Show that \((a+b)^2\le 2a^2+2b^2\,\).
Find the stationary points on the curve
$y=\big(a^2\cos^2\theta +b^2\sin^2\theta\big)^{\frac12}
+ \big(a^2\sin^2\theta +b^2\cos^2\theta\big)^{\frac12}\,$, where
\(a\) and \(b\) are constants. State, with brief reasons, which points are maxima and which are minima. Hence prove that
\[
\vert a\vert +\vert b \vert
\le \big(a^2\cos^2\theta +b^2\sin^2\theta\big)^{\frac12}
+ \big(a^2\sin^2\theta +b^2\cos^2\theta\big)^{\frac12}
\le \big(2a^2+2b^2\big)^{\frac12} \;.
\]
And so \(\frac{\pi}{4}, \frac{3\pi}{4}, \cdots\) are maxima, and the others minima. The maxima are where \(\sin^2 \theta = \cos^2 \theta = \frac12\), so \(y(\frac{\pi}{4}) = 2\left ( \frac{a^2+b^2}{2} \right)^{\frac12} = (2a^2+2b^2)^{\frac12}\) and the maxima are \(\cos^2 \theta = 1, \sin^2 \theta = 0\) and vice versa, ie \(y = |a| + |b|\), therefore we obtain our desired result.
Deduce that, when \(n\) is large,
\[
\sum_{k=0}^n \sin \left(\frac{k\pi}{n}\right) \approx \frac{2n}\pi\;.
\]
By differentiating \((*)\) with respect to \(\theta\), or otherwise,
show that, when \(n\) is large,
\[
\sum_{k=0}^n k \sin^2 \left(\frac{k\pi}{2n}\right)
\approx \left(\frac{1}4 +\frac{1}{\pi^2}
\right)n^2\;.
\]
Let
$$
{\rm f}(x)=\sin^2x + 2 \cos x + 1
$$
for \(0 \le x \le 2\pi\). Sketch the curve \(y={\rm f}(x)\), giving
the coordinates of the stationary points. Now let
$$
\hspace{0.6in}{\rm g}(x)={a{\rm f}(x)+b \over c{\rm f}(x)+d}
\hspace{0.8in}
ad\neq bc\,,\; d\neq -3c\,,\;
d\neq c\;.
$$
Show that the stationary points of \(y={\rm g}(x)\) occur at the same
values of \(x\) as those of \(y={\rm f}(x)\), and find the corresponding
values of \({\rm g}(x)\).
Explain why, if \(d/c <-3\) or \(d/c>1\), \(|{\rm g}(x)|\)
cannot be arbitrarily large.
By considering the maximum of \(\ln x-x\ln a\), or otherwise,
show that the equation
\(x=a^{x}\) has no real roots if \(a > e^{1/e}\).
How many real roots does the equation have if
\(0 < a < 1\)? Justify your answer.
\begin{align*}
&& y &= \ln x - x \ln a \\
\Rightarrow && y' &= \frac1x - \ln a \\
&& y'' &= -\frac{1}{x^2}
\end{align*}
Therefore the maximum is when \(x = \frac{1}{\ln a}\) and \(y_{max} = -\ln \ln a - 1\). If \(y_{max} < 0\) then \(y \neq 0\).
But that's equivalent to \(a > e^{1/e}\).
\begin{align*}
&& 0 &> -\ln \ln a - 1 \\
\Leftrightarrow && 1 &> - \ln \ln a \\
\Leftrightarrow && \ln \ln a &>-1 \\
\Leftrightarrow && \ln a &> e^{-1} \\
\Leftrightarrow && a & > e^{1/e}
\end{align*}
If \(0 < a < 1\) then, when \(x\) is small, \(\ln x - x \ln a\) is large and negative. When \(x\) is large and positive \(\ln x\) is positive and \(-x \ln a\) is positive. We also notice there is no turning point. Hence exactly one solution
A cylindrical biscuit tin has volume \(V\) and surface area \(S\) (including the ends). Show that the minimum possible surface area for a given value of \(V\) is \(S=3(2\pi V^{2})^{1/3}.\) For this value of \(S\) show that the volume of the largest sphere which can fit inside the tin is \(\frac{2}{3}V\), and find the volume of the smallest sphere into which the tin fits.
Solve the differential equation
\[
\frac{\mathrm{d}y}{\mathrm{d}x}-y-3y^{2}=-2
\]
by making the substitution \(y=-\dfrac{1}{3u}\dfrac{\mathrm{d}u}{\mathrm{d}x}.\)
Solve the differential equation
\[
x^{2}\frac{\mathrm{d}y}{\mathrm{d}x}+xy+x^{2}y^{2}=1
\]
by making the substitution
\[
y=\frac{1}{x}+\frac{1}{v},
\]
where \(v\) is a function of \(x\).
When the water level is \(x\) cm above the vertex, water leaves the
cone at a rate \(Ax\) \(\mathrm{cm}^{3}\mathrm{sec}^{-1},\) where \(A\)
is a positive constant. Suppose that the cone is initially filled
to a height \(h\) cm with \(0 < h < H.\) Show that it will take \(\pi a^{2}h^{2}/(2AH^{2})\)
seconds to empty.
Suppose now that the cone is initially filled to a height \(h\) cm,
but that water is poured in at a constant rate \(B\) \(\mathrm{cm}^{3}\mathrm{sec}^{-1}\)
and continues to drain as before. Establish, by considering the sign
of \(\mathrm{d}x/\mathrm{d}t\), or otherwise, what will happen subsequently
to the water level in the different cases that arise. (You are not
asked to find an explicit formula for \(x\).)
