SOMEONE ASKED ๐
Calculus Tangent Question?
HERE THE ANSWERS ๐

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 yvalue there is 50, so the top of the 25foot 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 ycoordinate 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
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