Slope and angle at theta

Question

telephone line hangs between two poles 8 meters apart in the shape of the catenary curve 𝑦 =

10 cosh(𝑥/10) − 5, where x and y are measured in meters.

A. Find the slope of this curve where it meets the right pole.

B. Find the angle 𝜃 between the line and the pole

Answer 1

cosh(x/10) = [e^x/10 + e^-x/10 ] / 2

So, y = 5e^x/10 + 5e^-x/10 - 5

dy/dx = (1/2)e^x/10 - (1/2)e^-x/10

Draw the x-axis and y-axis so that the base of the left pole is located at (-4,0) and the base of the right pole is located at (4,0). The lowest point on the telephone line is located at (0, 5).

Let P be the point where the telephone line intersects the right pole.

Slope of the curve at point P = (dy/dx evaluated at x = 4) = (1/2)e^0.4 - (1.2)e^-0.4 ≈ 0.41

Draw a right triangle with right angle at (4,0), vertical side is the right pole, hypotenuse PQ where Q is on the x-axis and is also on the tangent line, and horizontal side has endpoints Q and (4,0).

Let α be the interior angle of the right triangle with vertex at Q.

Since the slope of the tangent line is 0.41, tanα = 0.41. So, α = Tan^-1(0.41) = 0.39 radians.

θ = π/2 - α = 1.18 radians = 67.65°

Answer 2

See the video.