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STEP maths I, II, III 1991 solutions

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insparato

I think the problem is, the head could melt slower or faster than the body, which means you could reach half the initial height without the head or body necessarily being exactly half the respective radius.

Exactly.

When i attempted this question, this is what plagued me... I don't think you can assume that the two spheres are going to amount to half the initial height and be half the initial radii.

Form a differential equation for R as a function of time. Solve it for both spheres. Add the answers to get a formula for height as a function of time. Etc...

Reply 181

nota bene

insparato

Ah, of course! Thanks, this should set me off

Reply 182

nota bene

\frac{dV_U}{dt}=k16\pi R^2=k16\pi(\frac{3V_U}{4\pi})^{\frac{2}{3}}

and

\frac{dV_L}{dt}=k36\pi R^2=k36\pi(\frac{3V_L}{4\pi})^{\frac{2}{3}}

Separating variables

\int dt=\int \frac{1}{16k\pi} (\frac{3V_U}{4\pi})^{ -\frac{2}{3}}dV_U \Rightarrow t=\frac{1}{4k}(\frac{V_U}{4\pi})^{\frac{1}{3}}+c_1= \frac{1}{4k}(\frac{R^3}{3})^{\frac{1}{3}}+c_1= \frac{1}{4k \sqrt[3]{3}}R+c_1

Separating variables on the other one

\int dt=\int \frac{1}{36k\pi}(\frac{3V_L}{4\pi})^{-\frac{2}{3}}dV_L \Rightarrow t=\frac{1}{9k}(\frac{3V_L}{4\pi})^{\frac{1}{3}}+c_2=\frac{1}{9k}R+c_2

What do I do with the constants now? They're bugging me.

2\times4k\sqrt[3]{3}(t-c_1)

is the diameter (height) of the upper sphere and

3\times9k(t-c_2)

is the diameter of the lower sphere. So the height is

h(t)=27k(t-c_2)+8k\sqrt[3]{3}(t-c_1)

So 0.5=h(t) but don't see how this helps. I'm giving up, I have never been able to do rates of change:/

Reply 183

DFranklin

I'd do something like:

Consider the rate of evaporation of a sphere of radius r, volume v, surface area s.

v \varpropto r^3

\frac{dv}{dt} \varpropto 3r^2 \frac{dr}{dt} \varpropto r^2 \frac{dr}{dt}

And

s \varpropto r^2, \frac{dv}{dt} \varpropto -s

r^2 \frac{dr}{dt} \varpropto -r^2

, so

\frac{dr}{dt} = -k

for some constant k.

Subbing in for the initial values, we get

R_u(t) = 2R-kt, R_l(t)=3R-kt

.

Now find kt when R_u+R_l = 5R/2, etc...

Reply 184

brianeverit

DFranklin

I'd do something like:

Consider the rate of evaporation of a sphere of radius r, volume v, surface area s.

v \varpropto r^3

\frac{dv}{dt} \varpropto 3r^2 \frac{dr}{dt} \varpropto r^2 \frac{dr}{dt}

And

s \varpropto r^2, \frac{dv}{dt} \varpropto -s

r^2 \frac{dr}{dt} \varpropto -r^2

, so

\frac{dr}{dt} = -k

for some constant k.

Subbing in for the initial values, we get

R_u(t) = 2R-kt, R_l(t)=3R-kt

.

Now find kt when R_u+R_l = 5R/2, etc...

This question is number 7 in Advanced Problems in Mathematics

II/6:

STEP II Q16

Part 1a

Part 1b

2 Part

Reply 187

steppity-step

On the last part of question 15 of STEP II I got

\frac{d(m+1)}{6m}[2N(m-1) + 3m]

instead of

\frac{d(m-1)}{6m}[2N(m+1) + 3m]

. Did anybody else get this? What are the chances they made a typo? I may type up my solution in a bit, but I just typed up 16 and I'm latexed out.

