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A particle is moving along the curve $y=3\sqrt{5x+6}$. As the particle passes through the point $(2,12)$, 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.

I understand that $\dfrac{\mathrm dx}{\mathrm dt} = 3$, that $\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\mathrm dy/\mathrm dt}{\mathrm dx/\mathrm dt}$, and that $\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\mathrm dy}{\mathrm dt} \cdot \dfrac{1}{3}$.

I also know that the derivative of $y = \sqrt{x}$ is $\dfrac{1}{2\sqrt{x}}$. I then get $\dfrac{dy}{dt} = \dfrac{15}{2\sqrt{5x+6}} \cdot \dfrac{dx}{dt}$. What steps do I take next? Did I make any errors in reasoning? Would 45/8 be correct as the $\dfrac{dy}{dt}$ term? I tried the suggestion in the answers, but it doesn't work. What do I do next? If someone provides a full step-by-step summary of how to solve, I might even catch my own mistake.

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up vote 3 down vote accepted

If the question was asking about the rate of change in the distance from the particle to the $x$-axis, you'd be right on track. That's not what they want, though.

The distance from the particle to the origin is $$D=\sqrt{x^2+y^2}=\sqrt{x^2+\left(3\sqrt{5x+6}\right)^2}=\sqrt{x^2+9(5x+6)}=\sqrt{x^2+45x+54}.$$ The question wants $\frac{dD}{dt}$, i.e. $$\frac{dD}{dx}\cdot\frac{dx}{dt},$$ when $x=2$. Can you take it from there?

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I tried. Maybe I made a simplification error. I multiplied my answer by 3, the value of dx/dt, but it is still incorrect. (at least WebWorK marks it as incorrect). –  cuabanana Mar 29 '13 at 19:19
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We should get $$\frac{dD}{dx}=\frac{2x+45}{2\sqrt{x^2+45x+54}},$$ and so when $x=2,$ we have $$\frac{dD}{dt}=\frac{dD}{dx}\cdot\frac{dx}{dt}=\frac{49}{4\sqrt{37}}\cdot 3=\frac{147\sqrt{37}}{148}.$$ –  Cameron Buie Mar 29 '13 at 20:14
    
Thank's I tried it . It's the correct answer. –  cuabanana Mar 30 '13 at 23:28

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