2. As shown in the figure above a person whose eye level is 5 feet above the ground stands on the top of a hill overlooking a valley. The shape of the valley is modeled by the graph of f(x)= 50 cos (x/100). The person’s line of sigh is tangent to the side of the hill at point A.

The correct answer to (a) should read

y= -1/2(sin(a/100))x + (a/2)sin(a/100) + 50 cos(a/100).

(b) Since the shape of the valley is modeled by the graph of f(x)= 50 cos (x/100), the top of the hill is at (0, 50). Since the eye level is 5 feet above the top of the hill, the line of sight (tangent line) also contains the point (0, 55).

So the result of part (a) leads to (a/2)sin(a/100) + 50 cos(a/100) = 55.

I do not see a way to solve this algebraically, so a numerical approximation technique or a graphing calculator is needed to solve for a. The smallest positive solution for a turns out to be a = 45.935 .

(c) Since the shape of the valley is modeled by the graph of f(x)= 50 cos (x/100), the lowest point of the valley first occurs at x/100 = pi, or equivalently x = 100pi. The y-value there is -50, so the top of the 25-foot tall flagpole is at height -50 + 25 = -25.

If the top of the flagpole is below the tangent line, then it cannot be seen; if the top of the flagpole is at or above the tangent line, then it can be seen.

The y-coordinate of the point on the tangent line where x = 100pi is

-1/2(sin(45.935/100))(100 pi) + (45.935/2)sin(45.935/100) + 50 cos(45.935/100)

= -1/2(sin(45.935/100))(100 pi) + 55

= -14.644.

Since -25 < -14.644, the top of the flagpole is below the tangent line (line of sight) and so it cannot be seen. Lord bless you today!

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