Question

Suppose that a camera is fixed at (0,0) in the coordinate plane (measured in feet). An actor starts at (10,10) and moves down (negative y-direction) at 1 foot per second, and moves right (positive x-direction) at 1 foot per second, so that his position at time t is (10+t,10−t).

Let θ be the angle between the positive x-direction and the line of sight from the camera to the actor as a function of t. Find the rate of change of θ as a function of time t.

(Type ∗ for multiplication; / for division; ∧ for exponentiation. The functions sqrt(x), ln(x), sin(x), etc. are known. Type e and pi for the mathematical constants e and π.)
dθ/dt = ?

please help me that I don't know how to start? Thank you in advance!

Answer 1

at time t,
tanθ = y/x = (10-t)/(10+t)
sec^2θ dθ/dt = -20/(10+t)^2
dθ/dt = -20/(10+t)^2 * 1/(1 + tan^2θ)
= -20/(10+t)^2 * 1/(1 + ((10-t)/(10+t))^2)
= -10/(t^2+100)