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SERIOUS: Urgent Help Needed on a Calculus 1 Project :( - Details Inside

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Jul 22, 2008 8:32 AM GMT

EDIT: The Project is finished, you can find a link to the project and my thank yous at http://www.realjock.com/gayforums/238057/

Edit: I hurt my neck, so I'll be going to sleep and resting today - no class, no gym, no anything

. In the meanwhile: I'm going to hand in the project tomorrow evening (7/23 5:45PM EDT).

We were given this project to do by our Calculus 1 Professor. I've tried to get a grasp of how to do this problem for a while now, and I'm still at a severe loss. The project is due today (July 22, 5:45PM EDT// I'm in NYC) and I've blown every brain cell and nerve looking up down, mirrored and backwards to get a solution . . . sigh - any help guys? I'm hoping some really smart (and possibly cute) RealJock will give a fellow jocker a real hand (no pun intended hehe)...

1) The project

2) What I know so far...

Thanks so much in advance

REALLY

THANK YOU

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Jul 22, 2008 10:56 AM GMT

Where should a pilot start decent?

IN MY BLOODY PANTS, MO FOs!!! LOLLLLLLLLLL!!!
I love 'em in uniform... aviator shades and undies, too!!!

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Jul 22, 2008 2:02 PM GMT

Paging Boeing_Flyer. You're needed on this thread.

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Jul 22, 2008 2:14 PM GMT

I could probably solve that for you but I can't really make out the scan pic. Sorry bud. From what I can see, to find the cubic you need to sub h and l in P(x), get one equation, sub l into the derivative which equals zero when the plane starts decent (horizontal), get another equation, and sub zero in again for when the place touches down and again has slope zero. You now have three equations and three unknowns. You can solve that. d = 0 because there is no vertical translation from a regular cubic function. For the next part, show the inequality using the facts that d2P/dl (horizontal accelertation should be a constant, c, and that d2P/dh should not exceed k which is less than 9.8m/s^2. Once you have all that the rest is just looks like an application of the cubic. Sorry if that wasn't clear but I'm driving at the moment too. G'luck.

MSUBioNerd

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Jul 22, 2008 2:24 PM GMT

I can solve this for you, but I'm not going to just give you the answer. The key piece of information you might be missing is the relationship between position, velocity, and acceleration. The derivative of the position is the velocity of the object. The derivative of the velocity is the acceleration. Conditions 2 and 3 give you limitations on the velocity and acceleration that are allowed in the problem. This will give you a system of equations, which you can then solve to get your answer.

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Jul 22, 2008 2:31 PM GMT

1.
Start with P(x) = ax^3 + bx^2 + cx + d and P'(x) = 3ax^2 + 2bx + c
You know:
when x = 0, P = 0
when x = l, P = h
when x = 0, P' = 0
when x = l, P' = 0
So you have 4 equations and 4 unkowns. Solve for a, b, c, d

2.
Vertical velocity = dy/dt = (dy/dx)(dx/dt) = (P')(v)
To get vertical acceleration, differentiate that WITH RESPECT TO TIME, so you get (P'')(v^2). So the absolute value of vertical acceleration is |(P'')(v^2)|.
You'll notice that, according to the equation, as x gets bigger, absolute value of vertical acceleration gets bigger. But this is only true if the airplane follows the entire cubic curve. In fact, it doesn't: beyond x = l, the airplane follows a horizontal line (during which vertical acceleration = 0). So in fact, absolute value of vertical acceleration is max when x = l (and when x = 0). So let x = l (or let x = 0). Then, following iii, set this to be less than or equal to k. You should get 6hv^2/l^2 <= k

3. Plug in the values and solve for l. Don't forget to convert units when necessary.

4. Graph it!

CuriousJockAZ

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Jul 22, 2008 2:44 PM GMT

This is sooooo not my area of expertise

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Jul 22, 2008 2:45 PM GMT

I smell trouble like that facebook school incident.

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Jul 22, 2008 3:22 PM GMT

MSUBioNerd saidI can solve this for you, but I'm not going to just give you the answer.

Not a problem, I'd rather talk it out till the last breath lol.

MSUBioNerd saidThe key piece of information you might be missing is the relationship between position, velocity, and acceleration. The derivative of the position is the velocity of the object.

Y is the Altitude and X is the Distance from touchdown... so how could the Derivative of the position be the velocity of the plane? I thought the Velocity was (ds/dt) and since this one-dimensional motion... now what? Is it parametric... if so that's not going to make me a very happy camper...

MSUBioNerd saidConditions 2 and 3 give you limitations on the velocity and acceleration that are allowed in the problem.

According to the problem... Condition (i) applies to only Question 1, Conditions (ii) and (iii) are specific to Question 2.

MSUBioNerd saidThis will give you a system of equations, which you can then solve to get your answer.

How so...

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Jul 22, 2008 3:26 PM GMT

I thought you are going to become a PhD in Math. YOu cant even even wrap your mind around Calc I ...

