# Thread: Question on Parametric Equations

1. ## Question on Parametric Equations

Hey everyone, I am having a hard time understanding a question regarding parametric equations. Basically:

A ladder 16 feet in length slides down a wall as its bottom is pulled away from the wall, using the angle theta as parameter, find the parametric equations for the path followed by the point P located 5 feet from the top of the ladder.

I am unsure as to how to approach/solve this problem. Any feedback and help is appreciated.

Thanks

2. Where is angle theta located?

(By the way, only the top of the ladder is sliding down the wall.)

Introduce a coordinate system. Use Quadrant I.

The y-axis (wall), the x-axis (floor), and the 16-unit ladder form a bunch of right triangles, as the top of the ladder slides down the wall.

Draw a picture of one of these representative triangles. If you drop a vertical line segment from point P to the x-axis (or draw a horizontal line segment from point P to the y-axis), then you have similar triangles.

The hypotenuse of the bigger triangle is always 16 units, and you apparently know one of the acute angles (the parameter).

You know how to express the legs of those triangles in terms of theta, yes?

There is likely more than one approach to this exercise, but can you make a start now?

Please explain what you've thought about or tried, if you would like more guidance.

Cheers

3. ^^^
Angle theta is located at the base of the ladder sliding down the wall, across from the 90 degree angle in a right triangle.

Your method is a bit confusing. We just started learning about parametric equations and derivatives of parametric equations and I fail to see how this problem quite relates to the subject matter. I would appreciate a bit more guidance.

4. Originally Posted by Edder

the base of the ladder sliding down the wall
Perhaps, you mispoke? The base of the ladder is sliding along the floor -- away from the wall.

The top of the ladder is sliding down the wall.

Really, I think that we may freeze time, and just consider the triangle formed at that instant as representative of all the triangles.

Your method is a bit confusing.
I posted no method.

I posted some clues that I hoped would get you thinking.

I fail to see how this problem quite relates to the subject matter.
Ah, thank you for reminding me that you asked for an explanation of the exercise. You may always feel free to do that.

We assume that the bottom of the ladder does not sway to the left or right, but moves in a straight line away from the wall. Hence, point P moves within the plane that contains the ladder, and so we can describe the location of point P using an xy-coordinate system with a line segment representing the ladder. Point P will trace out a curved path, from start to finish.

For example, before the ladder moves, point P is located at (0,11). Do you agree? Likewise, when the ladder stops moving, the location of point P will be (5,0).

We may describe the path of point P with a system of two equations. Each equation will be a function definition, and these functions' input will be the measure of angle theta in radians (a Real number).

The output of the first function will be the x-coordinate of point P, for any given Real number theta within the domain.

The output of the second function will be the y-coordinate of point P, for the same Real number theta.

x(θ) = some expression containing θ

y(θ) = some expression containing θ

The two equations above are parametric equations. They give P's location (x(θ), y(θ)) in terms of the parameter theta. When plotted across the domain, the graph shows the path of P.

Do you remember Right-Triangle Trigonometry? Can you determine what the value of θ is before the ladder moves? When the ladder stops moving, it is obvious what θ equals, yes? (These beginning and ending values define the domain of θ. Stating the domain is a required part of the system of parametric equations, so you should report it.)

Have you drawn a diagram, yet? Do you understand how to draw a representative triangle in Quadrant I? Do you understand how to connect P to the x-axis with a vertical line segment?

In order to attempt clearing up confusion, I first need to know exactly where the confusion lies.

Cheers

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•