A particle is moving along the curve y=4 times square root of (4x+5).As the particle passes through the point (5, 20), 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.

Question

anonymous · Answer

Where did you get 0 from?

anonymous · Answer

Why was the coordinate point given? Extra information?

anonymous · Answer

am I plugging in 5 for x?

anonymous · Answer

The 0's come from the origin, $(0,0)$. Review the distance formula if you're still not sure.

The coordinate point gives you everything you need. However, if you were to use the y-coordinate you would have needed some information about the rate of change of that coordinate. You weren't, so we improvise; instead of using an exact number, we use the given equation.

And yes, $x=5$ and $\dfrac{dx}{dt}=3$.

anonymous · Answer

I am getting 27/250. It is coming out wrong

anonymous · Answer

Oops, tiny mistake.
$$y=4\sqrt{4x+5}$$
Let $(x,y)$ be a point that travels along the curve. The point will be equivalent to $\left(x,4\sqrt{4x+5}\right)$.

The distance between any point on the curve and the origin will be
$$D=\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+16(4x+5)}=\sqrt{x^2+64x+80}$$
Now do some implicit differentiation:
$$\frac{dD}{dt}=\frac{1}{2}\left(x^2+64x+80\right)^{-1/2}\left(2x+64\right)\frac{dx}{dt}$$
Plug in your given info.

anonymous · Answer

Now it worked