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?
 A: Bu definition,
$ z = \sqrt{\left(x-0\right)^2 +\left(y-0\right)^2 } = \sqrt{x^2 + y^2}$. 
First, let us compute $dx/dt:$
Since $y = x^2$ we can rewrite $z$ in the form $ z = z(x,y) = \sqrt{x^2 + x^4}$.
Then by chain rule we have 
$$
1 = \frac{d z}{d t} = \frac{d z}{d x} \,\frac{d x}{d t}  \implies 
\frac{d x}{d t} = \frac{1}{dz/dx} 
$$
Now,
$$
\frac{d z}{d x}  = \left(\sqrt{x^2 + x^4}\right)' 
=\frac12\, \frac{2x+4x^3}{\sqrt{x^2 + x^4}} = \frac{1+2x^2}{\sqrt{1+x^2}}
$$
Substituting $x=2$ into expression above will give you the speed of $x$ at the point $\left(2,4\right)$.
Second, $dy/dt$ can be computed in the same way. Hope you can pick it from here.

Alternatively, one can compute $dy/dt$ in the following manner:
$$
1 = \frac{d z}{d t} = \frac{d z}{d x} \,\frac{d x}{d t} + \frac{d z}{d y} \,\frac{d y}{d t}  
= \frac{x}{\sqrt{x^2 + y^2}}\,\frac{d x}{d t} + \frac{y}{\sqrt{x^2 + y^2}}\,\frac{d y}{d t}
\\ \implies 
\frac{d y}{d t} = \frac{\sqrt{x^2 + y^2}}{y} \left( 1 - \frac{x}{\sqrt{x^2 + y^2}}\,\frac{d x}{d t}\right) 
= \frac{\sqrt{x^2 + y^2}}{y} - \frac{x}{y}\,\frac{d x}{d t}, 
$$
where $dx/dt$ is known from the first part.
A: 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.
