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A bug is moving along the right side of the parabola y=x^2 at a rate such that its distance from the origin is increasing 3 cm/min. At what rate are the x- and y-coordinates of the bug increasing when the bug is at the point (1,1)?

D^2 = y + y^2

2 dD/dt = dy/dt + 2y dy/dt

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    $\begingroup$ Hint: $D^2=x^2+y^2$. Differentiate immediately with respect to $t$, using implicit differentiation. Alternately, use $D^2=x^2+x^4$ and $D^2=y+y^2$. $\endgroup$ Commented Oct 13, 2015 at 4:12

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You're on the right track, but the left hand side of your equation should be $2D\, (dD/dt)$. Solving for $dy/dt$, we get

$$\frac{dy}{dt} = \frac{2D}{1 + 2y}\frac{dD}{dt}\tag{*}$$

It's given $x = 1$, $y = 1$, and $dD/dt = 3$ and $y = 1$, so $D = \sqrt{2}$ and you can find the value of $dy/dt$. Since $x = 1$ and $dy/dt = 2x\, dx/dt$ by implicit differentiation, you can find the value of $dx/dt$.

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