516.0
3x3 matrices simultaneous equations linear systems Gaussian elimination consistency conditions parametric solution optimisation minimisation

Consider the equations \begin{alignat*}{2} ax-&y- \ z && =3 \;,\\ 2ax -&y -3z && = 7 \;,\\ 3ax-&y-5z && =b \;, \end{alignat*} where \(a\) and \(b\) are given constants.

  1. In the case \(a=0\,\), show that the equations have a solution if and only if \(b=11\,\).
  2. In the case \(a\ne0\) and \(b=11\,\) show that the equations have a solution with \(z=\lambda\) for any given number \(\lambda\,\).
  3. In the case \(a=2\) and \(b=11\,\) find the solution for which \(x^2+y^2+z^2\) is least.
  4. Find a value for \(a\) for which there is a solution such that \(x>10^6\) and \(y^2+z^2<1\,\).

Solution:

  1. If \(a = 0\), then then the LHS second equation is the average of the first and last equations, ie \(7 = \frac{b+3}{2}\) so \(b = 11\). This clearly has solutions, say \(x = 0, y = -1, z = -2\).
  2. If \(a \neq 0\) and \(b = 11\), it is still the case that the third equation a linear combination of the first two. Therefore we can consider the linear system: \begin{cases} ax - y &= 3 + \lambda \\ 2ax - y &= 7 + 3\lambda \end{cases} and since \(-a+2a = a \neq 0\) the solution has a unique solution for \(x\) and \(y\).
  3. \begin{align*} \begin{cases} 2x - y &= 3 + \lambda \\ 4x - y &= 7 + 3\lambda \end{cases} \Rightarrow x = 2 +\lambda, y = 1 + \lambda \\ x^2 + y^2 + z^2 &= (2 + \lambda)^2 + (1+\lambda)^2 + \lambda^2 \\ &= (4 + 1) + (4+2)\lambda + 3\lambda^2 \\ &= 5 + 3((\lambda+1)^2 - 1) \\ &= 3(\lambda + 1)^2 + 2 \end{align*} Therefore the solution is minimized when \(\lambda = -1, x = 1, y = 0, z = -1\)
  4. \begin{align*} \begin{cases} ax - y &= 3 + \lambda \\ 2ax - y &= 7 + 3\lambda \end{cases} \Rightarrow x = \frac{4 +2\lambda}{a}, y = 1 + \lambda \end{align*} We want say \(\lambda = -\frac12\) then we have \(y^2 + z^2 = \frac12\) and \(x = \frac{3}{a}\), so choose \(a < \frac{3}{10^6}\)

View
1997 Paper 1 Q9
D: 1500.0 B: 1484.0
vectors relative velocity optimisation trigonometric inequality crossing problem minimisation

A single stream of cars, each of width \(a\) and exactly in line, is passing along a straight road of breadth \(b\) with speed \(V\). The distance between the successive cars is \(c\).

\psset{xunit=0.9cm,yunit=0.9cm,algebraic=true,dotstyle=o,dotsize=3pt 0,linewidth=0.5pt,arrowsize=3pt 2,arrowinset=0.25} \begin{pspicture*}(-5.32,-1.36)(10.5,2.44) \psline(-5,2)(10,2) \psline(-5,-1)(10,-1) \psline(-4,1)(-4,0) \psline(-4,1)(-1,1) \psline(-1,1)(-1,0) \psline(-4,0)(-1,0) \psline(1,1)(1,0) \psline(1,1)(4,1) \psline(4,1)(4,0) \psline(4,0)(1,0) \psline(6,1)(6,0) \psline(6,1)(9,1) \psline(9,1)(9,0) \psline(9,0)(6,0) \psline{->}(-1,0.5)(1,0.5) \psline{->}(1,0.5)(-1,0.5) \psline{->}(4,0.5)(6,0.5) \psline{->}(6,0.5)(4,0.5) \psline{->}(-4.6,1)(-4.6,0) \psline{->}(-4.6,0)(-4.6,1) \psline{->}(10,2)(10,-1) \psline{->}(10,-1)(10,2) \rput[tl](-5,0.6){\(a\)} \rput[tl](-0.14,0.96){\(c\)} \rput[tl](4.86,0.98){\(c\)} \rput[tl](10.16,0.68){\(b\)} \end{pspicture*}
A chicken crosses the road in safety at a constant speed \(u\) in a straight line making an angle \(\theta\) with the direction of traffic. Show that \[u\geqslant \frac{Va}{c\sin\theta+a\cos\theta}.\] Show also that if the chicken chooses \(\theta\) and \(u\) so that it crosses the road at the least possible uniform speed, it crosses in time \[\frac{b}{V}\left(\frac{c}{a}+\frac{a}{c}\right) . \]