Reply 188

steppity-step

justinsh

I may have done mistakes but here it goes:

\ln(1+x+x^2+x^3)\\

=\ln(1+x)+\ln(1+x^2)\\

=(x-\frac{1}{2}x^2+\frac{1}{3}x^3+...+\frac{(-1)^{n+1}}{n}x^n)+(x^2-\frac{1}{2}x^4+\frac{1}{3}x^6+...+\frac{(-1)^{n+1}}{n}x^{2n})\\

=x+\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{3}{4}x^4+...+\left[\frac{(-1)^{n+1}}{n}-\frac{1}{2n}\right]x^{2n}+\frac{x^{2n+1}}{2n+1}

In my personal opinion, both solutions to STEP III Q1 part b on here fail to find a general term (although this second one is quite nice with two fairly simple general terms), so I am going to post the way I did it, which is quite similar to the above method except that I actually put a general term, rather than a set of general terms. I don't know whether they expect you to do this, but before looking at these I presumed you had to. Anyway, here goes:

ln(1+x+x^2+x^3) = ln(1+x)+ln(1+x^2) \\ \\ = (x-\frac{1}{2}x^2+\frac{1}{3}x^3+...+\frac{(-1)^{n+1}}{n}x^n+...)+(x^2-\frac{1}{2}x^4+\frac{1}{3}x^6+...+\frac{(-1)^{n+1}}{n}x^{2n}+...) \\ \\ = (x-\frac{1}{2}x^2+\frac{1}{3}x^3+...+\frac{(-1)^{n+1}}{n}x^n+...)+(x^2-\frac{1}{2}x^4+\frac{1}{3}x^6+...+\frac{(-1)^{\frac{n}{2}+1}}{n/2}\frac{1+(-1)^n}{2}x^{n}+...) \\ \\ = x+\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{3}{4}x^4+\frac{1}{5}x^5+...+\big[\frac{(-1)^{n+1}}{n}+ \frac{(-1)^{\frac{n}{2}+1}}{n}(1+(-1)^n)\big]x^n+...

It's just that this question would be too easy without this, I think that what they're testing is your ability to think of a way to merge the xⁿ and x²ⁿ terms, and finding a fairly simple function that returns 1 for even n and 0 for odd n.

Reply 189

steppity-step

Rabite

I should hope not. Not sure what I've done to make someone want to neg me though. Guess it's because I sound like a moron or something. Oh wait, I am a moron...

Anyway - I did III Q3. Thanks to Khai for pointing out my stupid errors. :p:

I updated the attachment.

Also, has anyone done the one in STEP III about some particle and two observers? Q4 I think it was...I thought I had the answer, but the direction vector has a 22 in it, so I'm guessing it's wrong.

1. There is a typo early on, you put 11 instead of 1.
2. No minimum at (1, -4) of any sort.
3. But there is a minimum at (2, -5), which is also necessary for the graph.
4. In the last part, you are defining f^-1(x) so the x we are talking about is a different x to before if that makes sense, ie it is what was on the y-axis before not the x-axis. So your table thing doesn't really make sense.
5. If we ignore point 4, the right-most box is wrong anyway, because the positive root is only valid to the right of 2.

So in the last part you need to consider

x<-5,-5 \leq x \leq -4, -4 \leq x \leq 1, 1 \leq x \leq 4, x \geq 4

. The first is undefined, all the others have two possibilities (one of which is the same for all four). It's seriously boring stuff. This question is not particularly challenging, but loads of cases to consider, and loads of room for little errors.

Reply 190

steppity-step

STEP III Q16

1 Part

2 Part

3 Part

4 Part

5 Part

Woooooooo! This has taken me almost all day and I've finally done it! I'm almost certain the answer is right as it works for a=2 and a=3, and everything follows nicely.

Reply 191

Glutamic Acid

This alternative to III/1 part (c) is quite cute.

e^{x \ln(1+x)} = (1+x)^x = 1 + x^2 + \dfrac{x(x-1)}{2!}x^2 + \dfrac{x(x-1)(x-2)}{3!}x^3 + ... = 1 + x^2 - \dfrac{x^3}{2} + \dfrac{5x^4}{6} + ...

from the binomial expansion.

