First runner runs 2x, second runs x
Angle of sight to first (faster) runner = arctan(2x)
Angle of sight to 2nd runner = arctan(x)
θ = arctan(2x) – arctan(x)
For a maximum dθ/dx =0
dθ/dx = 2/(4x2+1) – 1/(x2+1) = [2(x2+1) – 4x2-1]/[(4x2+1)(x2+1)] = 0 when:
2(x2+1) – 4x2-1 = 0 i.e. when 2x2=1 or x = 1/√2
Hence θ = arctan(√2) – arctan(1/√2) = 54.74° - 35.26° = 19.47°
UBC Math100? I have this question in my webwork too. I got atan(2sqrt(1/2))-atan(sqrt(1/2)) . Hope that helps! :)
There are two angles, theta_2 and theta_1, one for each runner. Theta is the difference between them. Use the formula for the sum (or difference) of two tangents:
tan theta = tan (theta_2 - theta_1)
= [tan(theta_2) - tan(theta_1)] / (1 + tan theta_1 x tan theta_2)
Then recognize that tan_1 = x_1 and tan theta_2 = x_2, the distances the two runners have run. But even more, x_2 = 2 x_1. So you could call them x and 2x.
Now the tangent equation for theta simplifies to:
tan(theta) = (2x - x)/(1 + x * 2x) = x/(1+2 x^2).
Take the derivative of that with respect to x and set equal to zero to find x. Plug that back in to get tan(theta). Then take the inverse tangent to get the actual angle.
Figure please? :)
An observer stands at a point P, one unit away from a track. Two runners start at the point S in the figure and run along the track. One runner runs 2 times as fast as the other. Find the maximum value of the observer's angle of sight θ between the runners. [Hint: Maximize tanθ]
I've done this question numerous times now and I keep getting 1/2sqrt(2), which is incorrect.
How do I do this question?
