A particle is moving along the curve y= 4 sqrt{2 x + 2}. As the particle passes through the point (1, 8), its x-coordinate increases at a rate of 5 units per second. Find the rate of change of the distance from the particle to the origin at this instant.

Answer 1

y = 4√(x+2)
The distance from the particle to (0,0) is

d = √(x^2+y^2) = √(x^2 + 16(x+2))
= √(x^2+32x+16)

At(1,8), d =√65

Since d^2 = x^2+32x+16,

2d dd/dt = (2x+32) dx/dt
At (1,8), then

2√65 dd/dt = 34*5 = 170
dd/dt = 85/√65

Answer 2

dy/dt = dy/dx * dx/dt

dy/dx = 4 (1/2) (2x+2)^-.5 (2)
= 4(2x+2)^-.5
dx/dt here = 5
so
dy/dt = 20 (2x+2)^-.5 = 20/2 = 10

so velocity vector at (1,8) is V =5 i + 10 j
the radius vector here is R =1 i + 8 j

I want the component of the velocity vector in the direction of the radius vector or |V|cos (angle T from V to R)
that is V dot R /|R| = |V||R|cos T/|R|

V dot R = 5 + 80 = 85
|R| = sqrt (1+64) = 8.06

so 85/8.06 = 10.5 units/second

Answer 3

LOL, math versus physics, round 4596