Reply 192

themaths

Original post by Rabite

Hi there just wanted to point out one final thing i noticed when doing this question about the inverse functions is you have to check the functions are still valid in the intervals you gave them:
For instance for -1<x<0

\sqrt\frac{x-1}{3}

is undefined
Also, in the interval x<-1

2- \sqrt{x+5}

is undefined once x<-5 so the interval should be -5<x<-1
and the inverse functions are undefined for the intervals -1<x<0 and x<-5

Reply 193

mikelbird

I think these solve 1991 Step 3 Q4 and Q9

Step1991Paper3Question4.pdf57.8KB

Step1991Paper3Question9.pdf79.3KB

Reply 194

brianeverit

STEP 1991 Paper II question 9

\text {Closure } \begin{pmatrix}a&b\\0&c \end{pmatrix}\begin{pmatrix}x & y \\ 0 & z \end{pmatrix}= \begin{pmatrix}ax&ay+bz\\0&cz \end{pmatrix}

\text{Associativity may be assumed}

\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \text{ is an obvious identity element}

A= \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} \implies A^{-1}= \dfrac {1}{ac} \begin{pmatrix}c & -b \\ 0 & a \end{pmatrix} \text{ and }ac\not=0 \implies \text{ existence of inverses}

\text{The order of G is the number of its elements}

a \text{ and }c \text{ can each take 4 values, 1,2,3, or 4 whilst }b \text{ can take 5, 0,1,2,3, or 4}

\text{so there are }4 \times 4 \times 5=80 \text{ distinct elements so the order of the group is }80

\text{G is not commutative, } \begin{pmatrix} x & y \\ 0 & z \end{pmatrix} \begin{pmatrix} a & b \\ 0 & c \end{pmatrix}= \begin{pmatrix} ax & bx+cy \\ 0 & cz \end{pmatrix} \text{ and } bx+cy \text { need not be the same as }ay+bz

\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \text{ is the only element of order 1, if } \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} \text{ is an element of order 2 then}

\begin{pmatrix} a & b \\ 0 & c \end{pmatrix}\begin{pmatrix} a & b \\ 0 & c \end{pmatrix}= \begin{pmatrix} a2 & ab+bc \\ 0 & c \end{pmatrix}= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\implies a^2 = c^2 =1 \text{ and } b(a+c)=0

\text{possibilities are } b=0,a=\pm 1,c=\pm 1, \text{ or } a+c=0 \implies a=1, c=-1,b=1,2,3 \text{ or }4

\text{or }a=-1,c=1,b=1,2,3 \text{ or }4 \text { giving a total of 12 elements}

\text{so there are 13 elements of order 1 or 2 hence this subset cannot be a subgroup since 13 is not a factor of 80}

Reply 195

brianeverit

\text {STEP 1991 paper II question 10}

An absolute swine to Latex but here goes

\text{Let the sun be S, the pole AB and the shadow AP with origin O}

Unparseable latex formula:

\mathbf {OS} = \begin{pmatrix} R \cos \theta \\ R \sin \theta \\ 0 \end{pmatrix}, \mathbf {OB}= \begin{pmatrix} 0 \\ b \cos \phi \\ b \sin \phi \end{pmatrix} \text{ so } \mathbf {SB} = \begin{pmatrix} -R \cos \theta \\ b \cos \phi-R \sin \theta \\ b \sin \phi \end{pmatrix}}

\text{horizontal plane through A has equation } (\cos \phi )y+( \sin \phi)z=a

Unparseable latex formula:

\text{so, taking equation of SB as }\mathbf r= \begin{pmatrix} 0 \\ b \cos \phi \\ b \sin \phi \end{pmatrix}+ \lambda \begin{pmatrix}-R \cos \theta \\ b \cos \phi-R \sin \theta \\ b \sin \phi \end{pmatrix}}

\text{P is the point where } \cos \phi (b \cos \phi + \lambda( b \cos \phi - R \sin \theta ))+b \sin^2 \phi + \lambda b \sin^2 \phi=a

\implies b+ \lambda b-\lambda R \sin \theta \cos \phi=a \implies \lambda= \dfrac{b-a}{R \sin \theta \cos \phi-b}= \dfrac{h} {R \sin \theta \cos \phi -b}

\mathbf{OP}= \begin{pmatrix} 0 \\ b \cos \phi \\ b \sin \phi \end{pmatrix}+ \left(\dfrac{h}{R \sin \theta \cos \phi -b}\right) \begin{pmatrix} -R \cos \theta \\ b \cos \phi - R \sin \theta \\ b \sin \phi \end{pmatrix}

