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A particle is moving along the curve $y=2\sqrt{5x+11}$. As the particle passes through the point $(5,12)$, its $x$-coordinate increases at a rate of 2 units per second. Find the rate of change of the distance from the particle to the origin at this instant.

I got $\frac{298}{13}$ which is incorrect. I think what I am having the most trouble with is finding the $\frac{dy}{dt}$ value.

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    $\begingroup$ try to include your attempt so that we can point out your mistake? $\endgroup$
    – Siong Thye Goh
    Jan 25, 2019 at 2:20

2 Answers 2

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Find r, the rate of change for the y axis and calculate the magnitude of the sum of those two vectors, $\sqrt{r^2 + 2^2}$.

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You are given $\frac{dx}{dt}$. You need $\frac{d\sqrt{x^2+y^2}}{dt}=\frac{1}{\sqrt{x^2+y^2}}\left(x\frac{dx}{dt}+y\frac{dy}{dt}\right)$ You want $\frac{dy}{dt}$ to be able to continue. You can compute $\frac{dy}{dx}$ and you know $\frac{dx}{dt}$. Can you put this together?

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