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
Best Answer
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 $dx/dt$.