View
1993 Paper 2 Q4
D: 1600.0 B: 1470.3
vectors shortest distance between lines cross product scalar triple product 3D geometry optimisation minimisation aircraft skew lines

Two non-parallel lines in 3-dimensional space are given by \(\mathbf{r}=\mathbf{p}_{1}+t_{1}\mathbf{m}_{1}\) and \(\mathbf{r}=\mathbf{p}_{2}+t_{2}\mathbf{m}_{2}\) respectively, where \(\mathbf{m}_{1}\) and \(\mathbf{m}_{2}\) are unit vectors. Explain by means of a sketch why the shortest distance between the two lines is \[ \frac{\left|(\mathbf{p}_{1}-\mathbf{p}_{2})\cdot(\mathbf{m}_{1}\times\mathbf{m}_{2})\right|}{\left|(\mathbf{m}_{1}\times\mathbf{m}_{2})\right|}. \]

  1. Find the shortest distance between the lines in the case \[ \mathbf{p}_{1}=(2,1,-1)\qquad\mathbf{p}_{2}=(1,0,-2)\qquad\mathbf{m}_{1}=\tfrac{1}{5}(4,3,0)\qquad\mathbf{m}_{2}=\tfrac{1}{\sqrt{10}}(0,-3,1). \]
  2. Two aircraft, \(A_{1}\) and \(A_{2},\) are flying in the directions given by the unit vectors \(\mathbf{m}_{1}\) and \(\mathbf{m}_{2}\) at constant speeds \(v_{1}\) and \(v_{2}.\) At time \(t=0\) they pass the points \(\mathbf{p}_{1}\) and \(\mathbf{p}_{2}\), respectively. If \(d\) is the shortest distance between the two aircraft during the flight, show that \[ d^{2}=\frac{\left|\mathbf{p}_{1}-\mathbf{p}_{2}\right|^{2}\left|v_{1}\mathbf{m}_{1}-v_{2}\mathbf{m}_{2}\right|^{2}-[(\mathbf{p}_{1}-\mathbf{p}_{2})\cdot(v_{1}\mathbf{m}_{1}-v_{2}\mathbf{m}_{2})]^{2}}{\left|v_{1}\mathbf{m}_{1}-v_{2}\mathbf{m}_{2}\right|^{2}}. \]
  3. Suppose that \(v_{1}\) is fixed. The pilot of \(A_{2}\) has chosen \(v_{2}\) so that \(A_{2}\) comes as close as possible to \(A_{1}.\) How close is that, if \(\mathbf{p}_{1},\mathbf{p}_{2},\mathbf{m}_{1}\) and \(\mathbf{m}_{2}\) are as in (i)?

View
1992 Paper 1 Q15
D: 1484.0 B: 1471.3
probability continuous uniform distribution optimisation expected value journey time piecewise linear minimisation

Trains leave Barchester Station for London at 12 minutes past the hour, taking 60 minutes to complete the journey and at 48 minutes past the hour taking 75 minutes to complete the journey. The arrival times of passengers for London at Barchester Station are uniformly distributed over the day and all passengers take the first available train. Show that their average journey time from arrival at Barchester Station to arrival in London is 84.6 minutes. Suppose that British Rail decide to retime the fast 60 minute train so that it leaves at \(x\) minutes past the hour. What choice of \(x\) will minimise the average journey time?

Solution: If you arrive between 12 to and 12 past, it will take 60 minutes + how many minutes you wait at the station. If you arrive between 12 past and 12 to, it will take 75 minutes plus waiting at the station. Let's say arrival time \(X \sim U(0,60)\) minutes past the hour, then travel time is. Let's say there are two random variables, \(X_{fast} \sim U(0,24)\) \(X_{slow} \sim U(0, 36)\). Then if you wait for a fast train your expected wait time is \(72\), if you wait for a slow time, your expected wait time is \(75 + 18 = 93\). There is a \(\frac{24}{60} = \frac{4}{10}\) chance of being in the first case, and \(\frac{6}{10}\) chance of the second, ie: \(\frac{4}{10} \cdot 72 + \frac{6}{10} \cdot 93 = \frac{846}{10} = 84.6\) expected wait time. Suppose the time the trains so the expected fraction of time waiting for the fast train is \(t\) and the slow train is \(1-t\). Then the expected time is: \begin{align*} t \l 30t + 60 \r + (1-t) \l 30(1-t) + 75 \r &= 60t^2 -75t + 105 \\ &= 60 \l t^2 - \frac{5}{4}t \r + 105 \\ &= 60 \l t - \frac{5}{8} \r^2 - ? + 105 \\ \end{align*} Threfore we should choose \(x\) such that \(t = \frac58\), which is \(~37.5\) minutes after the slower train, \(25.5\) minutes past.

