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Calculus 136.169 - Project Two

Submission Date: November 26^th, 1999    

Submitted To: Professor Woods    

Submitted By: Robert Young, Joemar Johnson, Kyle Davis

Objective: To find the curve that can be placed on a playing surface in which any point on the curve subtends the largest angle to the goal mouth, with respect to any other point on a line perpendicular to the goal mouth.

Solution:

First, we want to find an equation for q, in terms of x. We will declare the unknown constants, and variables:

    Let l be the perpendicular distance between the goals

    Let g be the width of the goal-mouth

w, and g are all positive real numbers, because of physical restraints.

    Let n be the perpendicular distance from the line down the field we are concerned with, to the closest lengthwise border

We make an imaginary line at a perpendicular distance, n, from the boards, we will refer to it as the "skating line". We can then imagine any point on that line would have to variables associated with it:

    Let x be the distance on the "skating line" from the point to the line extending widthwise from the goal-mouth.

    Let q be the angle subtended by x, with legs extending to the corners of the goal-mouth.

We can now make an equation relating and x, in terms of our constants: w, l, n, and g.

If you look at this graph, and using the trigonometry, we can make an equation involving the difference of two inverse tangent functions.

You will notice that using the two legs formed by the angle q, you can make two right angle triangles. Using basic trigonometry, we know that is equal to the larger triangle on the diagram, minus the smaller triangle, i.e.:

You will notice that if x is equal to zero that the function would be undefined. That relate physically to the problem, since the angle in which you would subtend while standing left or right of the goal would be zero.

If we multiply the numerators and denominators of the fractions in the brackets by 2, we can simplify our equation:

    D(q) = (0, ].

This is q in terms of x. For the next couple of steps, it would be easier if we simplified the above equation.

    Let a = (w + g - 2n)/2x.

    Let b = (w - g - 2n)/2x.

We now have:

    q = tan^-1(a) - tan^-1(b).

1. Endpoints

2. Singular Points (when q is defined, but q is not)

3. Critical Points (when q' =0)

There are no points in our function where q is defined, and is q' not, so we do not have to worry about singular points.

There is only one endpoint on our function and that is l. Intuitively, we know this is not a maximum because it is very far away from the goal-mouth and thus will subtend a small angle. However, just to be sure, we will consider this point, later in the paper.

    q' = 1/(1 + a²) * a - 1/(1 + b²) * b.

We want to find points where q' =0, i.e.:

      1/(1 + a²) * a - 1/(1 + b²) * b = 0.

i.e. 1/(1 + a²) * a = 1/(1 + b²) * b.

If we substitute [(w + g - 2n)/2x] for a and [(w - g - 2n)/2x] for b in the above equation, we have:

i.e.: [(-2w-2g+4n) / 4x²(1 + a²)] = [(-2w+2g+4n) / 4x²(1 + b²)].

Cross-multiplying, we have:

    (-2w+2g+4n)(4x²)(1+a²) = (-2w-2g+4n)(4x²)(1+b²).

Canceling 4x² from both sides, we have:

If we substitute [(w + g - 2n)/2x] for a and [(w - g - 2n)/2x] for b, and simplify, we have:

      4x²g + g³ - 4n²g - w²g + 4gnw = 0. {see Calculations, 2}

i.e.: 4x²g = -(g³ - 4n²g - w²g + 4gnw).

i.e.: 4x² = -(g² - 4n² - w² + 4nw).

i.e.: x² = -(g² - 4n² - w² + 4nw)/4.

i.e.: x = ±[-(g² - 4n² - w² + 4nw)/4], but since x is in the domain (0, l), thus cannot be negative, we discard the negative quantity. Thus:

      x = ½ * Ö[-(g² - 4n² - w² + 4nw)].

This is the critical point of q. That is, a point in which q could be at a local maximum or minimum.

To find out if it is a local max or min or neither, and also to compare it with the endpoint, we must give our unknown constants known quantities so that we can calculate values for at and near these points.

    x = ½ * Ö[-(3² - 4(3)² - 15² + 4(3)(15))].

    x = ½ * Ö[72]

    x @ 4.24264

The first-derivative test says that if is positive at M-d (increasing), and negative at M+ d (decreasing), then M is a local maximum.

If we get is increasing before M, and decreasing after M, we know M is a local maximum.

    q = tan^-1[(w + g - 2n)/2x] - tan^-1[(w - g - 2n)/2x].

Therefore, in this particular example,

      q = tan^-1[(15 + 3 - 6)/2x] - tan^-1[(15 - 3 - 6)/2x].

i.e.: q = tan^-1[6/x] - tan^-1[3/x].

    q(4.24) = tan^-1[6/4.24] - tan^-1[3/4.24].

                  @ 19.471

    q(3)= tan^-1[6/3] - tan^-1[3/3].

          = tan^-1[2] - tan^-1[1].

          @ 18.435

    q(5) = tan^-1[6/5] - tan^-1[3/5].

           @ 19.231

Since the value of q at 5 is less than at M, the function is decreasing on the interval (M ,l].

Therefore, M is a local maximum.

Since the function is decreasing on the entire interval (M, l], we know that there is no way q(l) could be greater than q (M), unless l < M. This is possible because doesn't change the function, it just is an upper limit of x. Generally, you will never find a playing surface where l is less than ½ * Ö[-(g² - 4n² - w² + 4nw)] but it could happen. Therefore, M the absolute maximum of q, as long as M < l. If l > M, then l is the maximum for that example.

The n values in which the "skating line" is within the goal posts is (w+g)/2 and (w-g)/2, using the first diagram as a guide. Therefore, we get the following function:

                                           
Result: The equation of the curve that can be placed on a playing surface in which any point on the curve subtends the largest angle to the goal mouth, with respect to any other point on a line perpendicular to the goal mouth, is the equation x = ½ * [-(g² - 4n² - w² + 4nw)], where g is the width of the goal-mouth, n the parallel distance from either of the length-wise borders of the playing surface, w is the width of the playing surface and x is the perpendicular distance from a line extending left-right from the goal posts. However if n is within the points (w-g)/2 and (w+g)/2, x = 0.

Attached: Graph of the above function, suitable for transposing onto a real playing surface 

                    Two Calculation sheets corresponding to certain parts of the solution

Copyright 1999: Robert Young, Joemar Johnston, Kyle Davis

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