Problems

Given that \(y = \ln ( x + \sqrt{x^2 + 1})\), show that \( \displaystyle \frac{\d y}{\d x} = \frac1 {\sqrt{x^2 + 1} }\;\). Prove by induction that, for \(n \ge 0\,\), \[ \l x^2 + 1 \r y^{\l n + 2 \r} + \l 2n + 1 \r x y^{\l n + 1 \r} + n^2 y^{\l n \r} = 0\;, \] where \(\displaystyle y^{(n)} = \frac{\d^n y}{\d x^n}\) and \(y^{(0)} =y\,\). Using this result in the case \(x = 0\,\), or otherwise, show that the Maclaurin series for \(y\) begins \[ x - {x^3 \over 6} +{3 x^5 \over 40} \] and find the next non-zero term.

Show that \( \cosh^{-1} x = \ln ( x + \sqrt{x^2-1})\). Show that the area of the region defined by the inequalities \(\displaystyle y^2 \ge x^2-8\) and \(\displaystyle x^2\ge 25y^2 -16 \) is \((72/5) \ln 2\).

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Solution: \begin{align*} && x &= \cosh y \\ \Rightarrow && x &= \tfrac12 (e^y + e^{-y} ) \\ \Rightarrow && 0 &= e^{2y} - 2xe^y + 1 \\ \Rightarrow && e^y &= \frac{2x \pm \sqrt{4x^2-4}}{2} \\ &&&= x \pm \sqrt{x^2-1} \\ \Rightarrow &&e^y &= x + \sqrt{x^2-1} \tag{by convention \(\cosh^{-1} > 0\)} \\ \Rightarrow && y &= \ln (x + \sqrt{x^2-1}) \end{align*}

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\begin{align*} && A &= 4 \left ( \int_0^3 \frac15\sqrt{16+x^2} \d x - \int_{2\sqrt{2}}^3\sqrt{x^2-8} \d x \right) \\ \\ x = 4\sinh u: && \int_0^3 \sqrt{4^2+x^2} \d x &= \int_{u=0}^{u=\sinh^{-1}(3/4)} \sqrt{4^2 (1+\sinh^2 u)} 4 \cosh u \d u \\ &&&= \int_0^{\sinh^{-1}(3/4)} 16 \cosh^{2}u \d u \\ &&&= 8\int_0^{\sinh^{-1}(3/4)} (1+\cosh 2u) \d u \\ &&&= 8 \left[u + \frac12 \sinh 2u\right]_0^{\sinh^{-1}(3/4)} \\ &&&= 8 \left (\sinh^{-1}(3/4) + \frac12 \sinh \left ( 2 \sinh^{-1}(3/4) \right) \right) \\ \\ && \sinh^{-1}(3/4) &= \ln\left ( \frac34 + \sqrt{\left ( \frac{3}{4} \right)^2 + 1} \right) \\ &&&= \ln \left ( \frac34 +\frac{5}{4} \right) \\ &&&= \ln 2 \\ \\ \Rightarrow && \int_0^3 \sqrt{4^2+x^2} \d x &= 8 \ln 2 + 4 \left ( \frac{e^{2 \ln 2} - e^{-2\ln2}}{2} \right) \\ &&&= 8 \ln 2 + 2 \cdot 4 - 2\cdot \frac{1}{4} \\ &&&= 8 \ln 2 + \frac{15}{2} \end{align*} \begin{align*} x = 2\sqrt{2} \cosh u: && \int_{2\sqrt{2}}^3\sqrt{x^2-8} \d x &= \int_{u=0}^{u = \cosh^{-1} \frac{3}{2\sqrt{2}}} \sqrt{8(\cosh^2 u - 1)} 2 \sqrt{2} \sinh u \d u \\ &&&= \int_0^{ \cosh^{-1} \frac{3}{2\sqrt{2}}} 8\sinh^2 u \d u \\ &&&= 4 \int_0^{ \cosh^{-1} \frac{3}{2\sqrt{2}}} 2\sinh^2 u \d u \\ &&&= 4 \int_0^{ \cosh^{-1} \frac{3}{2\sqrt{2}}} \cosh 2 u -1 \d u \\ &&&= 4 \left [\frac12 \sinh 2u - u \right]_0^{ \cosh^{-1} \frac{3}{2\sqrt{2}}} \\ \\ && \cosh^{-1} \frac{3}{2\sqrt{2}} &= \ln \left ( \frac{3}{2\sqrt{2}} + \sqrt{\left ( \frac{3}{2\sqrt{2}} \right)^2-1} \right) \\ &&&= \ln \left ( \frac{3}{2\sqrt{2}} + \sqrt{\frac{9}{8} - 1} \right) \\ &&&= \ln \left ( \frac{3}{2\sqrt{2}} + \sqrt{\frac{1}{8} } \right) \\ &&&= \ln \frac{4}{2\sqrt{2}} \\ &&&= \frac12 \ln 2 \\ \\ && \int_{2\sqrt{2}}^3\sqrt{x^2-8} \d x &= 4 \left ( \frac12 \frac{e^{\ln2} - e^{-\ln2}}{2} - \frac12 \ln 2\right) \\ &&&= 2 - \frac12 -2 \ln 2 \\ &&&= \frac32 - 2 \ln 2 \end{align*} \begin{align*} A &= 4 \left (\frac15\left(8\ln 2 + \frac{15}2 \right)- \left ( \frac32 - 2 \ln 2\right)\right) \\ &=4\cdot \left( \frac{8}{5} + 2 \right) \ln 2 \\ &= \frac{72}{5} \ln 2 \end{align*}

