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?


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

$$ 1 = \frac{d z}{d t} = \frac{d z}{d x} \,\frac{d x}{d t} \implies \frac{d x}{d t} = \frac{1}{dz/dx} $$


$$ \frac{d z}{d x} = \left(\sqrt{x^2 + x^4}\right)' =\frac12\, \frac{2x+4x^3}{\sqrt{x^2 + x^4}} = \frac{1+2x^2}{\sqrt{1+x^2}} $$

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.


I think this is simpler to understand. By definition


Since $y=x^2$


Taking the derivative of this equation using the chain rule gives


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


Then substitue $x=2$ and $\frac{dx}{dt}=\frac{\sqrt{20}}{18}$ to get your answer.


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