A particle is moving around the ellipse 4x^2+16y^2=64 . At any time t, its x- and y coordinates are given by x(t) = 4cos t and y(t)= 2sint . At what rate is the particle’s distance to the point (1,1) changing at any time t? At what rate is the distance changing when t=pi/6?

Answer 1

The distance z is
z^2 = (x-1)^2 + (y-1)^2

2z dz/dt = 2(x-1) dx/dt + 2(y-1) dy/dt
At t=π/6, z=√((4(√3/2)-1)^2+(2(1/2)-1)^2) = 2√3-1
Note that at t=π/6, the point on the curve is (2√3,1)

So, now we just have
dx/dt = -4sint = -2
dy/dt = 2cost = √3

(2√3-1) dz/dt = (2√3-1)(-2)+(1-1)(√3)
dz/dt = -2