Thread: Hard problem : Equation of ..

I have no idea where to start.

"A parabolic segment of a roller coaster track is to join two inclined segments of track. The incoming track has a slope of -4/3 and the slope at the end of the curve is to have a slope of 2/3. The parabolic portion of the curve must start at a height of 30 feet and end after 50 feet of horizontal distance."

A) Find the equation of the parabolic portion of the track.

B) Find the height of the track where the parabola joins the line with a slope of 2/3.


I know the equation for a parabola is y = ax^2 + bx + c.. &the derivative is y' = 2ax + b, but that's as far as I've gotten.

Originally Posted by JessicaWade

I have no idea where to start.

I know the equation for a parabola is y = ax^2 + bx + c.. &the derivative is y' = 2ax + b, but that's as far as I've gotten.

May I please point out that these statements are inconsistent?! Please, never use the first one again. You MUST have SOME idea. You demonstrated that you do.

1) Ponder your parabola on a coordiante axis. You'll have to decide on an orientation and a direction.

2) It seems prudent to decide (there is nothign magic about it) that the car will be moving generally from left to right as we consider this one seciton of track.

3) Having said that, on the left, the slope is -4/3 -- It is coming down. On the right, the slopr is 2/3 -- It is going back up. It's pretty clear, then that we have a cup-shape, of opening-up parabola.

4) Now you must decide one more thing. Where does the parabolic segment start? It's doesn't much matter, but you can make matters a little wors or a little better by choosing wisely. Since we get 50 horizontal feet, it may be wise to put the vertex on the origin. Maybe not. Frankly, it seemed easier to me at first to start the parabolic section on the y axis.


Given these definitions

y(x) = ax^2 + bx + c
y'(x) = 2ax + b

This gives:

"start at a height of 30 feet"

y(0) = 30

incoming track has a slope of -4/3

y'(0) = -4/3

the end of the curve is to have a slope of 2/3

y'(50) = 2/3

That should be enough.

Let me return to your opening statement. As soon as you believed that you could not do it, you lost track of your thinking process. Everything broke down ONLY when you gave up. Here's a plan. Don't do that.

Note: Really, it doesn't much matter what you assume. If we had decided to make it symmetrical about the y-axis, it would have looked like this...

y(-25) = 30

y'(-25) = -4/3

y'(25) = 2/3

It WILL produce a different solution, but the structure will be the same. Just make rational assumptions and WRITE THEM DOWN.

I still don't understand how to find an equation or find the height of the track.

Have you done a graph and tried to follow the hints and suggestions provided earlier? If so, what progress have you made?

Please be complete, so we can see where you're having trouble. Thank you!

Originally Posted by JessicaWade

I still don't understand how to find an equation or find the height of the track.

Before getting to derivative, you should have gotten a good background in the solution of linear systems of equations. Is this so?