# Rate of change of distance from particle (on a curve) to origin

A particle is moving along the curve $y = 2\sqrt{4x + 9}$ As the particle passes through the point (4, 10) its x-coordinate increases at a rate of 3 units per second. Find the rate of change of the distance from the particle to the origin at this instant.

Okay. Rate of change of the distance from the particle to the origin.

So the origin is going to be the point (0,y). So: $y = 2\sqrt{4(0)+9} = 6$. The point (0,6) is the origin, then.

Now the problem is asking us for the rate of change of the distance between these two points, we recall that:

$D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

So, now we have to differentiate:

$$d' = \frac{1}{2}\cdot ((x_2 - x_1)^2 + (y_2 - y_1)^2)^\frac{-1}{2} \cdot \left[ 2(x_2'-x_1') + 2(y_2'-y_1') \right]$$

Then we have to substitute.

But I don't know if I'm even correct thus so far. Can someone help?

The distance to the origin when the particle is at $(x,y)$ is given by $D(x,y)=\sqrt{x^2+y^2}$.
We want $\frac{dD}{dt}$ at a certain instant. I prefer to work with $D^2$. So we have $$D^2=x^2+y^2.$$ Differentiate, using the Chain Rule. We have $$2D\frac{dD}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}.\tag{1}.$$
We know that $y=2\sqrt{4x+9}$. So $$\frac{dy}{dt}=\frac{4}{\sqrt{4x+9}}\frac{dx}{dt}.\tag{2}$$
Now "freeze" things at the instant when $x=4$. We know $\frac{dx}{dt}$ at this instant. We also know $\frac{dy}{dt}$, by (2). We also know $y$ and therefore $D$. Now we can use (1) to find $\frac{dD}{dt}$ at this instant.
• My arithmetic is not necessarily reliable. I get $\frac{36}{\sqrt{116}}$. Oct 10, 2013 at 2:11