Consider the equation \[ x^2 - b x + c = 0 \;, \] where \(b\) and \(c\) are real numbers.

1. Show that the roots of the equation are real and positive if and only if \(b>0\) and \phantom{} \(b^2\ge4c>0\), and sketch the region of the \(b\,\)-\(c\) plane in which these conditions hold.
2. Sketch the region of the \(b\,\)-\(c\) plane in which the roots of the equation are real and less than \(1\) in magnitude.

In this question, the function \(\sin^{-1}\) is defined to have domain \( -1\le x \le 1\) and range \linebreak \( - \frac{1}{2}\pi \le x \le \frac{1}{2}\pi\) and the function \(\tan^{-1}\) is defined to have the real numbers as its domain and range \( - \frac{1}{2}\pi < x < \frac{1}{2}\pi\).

1. Let $$ \g(x) = \displaystyle {2x \over 1 + x^2}\;, \ \ \ \ \ \ \ \ \ \ -\infty
2. Let \[ \displaystyle \f \l x \r = \sin^{-1} \l {2x \over 1 + x^2} \r \;,\ \ \ \ \ \ \ \ \ -\infty < x < \infty\;. \] Show that $ \f(x ) = 2 \tan^{-1} x\( for \) -1 \le x \le 1\,\( and \)\f(x) = \pi - 2 \tan^{-1} x \( for \)x\ge1\,$. Sketch the graph of \(\f(x)\).

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.

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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*}

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The plane \[ {x \over a} + {y \over b} +{z \over c} = 1 \] meets the co-ordinate axes at the points \(A\), \(B\) and \(C\,\). The point \(M\) has coordinates \(\left( \frac12 a, \frac12 b, \frac 12 c \right)\) and \(O\) is the origin. Show that \(OM\) meets the plane at the centroid \(\left( \frac13 a, \frac13 b, \frac 13 c \right)\) of triangle \(ABC\). Show also that the perpendiculars to the plane from \(O\) and from \(M\) meet the plane at the orthocentre and at the circumcentre of triangle \(ABC\) respectively. Hence prove that the centroid of a triangle lies on the line segment joining its orthocentre and circumcentre, and that it divides this line segment in the ratio \(2 : 1\,\). [The orthocentre of a triangle is the point at which the three altitudes intersect; the circumcentre of a triangle is the point equidistant from the three vertices.]

Sketch the graph of the function \(\ln x - {1 \over 2} x^2\). Show that the differential equation \[ {\mathrm{d} y \over \mathrm{d} x} = {2xy \over x^2 - 1} \] describes a family of parabolas each of which passes through the points \((1,0)\) and \((-1,0)\) and has its vertex on the \(y\)-axis. Hence find the equation of the curve that passes through the point \((1,1)\) and intersects each of the above parabolas orthogonally. Sketch this curve. [Two curves intersect orthogonally if their tangents at the point of intersection are perpendicular.]

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Solution:

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\begin{align*} && y' &= \frac{2xy}{x^2-1} \\ \Rightarrow && \int \frac{1}{y} \d y &= \int \frac{2x}{x^2-1} \d x \\ \Rightarrow && \ln |y| &= \ln |x^2-1| + C \\ \Rightarrow && y &= A(x^2-1) \end{align*} which is a family of parabolas each passing through \((\pm1, 0)\) and with a vertex on the \(y\)-axis. The curve we seek must satisfy \begin{align*} && y' &= \frac{1-x^2}{2xy} \\ \Rightarrow && \int2 y \d y &= \int \left ( \frac{1}{x} - x \right) \d x \\ \Rightarrow && y^2 &= \ln x - \tfrac12 x^2 + C \\ (1,1): && 1 &= -\tfrac12+C \\ \Rightarrow && C &= \frac32 \\ \Rightarrow && y^2 &= \tfrac32 + \ln x - \tfrac12 x^2 \end{align*}

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1. Prove that the equations $$ \left|z - (1 + \mathrm{i}) \right|^2 = 2 \eqno(*) $$ and $$ \qquad \quad \ \left|z - (1 - \mathrm{i}) \right|^2 = 2 \left|z - 1 \right|^2 $$ describe the same locus in the complex \(z\)--plane. Sketch this locus.
2. Prove that the equation $$ \arg \l {z - 2 \over z} \r = {\pi \over 4} \eqno(**) $$ describes part of this same locus, and show on your sketch which part.
3. The complex number \(w\) is related to \(z\) by \[ w = {2 \over z}\;. \] Determine the locus produced in the complex \(w\)--plane if \(z\) satisfies \((*)\). Sketch this locus and indicate the part of this locus that corresponds to \((**)\).

