An object travelling on $x^2$ I have an object traveling along the curve $y=x^2$. $z$ is the distance from the origin and $dz/{dt}=1$ is the rate it's increasing per unit time. At what rate are my $x$ and $y$'s moving at the point (2,4)? In other words what is $dy/dt$ and $dx/dt$.
I've seen the geometry. Found $z$ as the hypotenuse of $x$ and $y$. Also, I infer that $dz=dt $ so they are interchangable. How do I solve this?

I think this is simpler to understand. By definition
$$z=\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+y^2}$$
Since $y=x^2$
$$z=\sqrt{x^2+x^4}$$
Taking the derivative of this equation using the chain rule gives
$$z^{'}=\left({(x^2+x^4)}^{1/2}\right)^{'}\implies\frac{dz}{dt}=1/2(x^2+x^4)^{1/2-1}(x^2+x^4)^{'}\implies\frac{dz}{dt}=(1/2)(x^2+x^4)^{-1/2}(2x+4x^3)\frac{dx}{dt}$$
Then substitue $\frac{dz}{dt}=1$ and $x=2$ into the previous equation
This gives $1=\frac{1}{2}\frac{1}{\sqrt{20}}(36)\frac{dx}{dt}\implies1=\frac{18}{\sqrt{20}}\frac{dx}{dt}\implies\frac{dx}{dt}=\frac{\sqrt{20}}{18}$
Then take the derivative of $y=x^2$ on both sides to get
$$\frac{dy}{dt}=2x\frac{dx}{dt}$$
Then substitue $x=2$ and $\frac{dx}{dt}=\frac{\sqrt{20}}{18}$ to get your answer.