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Jul 22, 2008 4:08 PM GMT

Coolrabbit said1.
Start with P(x) = ax^3 + bx^2 + cx + d and P'(x) = 3ax^2 + 2bx + c
You know:
when x = 0, P = 0
when x = l, P = h
when x = 0, P' = 0
when x = l, P' = 0
So you have 4 equations and 4 unkowns. Solve for a, b, c, d

I figured that P'(L)=0... I'm still skeptical of it because the purpose is to model at cubic function and a cubic function wouldn't have a P'(x)=0 anywhere [except when P'(0)=0 in this case]. Also wouldn't domain be in the interval [0,L]? - That would make it even more unlikely that the start of the model graph would have a P'(L)=0.

I figured that those were the 4 equations and 4 unknowns, but I have a few Q's about that.
1) if P(L)=H, then H = P(L) = aL^3 + bL^2 + cL + d?
2) if P'(x) = 3ax^2 + 2bx + c and P'(L)=0, then P'(L) = 0 = 3aL^2 + 2bL + c?
3) Since we have two cubic polynomials and two quadratic derivatives, could we add the four equations together in one big system of equations operation, or would I have to work with adding the two cubic polynomials first then work with the adding the two quadratic derivatives together later.

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Jul 22, 2008 4:10 PM GMT

Caslon5000 said

I would have given PETER.... 0.25 points for being clever. Seriously.

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Jul 22, 2008 4:11 PM GMT

Uhhhhhhh...

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Jul 22, 2008 6:04 PM GMT

HighVoltageGuy saidUhhhhhhh...

Edit Bump

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Jul 22, 2008 11:43 PM GMT

QUOTE: "I figured that P'(L)=0... I'm still skeptical of it because the purpose is to model at cubic function and a cubic function wouldn't have a P'(x)=0 anywhere [except when P'(0)=0 in this case]. Also wouldn't domain be in the interval [0,L]? - That would make it even more unlikely that the start of the model graph would have a P'(L)=0."

First, look at the graph: if you drew a tangent to the graph at x = l, the tangent would be a horizontal line; so P'(l) = 0.

Second, the derivative of any cubic function is a quadratic; and a quadratic can have zero/one/two "zeroes"; therefore the derivative of any cubic function can have zero/one/two "zeroes".

Third, yes the interval is [0,l]. But still, P'(l) = 0.

QUOTE: "I figured that those were the 4 equations and 4 unknowns, but I have a few Q's about that.
1) if P(L)=H, then H = P(L) = aL^3 + bL^2 + cL + d?
2) if P'(x) = 3ax^2 + 2bx + c and P'(L)=0, then P'(L) = 0 = 3aL^2 + 2bL + c?
3) Since we have two cubic polynomials and two quadratic derivatives, could we add the four equations together in one big system of equations operation, or would I have to work with adding the two cubic polynomials first then work with the adding the two quadratic derivatives together later."

{when x = 0, P = 0}: This tells you that d = 0
{when x = 0, P' = 0}: This tells you that c = 0
{when x = l, P = h}: This tells you that al^3 + bl^2 = h
{when x = l, P' = 0}: This tells you that 3al^2 + 2bl = 0. Multiply both sides of this by "l/2" and you get 3al^3/2 + bl^2 = 0, and you can subtract this from "al^3 + bl^2 = h" so you cancel out the bl^2. You end up with a = -2h/l^3 and b = 3h/l^2

By the way, the answer to the question is l = 64.5 miles.

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Jul 22, 2008 11:45 PM GMT

Lucasxxx saidI could probably solve that for you but I can't really make out the scan pic. Sorry bud. From what I can see, to find the cubic you need to sub h and l in P(x), get one equation, sub l into the derivative which equals zero when the plane starts decent (horizontal), get another equation, and sub zero in again for when the place touches down and again has slope zero. You now have three equations and three unknowns. You can solve that. d = 0 because there is no vertical translation from a regular cubic function. For the next part, show the inequality using the facts that d2P/dl (horizontal accelertation should be a constant, c, and that d2P/dh should not exceed k which is less than 9.8m/s^2. Once you have all that the rest is just looks like an application of the cubic. Sorry if that wasn't clear but I'm driving at the moment too. G'luck.

Okay, was any part of that in English?

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Jul 23, 2008 12:00 AM GMT

Coolrabbit said{when x = 0, P = 0}: This tells you that d = 0
{when x = 0, P' = 0}: This tells you that c = 0
{when x = l, P = h}: This tells you that al^3 + bl^2 = h
{when x = l, P' = 0}: This tells you that 3al^2 + 2bl = 0.

I always second guess myself when it comes to a function that goes through the origin. I guess that's a displaced precaution from "losing" an answer by introducing zeros or eliminating variables

D and C had to be 0 because P(0) = 0 and P'(0) = 0 will only leave the coefficient that is not multiplied with a variable.

I can do the math from that point on, I just had to make sure I was at least on the right track. I just wanted to really make sure of P'(L) = 0, and the issue with the domain. That's why I was a bit thrown off.

