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A particle is moving along the curve $y= 2 \sqrt{2 x + 2}$. As the particle passes through the point $(1, 4)$, 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.

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    $\begingroup$ Welcome to MSE! Do you have any thoughts and can share what you have tried? Regards $\endgroup$
    – Amzoti
    Commented Nov 5, 2013 at 16:43

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Hint: The distance from the origin is given in general by $$D=\sqrt{x^2+y^2}=\sqrt{x^2+4(2x+2)}=\sqrt{x^2+8x+4}.$$ You're trying to find $\frac{dD}{dt}$ when $x=1,$ given that $\frac{dx}{dt}=2$ units per second when $x=1$. Any ideas how you might do that?

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  • $\begingroup$ the next step could be 2DdDdt=4+2ydydt $\endgroup$
    – user105635
    Commented Nov 5, 2013 at 17:17
  • $\begingroup$ I think you are trying to say $$2D\frac{dD}{dt}=4+2y\frac{dy}{dt},$$ but I'm not sure why you're doing this, as we don't know anything about the rate of change of $y$, nor have we written $D$ in terms of $y$. Incidentally, if you want to see how something is formatted, just right-click on it, go to "Show Math As," then click "TeX Commands," and a window will pop up to show you how it was done. Put the result between $ $, and you'll get the same result. For example, $\frac{dD}{dt}$ gives you $\frac{dD}{dt}$. $\endgroup$ Commented Nov 5, 2013 at 17:22
  • $\begingroup$ im really bad at diffentiation $\endgroup$
    – user105635
    Commented Nov 5, 2013 at 18:00
  • $\begingroup$ we i got the value of $\frac{dy}{dt}$=2 now i have the value of $\frac{dx}{dt}$ and $\frac{dy}{dt}$ $\endgroup$
    – user105635
    Commented Nov 5, 2013 at 18:01
  • $\begingroup$ so how would i find the value of $\frac{dD}{dt}$ $\endgroup$
    – user105635
    Commented Nov 5, 2013 at 18:02

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