Solved: Urgent First-Year Calculus Word Problem Help Needed

[rockafella]

Urgent First-Year Calculus Word Problem! Help needed ASAP!

Homework Statement

Hey this is a first year calculus word problem:

A third design question involves our plan to provide interested tourists with a video of themselves on our biggest Ferris wheel. The idea is to place a camera on a platform facing the descending seats. The camera will have to tilt at a rate that will enable it to keep the paying tourist in focus. The wheel is 60 metres in diameter with its lowest point 3 metres from the ground; takes about 20 minutes to make one revolution. Our plan is to mount the camera 15 metres from the ground, at a distance of 40 metres from the base of the ride. For how long will the camera be able to keep an individual in view and how fast must the camera rotate in order to do so? If you do not find our plan reasonable, please recommend a better one.

Basically, I must use first year calculus to do the problem.

Homework Equations

Circumference=2pi*r
Area=pi*r^2

The Attempt at a Solution

I know the period above is 20minutes, or 1200seconds, however I am not sure how to use this information. Also, the rate we are given is revolutions per minute. The rate we are looking for is meters or cm per minute of the camera rotating. We are also looking for time that the camera will keep people in view. I am thinking that for the time the camera will keep people in view, it would be equal to half of the circumference of the ferris wheel (2pi*r/2). Am I on track for this part? And for the first section, any clue would be appreciated. Thanks!

 

[LCKurtz]

Presumably the camera will have a direct line to the passenger when the passenger is between the points where the lines from the camera are tangent to the wheel. Set up the equations and find those two points. Then you can figure the length of the portion of the circumference that is viewable and how long the passenger will be in view. Also notice that the camera obviously won't rotate at a constant rate if it stays aimed directly at the passenger.

 

[rockafella]

Presumably the camera will have a direct line to the passenger when the passenger is between the points where the lines from the camera are tangent to the wheel. Set up the equations and find those two points. Then you can figure the length of the portion of the circumference that is viewable and how long the passenger will be in view. Also notice that the camera obviously won't rotate at a constant rate if it stays aimed directly at the passenger.

Okay, I understand part of which you are saying, but I'm not sure which equations I should be using to find those two points? I know the two tangent lines are going to have different slopes (because the camera is mounted at a height of only 15metres, but midpoint of ferris wheel height is 30m+3=33metres). And I'm given distance from base of wheel to camera base, height of camera, and height of wheel off of ground, but I am just not sure how to go about finding point of tangent in this case.

 

[LCKurtz]

That is the calculus part of your problem. Set up a coordinate system and write the equation of your circle. Write the equation of a straight line through the camera with unknown slope m. Then use the fact that that straight line must intersect the circle at an unknown point where it has the same slope as the circle. You should find two points.

 

[rockafella]

So representing my straight line I have y=mx+b, where m is my slope. I'm assuming the equation of the circle that you're referring to is the circumference? (2pi*r). In that case, in order to find where they intersect, I set them equal to each other?

 

[LCKurtz]

So representing my straight line I have y=mx+b, where m is my slope. I'm assuming the equation of the circle that you're referring to is the circumference? (2pi*r). In that case, in order to find where they intersect, I set them equal to each other?

No, I don't mean the circumference. I mean the x-y equation of the circle representing your ferris wheel. And you might want the "point-slope" form of the straight line since you know it passes through the camera location.

 

[rockafella]

No, I don't mean the circumference. I mean the x-y equation of the circle representing your ferris wheel. And you might want the "point-slope" form of the straight line since you know it passes through the camera location.

Okay, so what I have so far is this: I'm going to make the circle (ferris wheel) centered at the point (0,33). Therefore, my equation of my circle in terms of x and y is as follows:
x^2+(y-33)^2=900
It can also be rewritten as +/-sqrt(900-x^2)+33

Now that I have my circle, I know my line(s) must be in form y-y1=m(x-x1), so they will look like this: y-15=m(x-40). Also, I know these two lines must pass through the point (40,15), aka the camera's location. I used pythagorous and length of line segment formula to find the lengths of the line segments from point(40,15) to the two points of tangent. I found them to be 32meters long. I now know the lengths of all three sides of both triangles in my circle, as well as the location of two points (centre(0,33) and camera (40,15)), however I am unsure how to calculate the exact locations of my tangent points.

Also, after I find the coordinates of these two points, how do I use that information to find arclength and a changing rate?

 

[LCKurtz]

Pretty good so far. So now you have unknown point(s) (a,b) on the circle and unknown slope m of the line. Three unknowns. Here are some other facts you know:

1. (a,b) must satisfy the equation of the circle.
2. (a,b) must satisfy the equation of the line
3. The slope of the tangent line must equal the slope of the circle. (I knew there would be some calculus in there somewhere ) I suppose you could alternatively use the fact that the tangent line is perpendicular to the radius since you are dealing with a circle but, hey, since you know calculus, might as well use it.

For your last question think about "polar coordinates" relative to the center of the circle.

 

[rockafella]

Pretty good so far. So now you have unknown point(s) (a,b) on the circle and unknown slope m of the line. Three unknowns. Here are some other facts you know:

1. (a,b) must satisfy the equation of the circle.
2. (a,b) must satisfy the equation of the line
3. The slope of the tangent line must equal the slope of the circle. (I knew there would be some calculus in there somewhere ) I suppose you could alternatively use the fact that the tangent line is perpendicular to the radius since you are dealing with a circle but, hey, since you know calculus, might as well use it.

For your last question think about "polar coordinates" relative to the center of the circle.

Okay, so basically I used the non-calculus method to find the time it takes for the ferris wheel to travel that arclength (312.32seconds), given the right triangles I created. I am on the second part of the question where I have to relate the rate of change of the angle inside my ferris wheel to the rate of change of the angle of the camera. I have to find an equation that relates these two things, differentiate it with respect to time, and then plug in our knowns to find out how fast the camera's angle needs to change. Am I right up to this point? If I am, this is the part I'm stuck on. I can't think of how to relate d(theta)/dt of the ferris wheel to d(phi)/dt of the camera's angle

 

[LCKurtz]

As the wheel rotates you can express (x,y) on the wheel in terms of your translated polar coordinates. So find the slope of the line from the camera to (x,y) as it moves. That will have theta in it. And remember that the slope of the line is tan(alpha) for the camera.

 

[rockafella]

As the wheel rotates you can express (x,y) on the wheel in terms of your translated polar coordinates. So find the slope of the line from the camera to (x,y) as it moves. That will have theta in it. And remember that the slope of the line is tan(alpha) for the camera.

Okay I'll give that a try. Thanks a lot for your help by the way!