Problems

Given that \[ 5x^{2}+2y^{2}-6xy+4x-4y\equiv a\left(x-y+2\right)^{2} +b\left(cx+y\right)^{2}+d\,, \] find the values of the constants \(a\), \(b\), \(c\) and \(d\). Solve the simultaneous equations \begin{align*} 5x^{2}+2y^{2}-6xy+4x-4y&=9\,, \\ 6x^{2}+3y^{2}-8xy+8x-8y&=14\,. \end{align*}

The curve \(\displaystyle y=\Bigl(\frac{x-a}{x-b}\Bigr)\e^{x}\), where \(a\) and \(b\) are constants, has two stationary points. Show that \[ a-b<0 \ \ \ \text{or} \ \ \ a-b>4 \,. \]

1. [(i)] Show that, in the case \(a=0\) and \(b= \frac12\), there is one stationary point on either side of the curve's vertical asymptote, and sketch the curve.
2. [(ii)] Sketch the curve in the case \( a=\tfrac{9}{2}\) and \(b=0\,\).

Show that \[ \sin(x+y) -\sin(x-y) = 2 \cos x \, \sin y \] and deduce that \[ \sin A - \sin B = 2 \cos \tfrac12 (A+B) \, \sin\tfrac12 (A-B) \,. \] Show also that \[ \cos A - \cos B = -2 \sin \tfrac12(A+B) \, \sin\tfrac12(A-B)\,. \] The points \(P\), \(Q\), \(R\) and \(S\) have coordinates \(\left(a\cos p,b\sin p\right)\), \(\left(a\cos q,b\sin q\right)\), \(\left(a\cos r,b\sin r\right)\) and \(\left(a\cos s,b\sin s\right)\) respectively, where \(0\le p < q < r < s <2\pi\), and \(a\) and \(b\) are positive. Given that neither of the lines \(PQ\) and \(SR\) is vertical, show that these lines are parallel if and only if \[ r+s-p-q = 2\pi\,. \]

Use the substitution \(x=\dfrac{1}{t^{2}-1}\; \), where \(t>1\), to show that, for \( x>0\), \[ \int \frac{1}{\sqrt{x\left(x+1\right) \; } \ }\; \d x =2 \ln \left(\sqrt x+ \sqrt{x +1} \; \right)+c \,. \] {\bf [Note} You may use without proof the result $\displaystyle \int \! \frac{1}{t^2-a^2} \, \d t = \frac{1}{2a} \ln \left| \frac{t-a}{t+a}\right| + \rm {constant}$. {\bf]} The section of the curve \[ y=\dfrac{1}{\sqrt{x}\; }-\dfrac{1}{\sqrt{x+1}\; } \] between \(x=\frac{1}{8}\) and \(x=\frac{9}{16}\) is rotated through \(360^{o}\) about the \(x\)-axis. Show that the volume enclosed is \(2\pi \ln \tfrac{5}{4}\,\). \(\phantom{\dfrac AB}\)

By considering the expansion of \(\left(1+x\right)^{n}\) where \(n\) is a positive integer, or otherwise, show that:

1. \[\binom{n}{0}+\binom{n}1+\binom{n}2 +\cdots +\binom{n}n=2^{n} \]
2. \[\binom{n}{1}+2\binom{n}2+3\binom{n}3 +\cdots +n\binom{n}n=n2^{n-1} \]
3. \[\binom{n}{0}+\frac12\binom{n}1+\frac13\binom{n}2 +\cdots +\frac1{n+1}\binom{n}n=\frac1{n+1}(2^{n+1}-1) \]
4. \[\binom{n}{1}+2^2\binom{n}2+3^2\binom{n}3 +\cdots +n^2\binom{n}n=n(n+1)2^{n-2} \]

Show that, if \(y=\e^x\), then \[ (x-1) \frac{\d^2 y}{\d x^2} -x \frac{\d y}{\d x} +y=0\,. \tag{\(*\)} \] In order to find other solutions of this differential equation, now let \(y=u\e^x\), where \(u\) is a function of \(x\). By substituting this into \((*)\), show that \[ (x-1) \frac{\d^2 u}{\d x^2} + (x-2) \frac{\d u}{\d x} =0\,. \tag{\(**\)} \] By setting \( \dfrac {\d u}{\d x}= v\) in \((**)\) and solving the resulting first order differential equation for \(v\), find~\(u\) in terms of \(x\). Hence show that \(y=Ax + B\e^x\) satisfies \((*)\), where \(A\) and \(B\) are any constants.

Relative to a fixed origin \(O\), the points \(A\) and \(B\) have position vectors \(\bf{a}\) and \(\bf{b}\), respectively. (The points \(O\), \(A\) and \(B\) are not collinear.) The point \(C\) has position vector \(\bf c\) given by \[ {\bf c} =\alpha {\bf a}+ \beta {\bf b}\,, \] where \(\alpha\) and \(\beta\) are positive constants with \(\alpha+\beta<1\,\). The lines \(OA\) and \(BC\) meet at the point \(P\) with position vector \(\bf p\) and the lines \(OB\) and \(AC\) meet at the point \(Q\) with position vector \(\bf q\). Show that \[ {\bf p} =\frac{\alpha {\bf a} }{1-\beta}\,, \] and write down \(\bf q\) in terms of \(\alpha,\ \beta\) and \(\bf {b}\). Show further that the point \(R\) with position vector \(\bf r\) given by \[ {\bf r} =\frac{\alpha {\bf a} + \beta {\bf b}}{\alpha + \beta}\,, \] lies on the lines \(OC\) and \(AB\). The lines \(OB\) and \(PR\) intersect at the point \(S\). Prove that $ \dfrac{OQ}{BQ} = \dfrac{OS}{BS}\,$.

