1531 problems found
The points \(S\), \(T\), \(U\) and \(V\) have coordinates \((s,ms)\), \((t,mt)\), \((u,nu)\) and \((v,nv)\), respectively. The lines \(SV\) and \(UT\) meet the line \(y=0\) at the points with coordinates \((p,0)\) and \((q,0)\), respectively. Show that \[ p = \frac{(m-n)sv}{ms-nv}\,, \] and write down a similar expression for \(q\). Given that \(S\) and \(T\) lie on the circle \(x^2 + (y-c)^2 = r^2\), find a quadratic equation satisfied by \(s\) and by \(t\), and hence determine \(st\) and \(s+t\) in terms of \(m\), \(c\) and \(r\). Given that \(S\), \(T\), \(U\) and \(V\) lie on the above circle, show that \(p+q=0\).
Solution:
The function \(\f(t)\) is defined, for \(t\ne0\), by \[ \f(t) = \frac t {\e^t-1}\,. \] \begin{questionparts} \item By expanding \(\e^t\), show that \(\displaystyle \lim _{t\to0} \f(t) = 1\,\). Find \(\f'(t)\) and evaluate \(\displaystyle \lim _{t\to0} \f'(t)\,\). \item Show that \(\f(t) +\frac12 t\) is an even function. [{\bf Note:} A function \(\g(t)\) is said to be {\em even} if \(\g(t) \equiv \g(-t)\).] \item Show with the aid of a sketch that \( \e^t( 1-t)\le 1\,\) and deduce that \(\f'(t)\ne 0\) for \(t\ne0\). \end{questionpart} Sketch the graph of \(\f(t)\).
Solution:
For any given (suitable) function \(\f\), the Laplace transform of \(\f\) is the function \(\F\) defined by \[ \F(s) = \int_0^\infty \e^{-st}\f(t)\d t \quad \quad \, (s>0) \,. \]
Solution:
The numbers \(x\), \(y\) and \(z\) satisfy \begin{align*} x+y+z&= 1\\ x^2+y^2+z^2&=2\\ x^3+y^3+z^3&=3\,. \end{align*} Show that \[ yz+zx+xy=-\frac12 \,.\] Show also that \(x^2y+x^2z+y^2z+y^2x+z^2x+z^2y=-1\,\), and hence that \[ xyz=\frac16 \,.\] Let \(S_n=x^n+y^n+z^n\,\). Use the above results to find numbers \(a\), \(b\) and \(c\) such that the relation \[ S_{n+1}=aS_{n}+bS_{n-1}+cS_{n-2}\,, \] holds for all \(n\).
Solution: \begin{align*} && (x+y+z)^2 &= x^2 + y^2 + z^2 + 2(xy+yz+zx) \\ \Rightarrow && 1^2 &= 2 + 2(xy+yz+zx) \\ \Rightarrow && xy+yz+zx &= -\frac12 \end{align*} \begin{align*} && 1 \cdot 2 &= (x+y+z)(x^2+y^2+z^2) \\ &&&= x^3 + y^3 + z^3 + x^2y+x^2z+y^2z+y^2x+z^2x+z^2y \\ &&&= 3 + x^2y+x^2z+y^2z+y^2x+z^2x+z^2y\\ \Rightarrow && -1 &= x^2y+x^2z+y^2z+y^2x+z^2x+z^2y \end{align*} \begin{align*} && (x+y+z)^3 &= x^3 + y^3 + z^3 + \\ &&&\quad \quad 3xy^2 + 3xz^2 + \cdots + 3zx^2 + 3zy^2 + \\ &&&\quad \quad \quad 6xyz \\ \Rightarrow && 1 &= 3 + 3(-1) + 6xyz \\ \Rightarrow && xyz &= \frac16 \end{align*} Since we have \(f(t) = (t-x)(t-y)(t-z) = t^3-t^2-\frac12 t - \frac16\) is zero for \(x,y,z\) we can notice that: \(t^{n+1} = t^n +\frac12 t^{n-1} + \frac16 t^{n-2}\) is also true for \(x,y,z\) (by multiplying by \(t^{n-2}\). Therefore: \(S_{n+1} = S_n + \frac12 S_{n-1} + \frac16 S_{n-2}\)
Show that $\big\vert \e^{\i\beta} -\e^{\i\alpha}\big\vert = 2\sin\frac12 (\beta-\alpha)\,\( for \)0<\alpha<\beta<2\pi\,$. Hence show that \[ \big\vert \e^{\i\alpha} -\e^{\i\beta}\big\vert \; \big\vert \e^{\i\gamma} -\e^{\i\delta}\big\vert + \big\vert \e^{\i\beta} -\e^{\i\gamma}\big\vert \; \big\vert \e^{\i\alpha} -\e^{\i\delta}\big\vert = \big\vert \e^{\i\alpha} -\e^{\i\gamma}\big\vert \; \big\vert \e^{\i\beta} -\e^{\i\delta}\big\vert \,, \] where \(0<\alpha<\beta<\gamma<\delta<2\pi\). Interpret this result as a theorem about cyclic quadrilaterals.
Let \(m\) be a positive integer and let \(n\) be a non-negative integer.
Solution:
A particle is projected under gravity from a point \(P\) and passes through a point \(Q\). The angles of the trajectory with the positive horizontal direction at \(P\) and at \(Q\) are \(\theta\) and \(\phi\), respectively. The angle of elevation of \(Q\) from \(P\) is \(\alpha\).
A light spring is fixed at its lower end and its axis is vertical. When a certain particle \(P\) rests on the top of the spring, the compression is \(d\). When, instead, \(P\) is dropped onto the top of the spring from a height \(h\) above it, the compression at time \(t\) after \(P\) hits the top of the spring is \(x\). Obtain a second-order differential equation relating \(x\) and \(t\) for \(0\le t \le T\), where \(T\) is the time at which \(P\) first loses contact with the spring. Find the solution of this equation in the form \[ x= A + B\cos (\omega t) + C\sin(\omega t)\,, \] where the constants \(A\), \(B\), \(C\) and \(\omega\) are to be given in terms of \(d\), \(g\) and \(h\) as appropriate. Show that \[ T = \sqrt{d/g\;} \left (2 \pi - 2 \arctan \sqrt{2h/d\;}\;\right)\,. \]
A comet in deep space picks up mass as it travels through a large stationary dust cloud. It is subject to a gravitational force of magnitude \(M\!f\) acting in the direction of its motion. When it entered the cloud, the comet had mass \(M\) and speed \(V\). After a time \(t\), it has travelled a distance \(x\) through the cloud, its mass is \(M(1+bx)\), where~\(b\) is a positive constant, and its speed is \(v\).
Solution:
Solution:
What does it mean to say that a number \(x\) is irrational? Prove by contradiction statements A and B below, where \(p\) and \(q\) are real numbers.
Solution:
The variables \(t\) and \(x\) are related by \(t=x+ \sqrt{x^2+2bx+c\;} \,\), where \(b\) and \(c\) are constants and \(b^2 < c\). Show that \[ \frac{\d x}{\d t} = \frac{t-x}{t+b}\;, \] and hence integrate \(\displaystyle \frac1 {\sqrt{x^2+2bx+c}}\,\). Verify by direct integration that your result holds also in the case \(b^2=c\) if \(x+b > 0\) but that your result does not hold in the case \(b^2=c\) if \(x+b < 0\,\).
Solution: \begin{align*} && t &= x+ \sqrt{x^2+2bx+c} \\ && \frac{\d t}{\d x} &= 1 + \frac{x+b}{\sqrt{x^2+2bx+c}} \\ &&&= \frac{x + \sqrt{x^2+2bx+c} + b}{\sqrt{x^2+2bx+c}} \\ &&&= \frac{t+b}{t-x} \\ \Rightarrow && \frac{\d x}{\d t} &= \frac{t-x}{t+b} \\ \\ && \int \frac{1}{\sqrt{x^2+2bx+c}} \d x &= \int \frac{1}{t-x} \frac{t-x}{t+b} \d t \\ &&&= \int \frac{1}{t+b} \d t \\ &&&= \ln (t + b) +C \\ &&&= \ln \left (x + \sqrt{x^2+2bx+c} + b \right) + C \end{align*} If \(b^2 = c\) then we have \(x^2+2bx+b^2 = (x+b)^2\) so \(\sqrt{x^2+2bx+c^2} = x+b\) (if \(x+b>0\)), so \begin{align*} && \int \frac{1}{\sqrt{x^2+2bx+c}} \d x &= \int \frac{1}{x+b} \d x\\ &&&= \ln (x + b) + C \\ &&&= \ln(x+b) + \ln 2 + C' \\ &&&= \ln (2(x+b)) + C' \\ &&&= \ln \left(x + b + \sqrt{(x+b)^2} \right)+C'\\ &&&= \ln \left(x + b + \sqrt{x^2+2bx+c} \right)+C'\\ \end{align*} If \(x+b < 0\) then the antiderivative is \(\ln 0\). \begin{align*} && \int \frac{1}{\sqrt{x^2+2bx+c}} \d x &= -\int \frac{1}{x+b} \d x\\ &&&= -\ln |x + b| + C \\ \end{align*} which are clearly different.