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

Question

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.

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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.

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

Answer 1

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.