\(B_1\) and \(B_2\) are parallel, thin, horizontal fixed beams. \(B_1\) is a vertical distance \(d \sin \alpha \) above \(B_2\), and a horizontal distance \(d\cos\alpha \) from \(B_2\,\), where \(0<\alpha<\pi/2\,\). A long heavy plank is held so that it rests on the two beams, perpendicular to each, with its centre of gravity at \(B_1\,\). The coefficients of friction between the plank and \(B_1\) and \(B_2\) are \(\mu_1\) and \(\mu_2\,\), respectively, where \(\mu_1<\mu_2\) and \(\mu_1+\mu_2=2\tan\alpha\,\). The plank is released and slips over the beams experiencing a force of resistance from each beam equal to the limiting frictional force (i.e. the product of the appropriate coefficient of friction and the normal reaction). Show that it will come to rest with its centre of gravity over \(B_2\) in a time \[ \pi \left(\frac{d}{g(\mu_2-\mu_1)\cos\alpha }\right)^{\!\frac12}\;. \]

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Solution:

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\begin{align*} \overset{\curvearrowright}{B_2} : && mgx\cos \alpha - R_1d &= 0 \\ && \frac{mgx \cos \alpha}{d} &= R_1 \\ \overset{\curvearrowright}{B_1} : && -mg(d-x)\cos \alpha + R_2d &= 0 \\ && \frac{mg(d-x) \cos \alpha}{d} &= R_2 \\ % \text{N2}(\perp B_1B_2): && R_1 + R_2 - mg\cos \alpha &=0 \\ \text{N2}(\parallel B_1B_2): && mg\sin \alpha - \mu_1R_1 - \mu_2R_2 &= m\ddot{x} \\ && mg \sin \alpha - \mu_1 \frac{mgx \cos \alpha}{d} - \mu_2\frac{mg(d-x) \cos \alpha}{d} &= m \ddot{x} \\ && gd \sin \alpha - \mu_2 gd \cos \alpha - (\mu_1 - \mu_2) x g \cos \alpha &= d \ddot{x} \\ && gd \frac12 \l \mu_1 + \mu_2 \r \cos \alpha - \mu_2 gd \cos \alpha - (\mu_1 - \mu_2) x g \cos \alpha &= d \ddot{x} \\ && gd \frac12 \l \mu_1 - \mu_2 \r \cos \alpha - (\mu_1 - \mu_2) x g \cos \alpha &= d \ddot{x} \\ && \frac12 d C &= d \ddot{x} + Cx \\ && \Big ( C &= g(\mu_1 - \mu_2) \cos \alpha \Big ) \\ \end{align*} We can recognise this differential equation from SHM as having the solution: \[x = A\sin \l \l \frac{d}{C} \r^{\frac12} t \r + B\cos \l \l \frac{d}{C} \r^{\frac12} t \r + \frac12 d\] Since when \(t = 0, x = d, \dot{x} = 0, A = 0, B = \frac{1}{2}d\). We will reach \(B_2, (x = 0)\) when \(\cos \l \l \frac{d}{C} \r^{\frac12} T \r = -1\) (at which point the speed will be zero) and \begin{align*} && \l \frac{d}{C} \r^{\frac12} T &= \pi \\ \Rightarrow && T&= \pi \l \frac{d}{g(\mu_1 - \mu_2) \cos \alpha} \r^{\frac12} \end{align*}

Three ships \(A\), \(B\) and \(C\) move with velocities \({\bf v}_1\), \({\bf v}_2\) and \(\bf u\) respectively. The velocities of \(A\) and \(B\) relative to \(C\) are equal in magnitude and perpendicular. Write down conditions that \(\bf u\), \({\bf v}_1\) and \({\bf v}_2\) must satisfy and show that \[ \left| {\bf u} -{\textstyle\frac12} \l {\bf v}_1 + {\bf v}_2 \r \right|^2 = \left|{\textstyle\frac12} \l {\bf v}_1 - {\bf v}_2 \r \right|^2 \] and \[ \l {\bf u} -{\textstyle\frac12} \l {\bf v}_1 + {\bf v}_2 \r \r \cdot \l {\bf v}_1 - {\bf v}_2 \r = 0 \;. \] Explain why these equations determine, for given \({\bf v}_1\) and \({\bf v}_2\), two possible velocities for \(C\,\), provided \({\bf v}_1 \ne {\bf v}_2 \,\). If \({\bf v}_1\) and \({\bf v}_2\) are equal in magnitude and perpendicular, show that if \({\bf u} \ne {\bf 0}\) then \({\bf u} = {\bf v}_1 + {\bf v}_2\,\).