A particle \(P\) is projected from a point \(O\) on horizontal ground with speed \(u\) and angle of projection \(\alpha\), where \(0 < \alpha < \frac{1}{2}\pi\).
Show that if \(\sin \alpha < \frac{2\sqrt{2}}{3}\), then the distance \(OP\) is increasing throughout the flight.
Show also that if \(\sin \alpha > \frac{2\sqrt{2}}{3}\), then \(OP\) will be decreasing at some time before the particle lands.
At the same time as \(P\) is projected, a particle \(Q\) is projected horizontally from \(O\) with speed \(v\) along the ground in the opposite direction from the trajectory of \(P\). The ground is smooth. Show that if
$$2\sqrt{2}v > (\sin \alpha - 2\sqrt{2} \cos \alpha)u,$$
then \(QP\) is increasing throughout the flight of \(P\).
Solution:
Notice that \(P = \begin{pmatrix} u \cos \alpha t\\ u \sin \alpha t - \frac12 g t^2 \end{pmatrix}\), so
\begin{align*}
&& |OP|^2 &= u^2 \cos^2 \alpha t^2 + \left (u \sin \alpha t - \frac12 g t^2 \right)^2 \\
&&&= u^2 \cos^2 \alpha t^2 +u^2 \sin^2 \alpha t^2 - u \sin \alpha g t^3 +\frac14 g^2 t^4 \\
&&&= u^2 t^2 -u\sin \alpha g t^3 + \frac14g^2t^4 \\
&& \frac{\d |OP|^2}{\d t} &= 2u^2 t - 3u \sin \alpha g t^2+g^2 t^3 \\
&&&= t \left (2u^2 - 3u \sin \alpha (gt)+(gt)^2\right) \\
&& \Delta &= 9u^2 \sin^2 \alpha -4 \cdot 2u^2 \cdot 1 \\
&&&= u^2 (9\sin^2 \alpha -8) \\
\end{align*}
Therefore if \(\sin \alpha < \frac{2\sqrt{2}}3\) the discriminant is negative, the quadratic factor is always positive and the distance \(|OP|\) is always increasing.
Similarly, if \(\sin \alpha > \frac{2 \sqrt{2}}3\) then the derivative has a root. This means somewhere on its (possibly extended) trajectory \(OP\) is decreasing. This must be before it lands, since if it were after it 'landed' then both the \(x\) and \(y\) distances are increasing, therefore it cannot occur after it 'lands'.
Note that \(Q = \begin{pmatrix} -v t \\0 \end{pmatrix}\)
\begin{align*}
&& |QP|^2 &= (u \cos \alpha t+vt)^2 + \left (u \sin \alpha t - \frac12 g t^2 \right)^2 \\
&&&= u^2 \cos^2 \alpha t^2+2u\cos \alpha v t^2 + v^2 t^2 +u^2 \sin^2 \alpha t^2 - u \sin \alpha g t^3 +\frac14 g^2 t^4 \\
&&&= (u^2+2u v \cos \alpha+v^2) t^2 - u \sin \alpha g t^3 + \frac14 g^2 t^4 \\
\\
\Rightarrow && \frac{\d |QP|^2}{\d t} &= 2(u^2+u v \cos \alpha+v^2) t - 3u \sin \alpha g t^2 + g^2 t^3 \\
&&&= t \left ( 2(u^2+2u v \cos \alpha+v^2) - 3u \sin \alpha (g t) + (g t)^2\right) \\
&& \Delta &= 9u^2 \sin^2 \alpha - 8(u^2+2u v \cos \alpha+v^2) \\
&&&= (9 \sin^2 \alpha -8)u^2 - 16v \cos \alpha u - 8v^2 \\
&&&= \left (( \sin \alpha-2\sqrt{2}\cos \alpha)u-2\sqrt{2} v \right) \left ( ( \sin \alpha+2\sqrt{2}\cos \alpha)u+2\sqrt{2} v \right)
\end{align*}
Since the second bracket is clearly positive, the first bracket must be negative (for \(\Delta < 0\) and our derivative to be positive), ie \(2\sqrt{2} v > ( \sin \alpha-2\sqrt{2}\cos \alpha)u\)
In this question, you may assume without proof that
any function \(\f\) for which \(\f'(x)\ge 0\) is increasing; that is,
\(\f(x_2)\ge \f(x_1)\) if \(x_2\ge x_1\,\).
Let \(\f(x) =\sin x -x\cos x\).
Show that \(\f(x)\) is
increasing
for \(0\le x \le \frac12\pi\,\)
and deduce that \(\f(x)\ge 0\,\) for
\(0\le x \le \frac12\pi\,\).
Given that \(\dfrac{\d}{\d x} (\arcsin x) \ge1\) for
\(0\le x< 1\),
show that
\[
\arcsin x\ge x
\quad (0\le x < 1).
\]
Let \(\g(x)= x\cosec x\, \text{ for }0< x < \frac12\pi\). Show that
\(\g\) is increasing and deduce that
\[
({\arcsin x})\, x^{-1} \ge x\,{\cosec x}
\quad (0 < x < 1).
\]
Given that $\dfrac{\d}{\d x}
(\arctan x)\le 1\text{ for }x\ge 0$, show
by considering the function \(x^{-1} \tan x\)
that
\[
(\tan x)( \arctan x) \ge x^2
\quad
(0< x < \tfrac12\pi).
\]
Solution:
Given \(\frac{\d}{\d x} (\arctan x) \leq 1\) we must have \(\frac{\d}{ \d x} (x-\arctan x) \geq 0\) for \(x \geq 0\), but since \( 0 - \arctan 0 = 0\) this means that \(x - \arctan x \geq 0\), ie \( \arctan x \geq x\) for \(x \geq 0\)
\(g(x) = x^{-1} \tan x \Rightarrow g'(x) = -x^{-2}\tan x +x^{-1} \sec^2 x\). If we can show \(f(x) = x \sec ^2 x - \tan x\) is positive that would be great. However \(f'(x) = x 2 \tan x \sec^2 x \geq 0\) and \(f(0) = 0\) so \(f(x)\) is positive and \(g'(x)\) is positive and hence increasing, therefore \(g(x) \geq g(\arctan x) \Rightarrow \frac{\tan x}{x} \geq \frac{x}{\arctan x}\) from which the result follows.
In the game of endless cricket the scores
\(X\) and \(Y\) of the two sides are such that
\[
\P (X=j,\ Y=k)=\e^{-1}\frac{(j+k)\lambda^{j+k}}{j!k!},\]
for some positive constant \(\lambda\), where \(j,k = 0\), \(1\), \(2\), \(\ldots\).
Find \(\P(X+Y=n)\) for each \(n>0\).
Show that \(2\lambda \e^{2\lambda-1}=1\).
Show that \(2x \e^{2x-1}\) is an increasing function
of \(x\) for \(x>0\) and deduce that the equation in (ii) has
at most one solution and hence determine \(\lambda\).
Consider \(f(x) = 2xe^{2x-1}\), then
\begin{align*}
&& f'(x) &= 2e^{2x-1} + 2xe^{2x-1} \cdot 2 \\
&&&= e^{2x-1} (2 + 4x) > 0
\end{align*}
Therefore \(f(x)\) is an increasing function of \(x\), which means \(f(x) = 1\) has at most one solution for \(\lambda\). Therefore \(\lambda = \frac12\)