1. Suppose that \(a\), \(b\) and \(c\) are integers that satisfy the equation \[ a^{3}+3b^{3}=9c^{3}. \] Explain why \(a\) must be divisible by 3, and show further that both \(b\) and \(c\) must also be divisible by 3. Hence show that the only integer solution is \(a=b=c=0\,\).
2. Suppose that \(p\), \(q\) and \(r\) are integers that satisfy the equation \[ p^4 +2q^4 = 5r^4 \,.\] By considering the possible final digit of each term, or otherwise, show that \(p\) and \(q\) are divisible by 5. Hence show that the only integer solution is \(p=q=r=0\,\).

A particle \(P\) moves so that, at time \(t\), its displacement \( \bf r \) from a fixed origin is given by \[ {\bf r} =\left( \e^{t}\cos t \right) {\bf i}+ \left(\e^t \sin t\right) {\bf j}\,.\] Show that the velocity of the particle always makes an angle of \(\frac{\pi}{4}\) with the particle's displacement, and that the acceleration of the particle is always perpendicular to its displacement. Sketch the path of the particle for \(0\le t \le \pi\). A second particle \(Q\) moves on the same path, passing through each point on the path a fixed time \(T\) after \(P\) does. Show that the distance between \(P\) and \(Q\) is proportional to \(\e^{t}\).

Show Solution

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Solution: \begin{align*} && {\bf r} &=\left( \e^{t}\cos t \right) {\bf i}+ \left(\e^t \sin t\right) {\bf j} \\ \Rightarrow && \dot{\bf r} &= \left( \e^{t}\cos t -\e^t \sin t\right) {\bf i}+ \left(\e^t \sin t+\e^t \cos t\right) {\bf j} \\ \Rightarrow && \mathbf{r}\cdot\dot{ \mathbf{r}} &= e^{2t}(\cos^2 t - \sin t \cos t) + e^{2t}(\sin^2 t+ \sin t \cos t) \\ &&&= e^{2t} (\cos^2 t + \sin ^2 t)\\ &&&= e^{2t} \\ \\ && | {\bf r}| &= e^{t} \\ && |{\bf \dot{r}}| &= e^t \sqrt{(\cos t - \sin t)^2 + (\sin t + \cos t)^2} \\ &&&= e^t \sqrt{2 \cos^2 t + 2 \sin^2 t} \\ &&&= \sqrt{2} e^t \\ \\ \Rightarrow && \frac{\mathbf{r}\cdot\dot{ \mathbf{r}}}{ |{\bf {r}}| |{\bf \dot{r}}|} &= \frac{e^{2t}}{\sqrt{2}e^te^t} \\ &&&= \frac{1}{\sqrt{2}} \end{align*} Therefore the angle between the velocity and displacement is \(\frac{\pi}{4}\). \begin{align*} && \ddot{\bf{r}} &= \left( \e^{t}(\cos t - \sin t) - \e^t (\sin t + \cos t)\right) {\bf i}+ \left(\e^t (\sin t + \cos t) + \e^t(\cos t - \sin t)\right) {\bf j} \\ &&&= \left ( -2\e^{t} \sin t \right) {\bf i}+ \left ( 2\e^{t} \cos t \right) {\bf j} \\ \Rightarrow && {\bf r} \cdot \ddot{\bf{r}} &= 2e^{2t} \left ( -\sin t \cos t + \sin t \cos t \right) \\ &&&= 0 \end{align*} Therefore the acceleration is perpendicular.

+ - ↺

\(Q\) has position $\mathbf{r}' = \left( \e^{t-T}\cos (t-T) \right) {\bf i}+ \left(\e^{t-T} \sin (t-T)\right) {\bf j}\( for \)t > T$. \begin{align*} && {\bf r' \cdot r} &= e^{2t-T} \left (\cos t \cos (t-T) + \sin t \sin(t - T) \right) \\ &&&= e^{2t-T} \cos (t - (t-T)) \\ &&&= e^{2t-T} \cos T \\ \\ && |{\bf r'}- {\bf r} |^2 &= |{\bf r}|^2 + |{\bf r}'|^2 - 2 {\bf r' \cdot r} \\ &&&= e^{2t} + e^{2(t-T)} - 2e^{2t-T} \cos T \\ &&&= e^{2t} \left (1 - 2e^{-T} \cos T + e^{-2T} \right) \\ \Rightarrow && |{\bf r'}- {\bf r} | &= e^{t} \sqrt{1 - 2e^{-T} \cos T + e^{-2T} } \end{align*} as required