View
1987 Paper 3 Q11
D: 1500.0 B: 1500.0
optimisation differentiation minimisation river crossing relative velocity Pythagoras curve sketching

A woman stands in a field at a distance of \(a\,\mathrm{m}\) from the straight bank of a river which flows with negligible speed. She sees her frightened child clinging to a tree stump standing in the river \(b\,\mathrm{m}\) downstream from where she stands and \(c\,\mathrm{m}\) from the bank. She runs at a speed of \(u\,\mathrm{ms}^{-1}\) and swims at \(v\,\mathrm{ms}^{-1}\) in straight lines. Find an equation to be satisfied by \(x,\) where \(x\,\mathrm{m}\) is the distance upstream from the stump at which she should enter the river if she is to reach the child in the shortest possible time. Suppose now that the river flows with speed \(v\) ms\(^{-1}\) and the stump remains fixed. Show that, in this case, \(x\) must satisfy the equation \[ 2vx^{2}(b-x)=u(x^{2}-c^{2})[a^{2}+(b-x)^{2}]^{\frac{1}{2}}. \] For this second case, draw sketches of the woman's path for the three possibilities \(b>c,\) \(b=c\) and \(b< c\).

Solution:

TikZ diagram
The distance to where she enters the water is \(\sqrt{a^2+(b-x)^2}\) and the distance through the water is \(\sqrt{x^2+c^2}\). The total time will be \(\frac{\sqrt{a^2+(b-x)^2}}{u}+\frac{\sqrt{x^2+c^2}}{v}\). To minimise this, we can differentiate. \begin{align*} \frac{\d}{\d x}: && \frac{-(b-x)}{u\sqrt{a^2+(b-x)^2}} + \frac{x}{v \sqrt{x^2+c^2}} &= 0 \\ \Rightarrow && v(b-x)(x^2+c^2)^{\frac12} &= xu(a^2+(b-x)^2)^{\frac12} \end{align*} When she is in the water, she can will move with velocity \(\begin{pmatrix} v \cos \theta \\ v \sin \theta -v \end{pmatrix}\). She needs to travel a distance \(\begin{pmatrix} c \\ -x \end{pmatrix}\), so we must have that \begin{align*} && \frac{x}{c} &= \frac{1-\sin \theta}{\cos \theta} \\ \Rightarrow && \sec \theta - \tan \theta &= \frac{x}{c} \\ \Rightarrow && \sec \theta &= \tan \theta + \frac{x}{c} \\ \Rightarrow && \sec^2 \theta &= \tan^2 \theta + 2 \tan \theta \frac{x}{c} + \frac{x^2}{c^2} \\ \Rightarrow && 1 + \tan^2 \theta &= \tan^2 \theta + 2 \tan \theta \frac{x}{c} + \frac{x^2}{c^2} \\ \Rightarrow && \tan \theta &=\frac{c^2-x^2}{2xc} \\ \Rightarrow && \sin \theta &= \frac{c^2-x^2}{c^2+x^2} \\ && \cos \theta &= \frac{2xc}{c^2+x^2} \\ \end{align*} (where we have taken the positive value for \(\cos \theta\) since we must be heading towards the child). Since \(v \cos \theta t = c\) the time taken to reach the child in the water is \(\frac{c}{v} \frac{c^2+x^2}{2xc} = \frac{c^2+x^2}{2xv}\). So the total time is: \(\frac{\sqrt{a^2+(b-x)^2}}{u}+\frac{c^2+x^2}{2xv}\). To minimise this, we can differentiate. \begin{align*} \frac{\d}{\d x}: && \frac{-(b-x)}{u\sqrt{a^2+(b-x)^2}} -\frac{c^2}{2vx^2} + \frac{x^2}{2vx^2}&= 0 \\ \Rightarrow && u(x^2-c^2)\sqrt{a^2+(b-x)^2}&= 2vx^2(b-x) \end{align*} as required. When \(b = c\), the shortest path will be running directly to the bank (there's no quicker way to get to the bank) then swimming directly out (and letting the current take you downstream exactly as far as you need)). Therefore the path will be:
TikZ diagram
If \(b > c\) then she should run a little downstream first.
TikZ diagram
and if \(c > b\) she should actually run a little upstream to take advantage of the current:
TikZ diagram

View