Maximum Viewing Angle

Ron Grimes Gerri Burns

Class: Calculus

Materials: String (if the suggestion in the Teacher Note is taken)

Goals: A major goal of this project is for students to gain greater insight into the theory of placement with desired advantages. This is an opportunity for students to make connections between the concepts and constructs of calculus and the real world application of them.

Background: Students should be well versed in trigonometry, inverse trig. functions, trig. formulas, derivative forms, and extrema theory.

Technology Required: Technological assistance from DERIVE and graphing calculators enriches the pursuit of this project's solution.

Problem:

Joe's Ice Cream Palace will purchase a sign with a huge banana split on it for advertisement. The sign will be placed on a hill that is 20 ft. high with the bottom of the sign 4 ft above the ground on the hill. At what distance from a stopped car at road level should the construction firm place the posts upon which the sign will be fastened so that a maximum viewing angle will be achieved? What will the maximum viewing angle be? Refer to diagram.

0x01 graphic

Eye level

Θ = viewing angle (maximum when Θ is as large as possible)

Sign and intersection position:

hill

/\

\\

stop

|

North

sign

\\

car direction

\\

\\


Teacher Note:

It would be beneficial to design a small model of this project using string attached to a wall to demonstrate how the angles change as the distance position is moved from the wall. Use the floor as eye level and a protractor to estimate angles.

Expected Product:

A report which includes drawings, calculations, and identification of the optimal distances to achieve a maximum viewing angle.

Solution:

0x01 graphic

tan α = y/x

tan(Θ + α) = (w + y)/x

Thus tan Θ = (xw)/(x2 + y2 + wy)

Now: dΘ/dx = (-wx2 + wy2 + w2y)/((x2 + y2 + wy)2 +x2w2)

Setting equal to zero and solving we have:

Only the positive value is possible.

x = (y2 + wy)1/2

Checking d2Θ/dx2 we find it to be less than zero since w > 0 and y>0 must be true. Therefore, this is a maximum value.

Sample Solution:

Assume the sign is 6 ft. tall. Assume the average height of eye level is 4 ft. above the road in a car. Thus y = 20 ft. and w = 6 ft. as in our diagram above. Evaluating for x in the maximum value from above we find that x is 22.8035 ft. (approx.).

Now solving for Θ using the formula for tan Θ above we find the maximum viewing angle to be 7.4947 degrees.


Evaluation:

20% based on assumptions;

20% based on organizational approach;

20% based on the mathematical model;

20% based on the correct solution of the model;

20% based on analysis and presentation.

Extensions:

1. Given a 15 ft. sign with base 15 ft. above eye level. How far away should the sign be place to achieve optimum viewing angle?

2. Are most signs placed this way?

3. Is there another position that will give the same maximum viewing angle?

4. Turn the sign to give maximum viewing angles for both the South and West bound traffic if this is a 3-way stop.

5. What happens when eye level is horizontally between the top and bottom of the sign? What is the maximum viewing angle?


Maximize the View!

Problem:

Joe's Ice Cream Palace will purchase a sign with a huge banana split on it for advertisement. The sign will be placed on a hill that is 20 ft. high with the bottom of the sign 4 ft above the ground on the hill. At what distance from a stopped car at road level should the construction firm place the posts upon which the sign will be fastened so that a maximum viewing angle will be achieved? What will the maximum viewing angle be? Refer to diagram.

0x01 graphic

Eye level

Θ = viewing angle (maximum when Θ is as large as possible)

Sign and intersection position:

hill

/\

\\

stop

|

North

sign

\\

car direction

\\

\\

Funded in part by the National Science Foundation and Indiana University 1995