Roller Coaster Lab: Solve 2 Cubic Polynomials & Graph Solution

[NCHEVYHEVN]

Hey guys!

Long time reader, first time poster. I have a lab that I'm trying to get done in Calculus 1 and I'm having a some problems. The basic idea is to come up with 2 equations that compliment a graph and are continuous.

Homework Statement

You are to design a small section of a roller coaster with the following requirements.

1. The section begins at a height of 50ft with a slope of 6.8.
2. The first 70ft of horizontal distance will be the shape of a cubic polynomial.
3. The second 70ft of horizontal distance will also be in the shape of a cubic polynomial.
4. The two polynomials will join at a height of 110ft.
5. The second polynomial will end with a slope of .94 at a height of 50ft.
6. The two polynomials will join with continious first and second derivatives to assure a smooth path.

Questions to answer: Show your work to support your claims.
1. find the two cubic polynomials which model the shape of the roller coaster.
2. Graph the roller coaster path from 0ft to 140ft.

Homework Equations

The Attempt at a Solution

So after making a sketch of coaster I decided to start with the following equation for the first section of roller coaster.

y=ax³+bx²+cx+d

I know that d will equal the start height of the roller coaster which is 50ft. From there I don't know where to start other than knowing that the slope is 6.8.

Should I take the first derivative of the equation? I mostly just need help with finding out which numbers to plug in and start solving.

Thanks for all of the help.

 

[I like Serena]

The Attempt at a Solution

So after making a sketch of coaster I decided to start with the following equation for the first section of roller coaster.

y=ax³+bx²+cx+d

I know that d will equal the start height of the roller coaster which is 50ft. From there I don't know where to start other than knowing that the slope is 6.8.

Should I take the first derivative of the equation? I mostly just need help with finding out which numbers to plug in and start solving.

Thanks for all of the help.

Yes, you should take the first derivative, which will tell you the value of c.

Then you'll have to define a second cubic polynomial for the second section with different parameters. That will give you a total of 8 parameters.

Then you'll have to set up a set of equations.
Since you have 8 variables (the 8 parameters), you'll need 8 equations, although a couple of them are very simple (the first is simply d = 50 ft).

 

[NCHEVYHEVN]

Okay so I have been trying to figure this out and I still just don't get it.

I took the derivative of the first equation and then plugged in 0 for x and 50 for y. That tells me that c is also equal to 50. Now how am I able to figure out what a and b are as well as plugging in the slope.

I look online and some people will eliminate one of the variables but the variable x being a 0 is throwing me off.

Thanks for the help.

 

[I like Serena]

Okay so I have been trying to figure this out and I still just don't get it.

I took the derivative of the first equation and then plugged in 0 for x and 50 for y. That tells me that c is also equal to 50. Now how am I able to figure out what a and b are as well as plugging in the slope.

I look online and some people will eliminate one of the variables but the variable x being a 0 is throwing me off.

Thanks for the help.

First off, the slope is 6.8 and not 50, so c = 6.8.

Then you should not fill in 0 for x any more.
According to the 4th bullet the end of the first section (at 70 ft horizontally) should be at 110 ft.
This means that if need to fill in 70 for x, and 110 for y.

 

[King Cole]

I am working on a similar problem and I am stuck near this point where you left off. I can come up with any number of polynomials that join at (70,110), and have continuous first derivatives if I guess the derivative at x=70. But I cannot get continuous second derivatives. Using the numbers from this example, I know the following:

My interpretation is a piecewise function:
{ f(x) 0 <= x <= 70, g(x) 70 <= x <= 140

f(x) = ax^3 + bx^2 + cx + d
f(0) = 50, thus 'd' = 50

f'(x) = 3ax^2 + 2bx + c
f'(0) = 6.8, thus 'c' = 6.8

f''(x) = 6ax+2b
f''(70) = 6a(70) + 2b = ?

f'''(x) = 6a (not sure if needed)

f(70) = a(70)^3 + b(70)^2 + 6.8(70) + 50 = 110
f(70) = a(70)^3 + b(70)^2 = -426

-----------------------------------------------

g(x) = ax^3 + bx^2 + cx + d
g(70) = a(70)^3 + b(70)^2 + c(70) + d = 110
g(140) = a(140)^3 + b(140)^2 + c(140) + d = 50

g'(x) = 3ax^2 + 2bx + c
g'(70) = 3a(70)^2 + 2b(70) + c = f'(70) = ?
g'(140) = 3a(140)^2 + 2b(140) + c = 0.94

g''(x) = 6ax + 2b
g''(70) = 6a(70) + 2b = f''(70) = ?

g'''(x) = 6a (not sure if needed)

I have tried guessing numbers for the first and/or second derivative to define how to solve it, but cannot find a workable method for getting both the first and second derivatives to be continuous. Can someone please nudge me into the next step?

 

[I like Serena]

Welcome to PF, King Cole!