\text {and so } \mathbf {AP}= \begin{pmatrix} 0 \\ b \cos \phi \\ b \sin \phi \end{pmatrix} +\left( \dfrac{h}{R \sin \theta \cos \phi -b}\right) \begin{pmatrix} -R \cos \theta \\ b \cos \phi - R \sin \theta \\ b \sin \phi \end{pmatrix}- \begin{pmatrix} 0 \\ a \cos \phi \\ a \sin \phi \end {pmatrix}

= \dfrac{1}{R \sin \theta \cos \phi-b} \begin{pmatrix} -hR \cos \theta \\ b \cos \phi(R \sin \theta \cos \phi -b)+bh \cos \phi - hR \sin \theta - a \cos \phi (R \sin \theta \cos \phi -b) \\b \sin \phi (R \sin \theta \cos \phi -b)+bh \sin \phi-a \sin \phi (R \sin \theta \cos \phi -b) \end{pmatrix}

= \dfrac{1}{R \sin \theta \cos \phi-b} \begin{pmatrix}-hR \cos \theta \\ bR\sin \theta \cos^2 \phi -b^2 \cos \phi+bh \cos \phi - hR \sin \theta -aR\sin \theta \cos^2 \phi +ab \cos \phi \\ bRsin \theta \sin \phi \cos \phi -b^2 \sin \phi+bh \sin \phi -aR\sin \phi \sin \theta \cos \phi +ab \sin \phi \end{pmatrix}

=\dfrac{1}{R \sin \theta \cos \phi-b} \begin{pmatrix}-hR \cos \theta \\ hR\sin \theta \cos^2 \phi -b^2 \cos \phi+(b^2-ab) \cos \phi - hR \sin \theta +ab \cos \phi \\ hRsin \theta \sin \phi \cos \phi -ab \sin \phi+ab \sin \phi \end{pmatrix}

=\dfrac{1}{R \sin \theta \cos \phi-b} \begin{pmatrix}-hR \cos \theta \\ hR\sin \theta \cos^2 \phi - hR \sin \theta \\ hRsin \theta \sin \phi \cos \phi \end{pmatrix}

= \dfrac {Rh}{R \sin \theta \cos \phi-b} \begin{pmatrix}- \cos \theta \\ -\sin \theta \sin^2 \phi \\ sin \theta \sin \phi \cos \phi \end{pmatrix}

Reply 196

brianeverit

\text{STEP 1991 paper II question 12}

\text{When hanging in equilibrium we have }mg= \dfrac{\lambda c}{a}

\text {At time }t, mg= \dfrac{\lambda (c+x-b \sin pt)}{a}=m \dfrac{ \text{d}^2x}{ \text{d}t^2} \implies \dfrac{ \text{d}^2x}{ \text{d}t^2}=- \dfrac {g}{c}(x-b \sin pt)=n^2(x-b \sin pt) \text { as required}

Unparseable latex formula:

\text{rearranging, }\dfrac{ \text{d}^2x}{ \text{d}t^2}+n^2x=n^2 \sin pt \text{ which has C.F. } A \cos nt+B \sin nt \text { and a P.I. } x=C \sin pt}

\text {hence }-Cp^2 \sin pt+Cn^2 \sin pt=bn^2 \sin pt \implies C= \dfrac{bn^2}{n^2-p^2}

\text {so }x=A \cos nt+B \sin nt+ \dfrac{bn^2}{n^2-p^2} \sin pt \implies \dfrac{ \text{d}^2x}{ \text{d}t^2}=-n^2A \cos nt-n^2 B \sin nt- \dfrac{bn^2p^2}{n^2-p^2} \sin pt

\text{so }-n^2A \cos nt-n^2 B \sin nt- \dfrac {bn^2p^2}{n^2-p^2} \sin pt+A \cos nt + B \sin nt+ \dfrac{bn^2}{n^2-p^2} \sin pt=b \sin pt

\text{hence, }-n^2A+A=0 \implies A=0 \text{ and }-n^2B- \dfrac{bn^2p^2}{n^2-p^2}+B+ \dfrac{bn^2}{n^2-p^2}=b