Frosty the snowman is made from two uniform spherical snowballs, of initial radii \(2R\) and \(3R.\) The smaller (which is his head) stands on top of the larger. As each snowball melts, its volume decreases at a rate which is directly proportional to its surface area, the constant of proportionality being the same for both snowballs. During melting each snowball remains spherical and uniform. When Frosty is half his initial height, find the ratio of his volume to his initial volume.
If \(V\) and \(S\) denote his total volume and surface area respectively, find the maximum value of \(\dfrac{\mathrm{d}V}{\mathrm{d}S}\) up to the moment when his head disappears.
The above diagram is a plan of a prison compound. The outer square \(ABCD\) represents the walls of the compound (whose height may be neglected), while the inner square \(XYZT\) is the Black Tower, a solid stone structure. A guard patrols along segment \(AE\) of the walls, for a distance of up to 4 units from \(A\). Determine the distance from \(A\) of points at which the area of the courtyard that he can see is
as small as possible,
as large as possible.
[\(Hint. \)It is suggested that you express the area he \textit{cannot
}see in terms of \(p\), his distance from \(A\).]
Show Solution
The area he cannot see is
\begin{align*}
&&A &= \underbrace{8^2}_{\text{everywhere above}(4,4)} - \underbrace{4^2}_{\text{inner square}} - \underbrace{\frac12 \cdot 4 \cdot (\frac32(4-p)+p - 4)}_{\text{blue triangle}} - \underbrace{\frac12 \cdot 4 \cdot \frac{4(12-p)}{8-p}}_{\text{green triangle}} \\
&&&= 48 - 3(4-p)-2(p-4) - \frac{8(12-p)}{8-p} \\
&&&= 36-5p-\frac{32}{8-p} \\
\\
\Rightarrow && \frac{\d A}{\d p} &= -5 + \frac{32}{(8-p)^2} \\
&&&> 0 \text{ if } 0 \leq p \leq 4
\end{align*}
Since the area not visible is increasing as \(p\) increases, we would like \(p\) to be as large as possible, ie \(p = 4\).
According to the Institute of Economic Modelling Sciences, the Slakan
economy has alternate years of growth and decline, as in the following
model. The number \(V\) of vloskan (the unit of currency) in the Slakan
Treasury is assumed to behave as a continuous variable, as follows.
In a year of growth it increases continuously at an annual rate \(aV_{0}\left(1+(V/V_{0})\right)^{2}.\)
During a year of decline, as long as there is still money in the Treasury,
the amount decreases continuously at an annual rate \(bV_{0}\left(1+(V/V_{0})\right)^{2};\)
but if \(V\) becomes zero, it remains zero until the end of the year.
Here \(a,b\) and \(V_{0}\) are positive constants. A year of growth
has just begun and there are \(k_{0}V_{0}\) vloskan in the Treasury,
where \(0\leqslant k_{0} < a^{-1}-1\). Explain the significance of these
inequalities for the model to be remotely sensible.
If \(k_{0}\) is as above and at the end of one year there are \(k_{1}V_{0}\)
vloskan in the Treasury, where \(k_{1} > 0\), find the condition involving
\(b\) which \(k_{1}\) must satisfy so that there will be some vloskan
left after a further year. Under what condition (involving \(a,b\)
and \(k_{0}\)) does the model predict that unlimited growth will take
place in the third year (but not before)?
Prove that, for any integers \(n\) and \(r\), with \(1\leqslant r\leqslant n,\)
\[
\binom{n}{r}+\binom{n}{r-1}=\binom{n+1}{r}.
\]
Hence or otherwise, prove that
\[
(uv)^{(n)}=u^{(n)}v+\binom{n}{1}u^{(n-1)}v^{(1)}+\binom{n}{2}u^{(n-2)}v^{(2)}+\cdots+uv^{(n)},
\]
where \(u\) and \(v\) are functions of \(x\) and \(z^{(r)}\) means \(\dfrac{\mathrm{d}^{r}z}{\mathrm{d}x^{r}}\).
Prove that, if \(y=\sin^{-1}x,\) then \((1-x^{2})y^{(n+2)}-(2n+1)xy^{(n+1)}-n^{2}y^{(n)}=0.\)
The function \(\mathrm{f}\) and \(\mathrm{g}\) are related (for all real \(x\)) by
\[
\mathrm{g}(x)=\mathrm{f}(x)+\frac{1}{\mathrm{f}(x)}\,.
\]
Express \(\mathrm{g}'(x)\) and \(\mathrm{g}''(x)\) in terms of \(\mathrm{f}(x)\) and its derivatives.
If \(\mathrm{f}(x)=4+\cos2x+2\sin x\), find the stationary points of \(\mathrm{g}\) for \(0\leqslant x\leqslant2\pi,\) and determine which are maxima and which are minima.
Let \(\mathrm{f}(x)=\sin2x\cos x.\) Find the 1988th derivative of \(\mathrm{f}(x).\)
Show that the smallest positive value of \(x\) for which this derivative
is zero is \(\frac{1}{3}\pi+\epsilon,\) where \(\epsilon\) is approximately
equal to
\[
\frac{3^{-1988}\sqrt{3}}{2}.
\]
Find the stationary points of the function \(\mathrm{f}\) given by
\[
\mathrm{f}(x)=\mathrm{e}^{ax}\cos bx,\mbox{ }(a>0,b>0).
\]
Show that the values of \(\mathrm{f}\) at the stationary points with \(x>0\) form a geometric progression with common ratio \(-\mathrm{e}^{a\pi/b}\).
Give a rough sketch of the graph of \(\mathrm{f}\).
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 \(b< c\).
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:
If \(b > c\) then she should run a little downstream first.
and if \(c > b\) she should actually run a little upstream to take advantage of the current:
