Help forming this equation

[grantzam]

Hey all. I'm doing the following for an assignment in my Calc class, and I just need some help finishing up the problem. I'm having trouble with #1 of the first project here:

 

http://www.math.washington.edu/~m124/Stewart5Eprobs/appliedproject.pdf

 

I have looked at it, analyzed it, figured out a lot of stuff about the problem, but I just can't put it all together to get a final polynomial. I have things listed out such as the derivative of P(x) = 0 at x = L and x = 0 and other various conditions I believe to be true. I just can't put it all together.

 

If any of you can help with this last step in acquiring the answer to this problem, it would be greatly appreciated. Thanks.

[Dogtanian]

You are given a general cubic polynomial and have can find it's derivative.

 

You can find the values of c and d by sustituing x=0 into both P and P'.

 

by looking at the "shape" of the curve at x=L you can see what P'(L) will be.

You also know what P(L) equals. Use these facts to find 2 simultaneous equations and solve them to find a and b in terms of L and h.

 

I hope this helps.

[grantzam]

Thanks for replying.

 

You can find the values of c and d by sustituing x=0 into both P and P'.

 

So...

P(x) = ax^3 + bx^2 + cx + d and P'(x) = 3ax^2 + 2bx + c

so...

P(0) = d and P'(0) = c

What should this tell me though? Are the values 0? I was thinking that d was equal to the height.

 

by looking at the "shape" of the curve at x=L you can see what P'(L) will be.

You also know what P(L) equals.

 

I'm assuming that by looking at x=L that P'(L) is 0. P(L) is h?

 

Use these facts to find 2 simultaneous equations and solve them to find a and b in terms of L and h.

 

I guess this is the part that I'm not quite understanding. Or I might not be understanding how to use some of the previous info I know.

[Dogtanian]

Thanks for replying.

 

 

 

So...

P(x) = ax^3 + bx^2 + cx + d and P'(x) = 3ax^2 + 2bx + c

so...

P(0) = d and P'(0) = c

What should this tell me though? Are the values 0? I was thinking that d was equal to the height.

I think you have the picture back-to-front. From the graph/diagram given with the question the areplane is flying towards the origin' date=' not away from it. So P(0) will give you the height above the ground when the plane has landed.

(This idea seams rather counter-intuitive to me, like I think you have done, it would be better to assume the plane styarts at x=0 and heads towards x=L , but it so happens to be going the other way round )

 

I'm assuming that by looking at x=L that P'(L) is 0. P(L) is h?

 

 

 

I guess this is the part that I'm not quite understanding. Or I might not be understanding how to use some of the previous info I know.

Try this agian now you've re done the first part again...hopefully the answer should fall out of the sky to you...lets hope unlike the plane...

[grantzam]

That just confused me more :x

[bloodhound]

The gerenal cubic has 4 arbitary constants.

 

To find the particular soultion you thus require 4 initial (boundary) conditions.

 

The first two are obvious.

 

1)at distance x = l, the plane must be at height p(l)=h

2)at x=0 , the plane must be height 0

 

The second two are the bit subtle

 

The flight of the plane will have this function

 

f(x)= 0 for x<=0

f(x)=P(x) for 0<=x<=l

f(x)=h for x=>l

 

and since , the motion of the place has to be smooth (i.e f(x) must be differentiable everywhere ) it follows that f'(0)=P'(0)=0=f'(l)=P'(l)

 

[The above is a bit long winded, and the results should be clear just by intuition]

 

Once, you put in those conditions, you get two equations for two unknowns {a,b}

 

solve it, and you get [math]a=\frac{-2h}{l^3}[/math] and [math]b=\frac{3h}{l^2}[/math]

 

thus the particural solution for P(x) is

 

[math]P(x)=\frac{-2h}{l^3}x^3+\frac{3h}{l^2}x^2[/math]

[bloodhound]

I dont if the above solution is rite. can anyone check.... as i am not able to complete the second part of the question using P(x) from above

 

[edit: oops seems fine now.. sorted ]

 

[edit no.2] oops no fine at all.. redoing it again[/edit]

 

[edit no.3] everything is good now [/edit]

[grantzam]

Thanks a lot bloodhound. I appreciate the time you took

 

lol at the edits

[grantzam]

I've done this multiple times now to try and check this and I've been getting b to equal -h/l^2 instead of your 3h/l^2.

 

I did get the same answer for a however.

[bloodhound]

Ok . this is how it goes.

 

using P(0)=0 and P'(0)=0 you get c=d=0

 

and then

1)[math]P(l)=al^3+bl^2=h[/math] and

2)[math]P'(l)=3al^2+2bl=0[/math]

 

from 2)

[math](3)a=-\frac{2b}{3l}[/math] substituting a into 1) we get

[math]h=-\frac{2b}{3l}l^3+bl^2[/math]

[math]=\frac{-2bl^2+3bl^2}{3}[/math]

[math]=\frac{bl^2}{3}[/math]

Therefore,

[math]b=\frac{3h}{l^2}[/math]

put that back into (3) we have

 

[math]a=-\frac{2b}{3l}[/math]

[math]=\frac{\frac{-6h}{l^2}}{3l}[/math]

[math]=-\frac{2h}{l^3}[/math]

[Dogtanian]

Sorry about confusing you more Grantzam

 

I solved the equations in a slightly different way to Bloodhound and I got the same solutions for a and b. I'm pretty sure they are the correct ones.

[grantzam]

Thanks to both of you.

 

If anyone wants to take a stab at how to complete #2, you're more than welcome to You do not have to complete it, just a short explanation of how you would come to that conclusion would be fine for the type of assignment I have for this.

 

I already got #3.

[grantzam]

Thanks to both of you.

 

If anyone wants to take a stab at how to complete #2, you're more than welcome to You do not have to complete it, just a short explanation of how you would come to that conclusion would be fine for the type of assignment I have for this.

 

I already got #3.