\text{i.e. }B(1-n^2)+ \dfrac{bn^2}{n^2-p^2}(1-p^2)=b \implies B= \dfrac{b- \frac{bn^2}{n^2-p^2}(1-p^2)}{1-n^2}= \dfrac{b(n^2-p^2)-bn^2+bn^2p^2}{(1-n^2)(n^2-p^2)}= \dfrac{bp^2(n^2-1)}{(1-n^2)(n^2-p^2)}

\text{so solution is }x=- \dfrac{bp^2}{n^2-p^2} \sin nt+ \dfrac{bn^2}{n^2-p^2} \sin pt= \dfrac{bn}{n^2-p^2}(n \sin pt-p \sin nt)

\text{The condition }0<p<n \text{ is required as otherwise the initial movement of the particle is upwards}

\text{which is clearly impossible}

\text{String remains taut throughout the motion if }x-b \sin pt>-c

\text{i.e. }c+ \dfrac{bn}{n^2-p^2}(n \sin pt-p \sin nt)>b \sin pt \implies cn^2-cp^2+bn^2 \sin pt-bnp \sin nt > bn^2 \sin pt-bp^2 \sin pt

\implies cn^2-cp^2-bnp \sin nt+bp^2 \sin pt>0 \text { but } \sin nt< \sin pt

\text{so } cn^2-cp^2+(bnp+bp^2) \sin pt>0 \implies c(n^2-p^2)+bp(n+p) \sin pt>0 \implies c(n-p)+bp \sin pt>0

\implies \sin pt> \dfrac{c(p-n)}{bp} \text { which is certainly true if } \dfrac{c(p-n)}{bp}<-1 \implies bp<c(n-p)

(edited 12 years ago)

Reply 197

brianeverit

\text{STEP 1991 Paper II question 13}

\text {For particle we have (1) } R-amg \sin \theta =amx \ddot \theta

\text{ and (2) }F-amg \cos \theta=amx \dot \theta^2

\text{For rod, we have (3) }mk^2 \ddot \theta =-Rx-dmg \sin \theta

\text{From (1) and (3) } mk^2 \ddot \theta+dmg \sin \theta=-ax^2 \ddot \theta-amgx \sin \theta

\text{ i.e. }m(k^2+ax^2) \ddot \theta=-mg \sin \theta(ax+d)

\text{so } \ddot \theta= -\dfrac{g \sin \theta(ax+d)}{k^2+ax^2} \implies 2 \dot \theta \ddot \theta =- \dfrac{2g \sin \theta(ax+d)}{k^2+ax^2 \dot \theta}

Unparseable latex formula:

\implies \dot \theta^2= \dfrac {2g(ax+d)}{k^2+ax^2} \cos \theta}

\text {substituting in (2) gives }F=amg \cos \theta+ \dfrac{2amxg(ax+d)}{k^2+ax^2} \cos \theta

=amg \cos \theta \left( \dfrac{k^2+ax^2+2x(ax+d)}{k^2+ax^2} \right) \text { and } R=amg \sin \theta- \dfrac{amxg \sin \theta(ax+d)}{k^2+ax^2}

=amg \sin \theta \left( \dfrac{k^2+ax^2-x(ax+d)}{k^2+ax^2} \right) = amg \sin \theta \left( \dfrac{k^2-dx}{k^2+ax^2}\right)

\text{hence, } \mu= \dfrac{F}{R}= \dfrac{k^2+ax^2+2x(ax+d)}{(k^2-dx) \tan \theta} \implies \mu \tan \theta= \dfrac{3ax^2+k^2+2xd}{k^2-dx} \text { as required}

Unparseable latex formula:

\text {(i) If }k^2=xd \text { then } \ddot \theta=- \dfrac{g \soin \theta}{x} \text { and particle moves as if it is a simple pendulum of length }x

\text{(ii) If }k^2<xd \text { then particle willm slide upwards towards }O.

