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!

Tangent Feet

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!

Tangent Feet