STEP Maths I,II,III 1987 Solutions

(I've posted some solutions in the 1992 thread if you care)
STEP III Q11

Call the position of the mother A, the point where she enters the river B, and the position of the child C.
$AB=\sqrt{a^2+(b-x)^2}[br]BC=\sqrt{x^2+c^2}$
So, total time $T=\frac{\sqrt{a^2+(b-x)^2}}{u}+\frac{\sqrt{x^2+c^2}}{v}$
Differentiating, $T'=\frac{x-b}{u\sqrt{a^2+(b-x)^2}}+\frac{x}{v\sqrt{x^2+c^2}}$
set the derivative to zero to find the minimum time. x must satisfy that.
For the next part, $AB$ is the same but the velocity in $BC$ has an extra $v$ in it's horizontal component. Think of it's velocity in terms of her swimming vector incllned at an angle $\theta$ and then add on the horizontal component due to the river at the end. Because she can only swim in straight lines:
$\frac{x}{c}=\frac{vcos\theta +v}{vsin\theta}$. Using the identity:
and rearranging we get:
$(c^2-x^2)sec^2\theta+2c^2sec\theta+c^2+x^2=0$ which is a quadratic in $sec\theta$, so, using the quadratic formula:
$sec\theta=\frac{-2c^2\pm\sqrt{4c^4-4(c^2-x^2)(c^2+x^2)}}{2(c^2-x^2)}=\frac{-c^2 \pm x^2}{c^2-x^2}$. We can discard the 'plus' root as it gives theta to be $\pi$ which makes no sense because the mother would never reach the child if she had no vertical component in her velocity.So:
$sec\theta=\frac{-c^2 - x^2}{c^2-x^2} \Rightarrow cos\theta=\frac{c^2-x^2}{-c^2-x^2}$
To find the speed for $BC$ we use pythagoras:
$speed_{BC}=\sqrt{v^2sin^2\theta+(vcos\theta+v)^2}=\sqrt{v^2(2cos \theta+2)}$, substituting in and simplifying:
$speed_{BC}=\frac{2vx}{\sqrt{c^2+x^2}}$
so, just like in the previous part:
$T=\frac{\sqrt{a^2+(b-x)^2}}{u}+\frac{\sqrt{c^2+x^2}}{\frac{2vx}{\sqrt{c^2+x^2}}}= \frac{\sqrt{a^2+(b-x)^2}}{u}+\frac{c^2+x^2}{2vx}$ and differentiating we get:
$T'= \frac{x-b}{u\sqrt{a^2+(b-x)^2}}+\frac{4vx^2-2vx^2-2vc^2}{4v^2x^2}= \frac{x-b}{u\sqrt{a^2+(b-x)^2}}+\frac{x^2-c^2}{2vx^2}$
letting $T'=0$ to find minima and rearranging:
$\frac{b-x}{u\sqrt{a^2+(b-x)^2}}= \frac{x^2-c^2}{2vx^2} \Rightarrow 2vx^2(b-x)=u(x^2-c^2)[a^2+(b-x)^2]^{1/2}$ as required.
Can't really upload the sketches so if anyone wants to do it be my guest

198987 STEP Fma numbers 8,9 and 10

STEP III, Q6.

$x' + 2x - 5y = 0 \Rightarrow x'' + 2x' - 5y' = 0$

$y' = 2\cos t + 2y - ax$

$x'' + 2x' - 10\cos t - 10y + 5ax = 0$

$5y = x' + 2x$

$x'' + 2x' - 10\cos t - 2x' - 4x + 5ax = 0$

$x'' + (5a - 4)x = 10\cos t$

$a = 1 \Rightarrow x'' + x = 10\cos t$

CF: $\lambda^{2} + 1 = 0 \Rightarrow \lambda = \pm i$

$x = A\cos t + B\sin t$

PI: Normally we would suggest a solution of the form $P\cos t + Q\sin t$ However, both of these are part of the CF, so instead suppose $x = Pt\cos t + Qt\sin t$

$x' = P\cos t - Pt\sin t + Q\sin t + Qt\cos t$
$x'' = -P\sin t - P\sin t - Pt\cos t + Q\cos t + Q\cos t - Qt\cos t$

Putting these into the equation we get

$-2P = 0, \; 2Q= 10 \Rightarrow P = 0, \; Q=5$

Therefore the solution is $x = A\cos t + (B + 5t)\sin t$

Using the initial conditions, we get $A = 0, \; B=-5$

Therefore the particular solution is $x = 5(t - 1)\sin t, \; y=t(2\sin t + \cos t) - (\sin t + \cos t)$

If a > 1, the complementary function would be of the form $A\cos(\alpha t) + B\sin(\alpha t), \; \alpha > 1$. Therefore, the particular integral would not have to include t, and the solutions for x and y would also not include it.

STEP I, Q2: (spoilered as people seem to be still working on it...)

STEP I, question 4.

Let $S = \mathrm{log}_2 e - \mathrm{log}_4 e + \mathrm{log}_{16} e...$

Take each term to logarithm base 2 (change the base rule)

$\displaystyle S = \frac{\mathrm{log}_2 e}{1} - \frac{\mathrm{log}_2 e}{2} + \frac{\mathrm{log}_2 e}{4}...$



Split the second brackets into two parts.

P = 1 + 1/4 + 1/16...
N = -1/2 - 1/8 - 1/32...

So



P is a geometric series with first term 1 and common ratio 1/4. Its sum to infinity is:

$P = 1/(1-1/4) = 4/3$

N is a geometric series with first term -1/2 and common ratio 1/4. Its sum to infinity is:

$N = (-1/2)/(1-1/4) = -2/3$





Change the base to e, so we have the natural logarithm.


$\displaystyle S = \frac{2}{3\mathrm{ln} 2}$

Using the power rule

$\displaystyle S = \frac{2}{3\mathrm{ln} 2} = \frac{2}{\mathrm{ln} 8}$

$\displaystyle S = \frac{2*\frac{1}{2}}{\frac{1}{2}\mathrm{ln} 8}$ (multiplying top and bottom by 1/2)

$\displaystyle S = \frac{1}{\frac{1}{2}\mathrm{ln} 8} = \frac{1}{\mathrm{ln} \sqrt 8}$

$\displaystyle S = \frac{1}{\mathrm{ln} 2\sqrt 2}$

WWWWW.