Please explain how to solve this problem

Question

A point P is moving along the line whose equation is y=2x. How fast is the distance between P and the point

(3, 0) changing at the instant when P is at (3, 6) if x is decreasing at the rate of 2 units/s at that instant?

Answer 1

The distance between P on the curve and the point (3,0) is D and

D = sqrt{ (y-0)² + (x-3)² } = sqrt{ y² + (x-3)² } = sqrt{ 4x² + x² -6x +9 } = sqrt(5x² -6x + 9 }

The rate of change dD/dt = { (10x-6)/(2sqrt(5x²-6x+9)) }dx/dt

When x = 3, & dx/dt = -2 then dD/dt = -4 units/sec

Answer 2

The distance, D, from a point (m.n) and a line Ax+By+C=0 is given by:

D

The distance between any two points (x₁,y₁) and (x₂,y₂) points is

√[(x₂-x₁)²+(y₂-y₁)²]

In this case (x,y)=(x,2x)=(x₁,y₁) is P and (x₂,y₂)=(3,0).

So

D=√[(3-x)²+(0-2x)²]=√[(3-x)²+4x²]

=√[9-6x+x²+4x²]=

=√[5x²-6x+9]

Now dD/dt=(1/2)(5x²-6x+9)^-1/2(10x-6)dx/dt

=[1/((5x²-6x+9)^1/2](5x-3)dx/dt=[(5x-3)/(5x²-6x+9)^1/2](dx/dt)

We know dx/dt = -2 units/sec, and, if we substitute x = 3

dD/dt=(-2)(15-3)/[5(9)-18+9]^1/2=-2(12)/√(45-18+9)=-24/√36=-24/√[(4)(9)]=-24/6=-4 units/sec

Solution: Dd/dt=-4 units/sec