First off, you have defined g(x) using the parameters a,b,c,d.
But these are not the same parameters as you have for f(x)!
So you should use for instance A,B,C,D instead (giving you a total of 8 variables).

The equations that you have are:
f(0)=
f(70)=
f'(0)=
f'(70)=g'(70)
f''(70)=g''(70)
g(70)=f(70)=
g(140)=
g'(140)=

These are 8 equations from which you can solve a,b,c,d,A,B,C,D.

 

[King Cole]

Brilliant. Just the nudge I needed. Thank you.

 

[I like Serena]

Thanks!

At least some good came from this thread, since the OP never came back. :)

 

[NCHEVYHEVN]

I'm still here and still confused. I think the thing that confuses me is how to solve for a and b. I have the same lab again but with different variables and I still can't figure it out. I know what d and c is for both equations but I can not solve for a and b.

My starting point is (0,50) and goes to a max of (80,120) and ends at (160,60). The slope of the first equation is 5.4 and the slope of the second equation is .82

I now have 8 equations but I'm not sure how to solve for some of them.1. f(0)=d=50
2. f(80)=512,000a+6,400b+80c+d
3. f'(0)=c=5.4
4. f'(80)=g'(80)
19,200a+160b+c=19,200A+160B+C
5. f(80)=g(80)
19,200a+6,400b+80c+d=19,200A+6,400B+80C+D
6. f''(80)=g''(80)
4,800a+2b=4,800A+2B
7. g(160)=4,096,000A+25,600B+160C+D=60
8. g'(160)=76,800A+320B+C=.82

What am I supposed to do with the above equations? My calculus teacher said I needed to use a program on the Ti-89 and that I should input my 8 equations. I have 8 equations but some are not solved for and look like the one above.

Any help?

 

[I like Serena]

Hi NCHEVYHEVN, welcome back!

It seems you forgot to fill in a couple of numbers.
Equation 2 and 5 are supposed to be equal to 120.
With that your calculator should be able to solve the system.

 

[kyuu]

I don't know if anyone is still browsing this thread but I was given a lab similar to this and I'm not entirely sure where to go after this. I got to NCHEVYHEVN's last post but I have a Ti-84 and was told it only works via matrices rref function.

My point starts 70 feet up with a slope of 6.2 and meets at the point (70,120) ending at (140,50) with a slope of 2.94

1. f(0)=d=70
2. f(70)=343,000a+4900b+70c+d=120
3. f'(0)=c=6.2
4. f'(70)=g'(70) 14700a+140b+c=14700A+140B+C
5. f(70)=g(70) 14700a+4900b+70c+d=14700A+4900B+70C+D=120
6. f''(70)=g''(70) 420a+2b=420A+2B
7. g(140)=2,744,000A+19600B+140C+D=50
8. g'(140)=58,800A+280B+C=2.94

I tried to put it in matrix A and do the "rref([A])" function but it gives me obviously incorrect answers. I am stumped and appreciate any help.

 

[I like Serena]

Welcome to PF, kyuu!

What did you try to put in your matrix A?

 

[kyuu]

I haven't messed around with the ti-84 too much so I'm not that confident but I put in:
[[343,000 4,900 70 1 0 0 0 0 120]
[14,700 140 1 0 0 0 0 0 70]
[0 0 1 0 0 0 0 0 6.2]
[0 0 0 1 0 0 0 0 70
[0 0 0 0 14,7000 140 70 1 120]
[0 0 0 0 420 2 0 0 70]
[0 0 0 0 58,800 280 1 0 2.94]
[0 0 0 0 2,744,000 19,600 140 1 50]]

It doesn't look right to me but today my professor did some practice problems(None this big most only like 3x4's or so) and I feel like it being so big, I'm missing something or putting it in the wrong place.
Thanks for your quick response.
Edit: Sorry I don't know how to make the matrix neater

 

[I like Serena]

You're in the right direction, but your matrix isn't right yet.

If you look for instance at equation 4:
4. f'(70)=g'(70) 14700a+140b+c=14700A+140B+C

you should rewrite it to:
14700a+140b+c-14700A-140B-C=0

the corresponding line in your matrix becomes:
[14700 140 1 0 -14700 -140 -1 0 0]

And yes, I'm afraid it does have to be this big.
If you want, you can reduce it a little to 6 equations, since you already have c=6.2 and d=70, which you can substitute, but it hardly seems worth the effort.

 

[kyuu]

Oh, I would do the same thing with line 5and 6 right since I only set them equal to each other and they would need to equal 0, right?

Just wanted to say thanks, I was stuck before this and saw this thread and it helped me get pass y first roadblock.

 

[I like Serena]

Line 5. should be equal to 120 instead of 0.
But yes, I believe you understand!

And you're welcome!

 

[kyuu]

I've browsed this site occasionally but never actually asked questions here myself, if I have problems I'll come here from now on. :D
Thanks for being so helpful.

 

[I like Serena]

Do so!

For me it certainly helps that you appreciate my help and say thanks. :)