1991 II 13(a).jpg19.9KB

1991 II 13(b).jpg16.4KB

Reply 198

brianeverit

\text{1991 STEP Paper II question 14}

\text{Velocity vector of boat is } \begin{pmatrix} -v \cos \theta \\ av-v \sin \theta \end{pmatrix}

\text{Using polar coordinates, }x=r \cos \theta, y=r \ sin \theta

\text{we have } \dfrac{ \text{d}x}{\text{d}t}=-r \sin \theta \dot \theta+ \dot r \cos \theta \dfrac{ \text{d}y}{ \text{d}t}=r \cos \theta \dot \theta+ \dot r \sin \theta

Unparseable latex formula:

\terxt{so } \tan \theta \dfrac{ \text{d}x}{ \text{d}t} + x \sec^2 \theta \dfrac{ \text{d}y}{ \text{d}t} = \dfrac{\sin \theta}{\cos \theta} (-r \sin \theta \dot \theta+ \dot r \cos \theta)+ \dffrac{r \cos \theta}{ \cos^2 \theta} \dfrac { \text{d} \theta}{ \text{d}t}=\dot r \sin \theta- \lefty( \dfrac{r \sin^2 \theta}{ \cos \theta}- \dfrac{r}{ \cos \theta} \right) \dfrac{ \text{d} \theta}{ \text{d}t}

Unparseable latex formula:

= \dot r \sin \theta+r \cos \theta \dfrac { \text{d} \thewta}{ \text{d}t}=\dfrac{ \text{d}y}{ \text{d}t}

Unparseable latex formula:

\text{so velocity vector of boat is } \begin{pmatrix} \dfrac { \text{d}x}{ \text{d}t} \\ \tan \theta \dfrac {\tyext{d}x}{ \text{d}t}+ x \sec^2 \theta \dfrac{ \text{d} \theta}{ \text{d}t} \end{pmatrix}

\text {comparing with previous expression we have } \dfrac{ \text{d}x}{ \text{d}t}=-v \cos \theta \text{ and } \tan \theta \dfrac{ \text{d}x}{ \text{d}t}+ x \sec^2 \theta \dfrac{ \text{d} \theta}{ \text{d}t}=av-v \sin \theta

Unparseable latex formula:

\implies \dfrac{ \cos \theta}{a- \sin \theta}= \dfrac{ \frac{ \text{d}x}{ \text6{d}t}}{ \tan \theta \frac{ \text{d}x}{ \text{d}t}+ x \sec^2 \theta \frac { \text{d} \theta}{} \text{d}t}}= \dfrac{1}{ \tan \thewta+x \sec^2 \theta \frac { \text{d} \theta}{} \text{d}x}}

\implies \dfrac{ \text{d}x}{ \text{d} \theta}= \dfrac{x}{a \cos \theta} \text{ or } a \dfrac{ \text{d}x}{ \text{d} \theta}=-x \sec \theta

Unparseable latex formula:

\text{integrating we have } \int \dfrac{a}{x} \terxt{d}x=-\int \sec \theta \text{d} \theta \implies a \ln x=- \ln( \sec \theta+ \tan \theta)+C

x=h \text{ when } \theta=0 \text{ so } C=a \ln h \implies a \ln \left( \dfrac{x}{h} \right) = \ln \left( \dfrac {1}{ \sec \theta+ \tan \theta} \right) \implies \left( \dfrac{x}{h} \right)^a =( \sec \theta+ \tan \theta)^{-1}

x=h \text{e}^{-s} \implies \dfrac{ \text{d}x}{ \text{d}t}=-h \text{e}^{-s} \dfrac{ \text{d}s}{ \text{d}t} \implies-h \text{e}^{-s} \dfrac{ \text{d}s}{ \text{d}t}=-v \cos \theta \implies \dfrac{h \text{e}^{-s}}{ \cos \theta} \dfrac { \text{d}s}{ \text{d}t}=v

\text{but } \dfrac{1}{ \cos \theta}= \sqrt{\sec^2 \theta} = \sqrt{1+ \tan^2 \theta}= \sqrt{1+ \sinh^2(as)}= \cosh(as)

\text{hence, }h \text{e}^{-s} \cosh(as) \dfrac { \text{d}s}{ \text{d}t} =v \text { as required}

Unparseable latex formula:

\text{integrating } t= \dfrac{h}{v} \displaystyle\int_0^\infty \text{e}^{-s} \cosh(as) \text{d}s \impliest= \dfrac{h}{2v} \displaystyle\int_0^\infty \texy{e}^{-s}( \text{e}^{as}+ \text{e}^{-as}) \text{d}s= \dfrac{h}{2v} \left[ \dfrac{ \text{e}^{-(1-a)s}}{a-1}- \dfrac{ \text{e}^{-(a+1)s}}{a+1} \right]_0^\infty