The second question I feel I can do on my own... The third question relies on the validity of 1 and 2, and the 4th question is just playing with the Maple Plot feature

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Jul 23, 2008 12:01 AM GMT

blind2limits said
Coolrabbit said{when x = 0, P = 0}: This tells you that d = 0
{when x = 0, P' = 0}: This tells you that c = 0
{when x = l, P = h}: This tells you that al^3 + bl^2 = h
{when x = l, P' = 0}: This tells you that 3al^2 + 2bl = 0.
I always second guess myself when it comes to a function that goes through the origin.

D and C had to be 0, but I second guess myself because finding coeffients of functions starting with the origin doesn't always lead to the best answer (Call it the stigma of bunch of wrong answers in the past lol).

I can do the math to get the answer, I just wanted to really make sure of P'(L) = 0, and the issue with the domain. That's why I was a bit thrown off.

The second question I feel I can do on my own... The third question relies on the validity of 1 and 2, and the 4th question is just playing with the Maple Plot feature

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Jul 23, 2008 12:23 AM GMT

I am very well-acquainted too with matters mathematical
I understand equations both the simple and quadratical
About binomial theorem I am teeming with a lot o' news
With many cheerful facts about the square of the hypotenuse!

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Jul 23, 2008 1:38 AM GMT

jprichva saidI am very well-acquainted too with matters mathematical
I understand equations both the simple and quadratical
About binomial theorem I am teeming with a lot o' news
With many cheerful facts about the square of the hypotenuse!

Well cheerful is me

at the moment... despite me being stuck in bed lol

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Jul 23, 2008 3:19 AM GMT

Coolrabbit said{when x = 0, P = 0}: This tells you that d = 0
{when x = 0, P' = 0}: This tells you that c = 0
{when x = l, P = h}: This tells you that al^3 + bl^2 = h
{when x = l, P' = 0}: This tells you that 3al^2 + 2bl = 0. Multiply both sides of this by "l/2" and you get 3al^3/2 + bl^2 = 0, and you can subtract this from "al^3 + bl^2 = h" so you cancel out the bl^2. You end up with a = -2h/l^3 and b = 3h/l^2

I solved question number one by substitution rather than system of equations... but that's okay because I got the same answer, and it works in Question 2

Since I don't really know the relationship between velocity and acceleration, and it looks like I really need to if I want to get through questions 2 and 3.This is what I've gathered so far.

After differentiating P(x) twice I got acceleration = P''(x) = ([6H(V^2)]/[L^2])*[(-2x/L)+1].

The first part ([6H(V^2)]/[L^2]) is exactly what's given on the left side of the equation shown in Question 2. Where does this "less than or equal to K" come from? [(-2x/L)+1] isn't be equal to k...? |P''(x)| <=k is what we're working with...?

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Jul 23, 2008 3:49 AM GMT

Your equation for acceleration is correct:
acceleration = ([6H(V^2)]/[L^2])*[(-2x/L)+1]

Notice how the acceleration depends on the x value; that is, acceleration is changing as the airplane moves along the curve.

The question says the absolute value of acceleration should not exceed k. That means, the MAX absolute value of acceleration should not exceed k. So when is the absolute value of acceleration at it's max?

Well, let's begin by asking, when is the absolute value of the accleraion the lowest? Solve for acceleration = 0. You get x = l/2. Then, notice that, as x moves away from l/2, in either direction, the absolute value of acceleration increases. But you can't just keep moving x away from l/2 forever, because when x < 0 and when x > 0, the acceleration is 0 (remember, the airplane doesn't follow the cubic in those regions, but rather a horizontal line).

So what is the absolute value of the acceleration at the limits of our domain? When x = 0, absolute value of acceleration = 6hv^2/l^2. When x = l, absolute value of acceleration is also 6hv^2/l^2. This is the maximum absolute value of acceleration during the descent.

According to the question, this max absolute value of acceleration should not exceed k. So we have: 6hv^2/l^2 <= k

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Jul 23, 2008 3:59 AM GMT

Coolrabbit saidYour equation for acceleration is correct:
acceleration = ([6H(V^2)]/[L^2])*[(-2x/L)+1]

Notice how the acceleration depends on the x value; that is, acceleration is changing as the airplane moves along the curve.

So what is the absolute value of the acceleration at the limits of our domain? When x = 0, absolute value of acceleration = 6hv^2/l^2. When x = l, absolute value of acceleration is also 6hv^2/l^2. This is the maximum absolute value of acceleration during the descent.

According to the question, this max absolute value of acceleration should not exceed k. So we have: 6hv^2/l^2 <= k

AHHHHHHHHH

It's soooo simple...

WHY DIDNT I SEE THAT BEFORE!!!! I was focusing soo much on the absolute value inequality, that I didn't realize the problem really had nothing to with that, it had to do with the values of x at P''(x).

::cries::

Now I have to tie the three pieces of information together to get the answer to question 3 and 4.

metalxracr

Posts: 761

Jul 23, 2008 4:06 AM GMT

Wow, next time I need help on my homework I know where to come!

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Jul 23, 2008 4:11 AM GMT

metalxracr saidWow, next time I need help on my homework I know where to come!

This was a real long shot, I thought this thread was going to be absolutely ignored, but people came to the rescue, I owe my grade to yall

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SERIOUS: Urgent Help Needed on a Calculus 1 Project :( - Details Inside