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joe2812
13th Oct 2004, 18:08
http://www.members.aol.com/jsimon456/maths.JPG

Can anyone help with this please?! Really struggling so thought i'd bring it to the most knowledgable people I know (brown nosing achieved).

Mods can delete this thread once we've figured it out!

Will owe the winner big time!
Thanks.

<ps this is what i have so far..

ln both sides and drop the power of -(x/B)^2 down

lnH=-(x/B)2lnHoe

lne = 1

lnH=-(x/B)2Ho

corky83
13th Oct 2004, 19:55
Funny things, logarithms. Remember that log(ab) = log(a) + log(b), not the product of the two, hence your statement is untrue.

Here's my attempt (sorry for the awkward notation, but its hard writing it in text!):

In the region x<0, H is constant, so there is no gradient. Therefore the steepest point must be in the positive x region, so take the derivative of the curve with respect to x in this region:

dH/dx = -(2*h*x/B^2)*exp(-x^2/B^2)

The steepest point on the curve is the point where this gradient is a maximum, therefore we must differentiate the gradient with respect to x to get the second derivative of the curve wrt x, thus:

d2H/dx2 = -(2*h/B^2*)exp(-x^2/B^2)+(4*h*x^2/B^4)*exp(-x^2/B^2)

We must now set this to zero and solve for x to find the point of maximum slope. Doing so yields the position of the point of steepest slope:

x = (sqrt(2)/2)*B (There's also a negative value, but this occurs in the flat part of the plateau, so we can ignore it. Yay!

To find the height of the ground at this point, we must substitute this expression for x back into the original equation, thus:

H = h*exp(-(1/2*2^(1/2)*B

To find the slope at this position, we substitute the expression for x into the first derivative that we took, thus:

dH/dx = h*exp(-(1/2*(2^(1/2))*B)

So that's the first part done!

For the numerical values, just substitute in the numbers - I get these:

Position: x = 1061m

Height: H = 303m

Slope: dH/dx = 0.29

As for the shape of the curve, it would be flat, then would fall in a swept out backwards S shape and approaches zero. I'm not going to try to draw it here!

Did I pass? I'm sure someone will find a mistake somewhere, and I shall be suitably embarrassed.

Hope you can follow that, again sorry for not being very clear with the notation and not showing working. Hope I haven't missed and brackets out!

Oh, and I got my computer to do the nasty differentiation, so save my tired brain.

(Edited for spelling).

Fujiflyer
13th Oct 2004, 20:06
Nice question, Joe, looks like typical A level maths type stuff. My maths muscles are somewhat wasted away but looking at the applied side of the problem you must solve for dH/dx, and d^2H/dx^2 (2nd derivative) then place the latter equal to zero to find values for x which give greatest dH/dx (turning points). The slope gradient is obviously the dH/dx value and x the position.

Part (a) is the part where the hard work is (obtaining an expression for the derivatives), (b)&(c) are substitutions using the results from (a). I'm not sure at what slope the car is judged to not be able to climb but at the very least it would be where the slope is vertical or an overhang -> dH/dx is negative or approaching the 1/0 asymptote.

As for the algebra - not exactly sure at the moment but it looks like the chain rule might help (function of a function). Else try Laplace... :ouch: Will keep thinking about it,

Fuji

edited to say: Corky obvoiusly beat me but then my PC won't do maths :{

joe2812
13th Oct 2004, 20:07
corky... the minute I turn 18 theres a beer (or a few) waiting for ya in sunny Yarmouth!

thanks very very much! :p

Fujiflyer, same for you. :ok:

Foxy Loxy
13th Oct 2004, 20:09
Blimey, this isn't your coursework is it??
;)

Foxy

joe2812
13th Oct 2004, 20:18
No not coursework...but another bit of homework I can't afford not to hand in!

corky83
13th Oct 2004, 20:19
joe - I'll be down there to collect, just tell me the day.

I know it's cheating to use my computer to do it, but it's the method that counts, honest - plus i have to know how to make my computer do it. It's also cheating to hand my solution in for your homwork, but only if you don't understand it, so here's a test:

Why does taking the second derrivative lets you see the point where the slope is the greatest?

Oh, and thanks joe for getting me to do something I learnt at uni for the first time since I left!

(Edited to remove the semi-rant about coursework.)

Corky

joe2812
13th Oct 2004, 20:27
Well if i'm correct in comparing the plateau to a graph then then d2y/dx2 is used in finding a max or a min point (or turning point) on a graph, therefore it's used to find the max (and the min i suppose) of the cliff?

flyblue
13th Oct 2004, 20:49
Guys errrr...just leave me a minute or two to check this stuff but I guarantee you that if this is against the JB ROE you are in trouble!

joe2812
13th Oct 2004, 20:53
:ugh:

Anybody bail me out of this one again? :\

TimS
13th Oct 2004, 20:58
Since when was talking Martian against the JB ROE ??????

cringe
13th Oct 2004, 21:14
Well done Corky. Just a tiny correction:

x = (sqrt(2)/2)*B (There's also a negative value, but this occurs in the flat part of the plateau, so we can ignore it. Yay!

To find the height of the ground at this point, we must substitute this expression for x back into the original equation, thus:

H = h*exp(-(1/2*2^(1/2)*BYou get H = h*exp(-1/2)


To find the slope at this position, we substitute the expression for x into the first derivative that we took, thus:

dH/dx = h*exp(-(1/2*(2^(1/2))*B)Here you get dH/dx = -sqrt(2)*h*exp(-1/2)/B

dH/dx = -0.29 is negative, as we have a downward slope.

The maximum slope angle is about 16 deg (=arctg(slope)).

Slim20
13th Oct 2004, 23:05
I'm trying to work out how the **** you can work out whether a car can drive up it or not?

4 lousy marks for leaving the exam room, buying a car, setting up a ramp made from your sketch of the gradient, and then seeing if you can drive up it! I'm glad I finished maths a long time ago.....

Bre901
13th Oct 2004, 23:26
Looks like I missed something tonight. ;) :8 ;) :8

Just my 0.02 Euro worth about the car stuff : the steepest road I ever saw was Honister Pass in Lake District. The road signs said 1/4, which is 0.25 (give or take as I guess the 1/4 is the result of rounding). Driving downwards the revs went up to 5500 rpm in 1st gear, idle (before anyone says something, yes, this is some kind of repost)

I saw cars climb that slope, so the answer is probably yes, but not easily

BlueDiamond
14th Oct 2004, 02:35
My old 4WD would have romped up a little gradient like that. First three gears were for climbing walls.

:E