# 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.

## 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

## 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

## Well, well, well, looks like the particle is on a joyride on the curve! Let's see what we can do.

The distance from the particle to the origin can be found using the good old Pythagorean theorem. We'll call this distance "d" just to make it fancy.

Now, since the particle is moving along the curve, we need to find an equation that relates the x-coordinate to the distance d. We can do this by using the equation of the curve and some basic trigonometry.

We know that y^2 = 2dx + 2d, where d is the distance from the origin to the particle.

Plugging in the equation of the curve, y= 4 sqrt{2 x + 2}, we get (4 sqrt{2 x + 2})^2 = 2dx + 2d.

Simplifying that, we get 16(2 x + 2) = 2dx + 2d.

Now, let's differentiate both sides of this equation with respect to time, t, because why not? Since x is changing with time, it's the most logical choice.

So, we have 32(dx/dt) = 2(dx/dt)x + 2(d/dt).

Now, we need to find dx/dt. We know that dx/dt is increasing at a rate of 5 units per second. So, let's plug that in.

32(5) = 2(5)x + 2(dy/dt).

Simplifying that, 160 = 10x + 2(dy/dt).

Rearranging the equation, we get dy/dt = (160 - 10x)/2.

But hold your horses, we're not done yet. We need to find the rate of change of the distance from the particle to the origin at the point (1, 8).

So, let's substitute x = 1 into our equation.

dy/dt = (160 - 10(1))/2 = (150)/2 = 75.

So, the rate of change of the distance from the particle to the origin at this instant is 75 units per second.

Now, wasn't that a wild ride on the curve? I hope you enjoyed it!

## To find the rate of change of the distance from the particle to the origin, we need to find an expression for the distance between the particle and the origin and then differentiate it with respect to time.

The distance between two points (x1, y1) and (x2, y2) in a two-dimensional plane is given by the formula:

Distance = √[(x2 - x1)^2 + (y2 - y1)^2]

In this case, the particle is moving along the curve y = 4√(2x + 2) and we know that it passes through the point (1, 8). Therefore, we can consider the particle's position as (x, 4√(2x + 2)).

The distance between the particle's position and the origin (0, 0) is:

Distance = √[(x - 0)^2 + (4√(2x + 2) - 0)^2]

Simplifying this expression, we get:

Distance = √[x^2 + 16(2x + 2)]

= √[x^2 + 32x + 64]

Now, we can differentiate this expression with respect to time (t) using the chain rule:

d(Distance)/dt = d(√[x^2 + 32x + 64])/dt

= (1/2) * (x^2 + 32x + 64)^(-1/2) * d(x^2 + 32x + 64)/dt

= (1/2) * (x^2 + 32x + 64)^(-1/2) * (2x + 32) * dx/dt

Note that dx/dt represents the rate at which the x-coordinate is changing, which is given as 5 units per second.

Now, we need to evaluate this expression at the point (1, 8):

d(Distance)/dt = (1/2) * (1^2 + 32(1) + 64)^(-1/2) * (2(1) + 32) * 5

= (1/2) * (97)^(-1/2) * (34) * 5

= (17/97) * 34 * 5

= 17 * 2 * 5

= 170 units per second

So, the rate of change of the distance from the particle to the origin at this instant is 170 units per second.