Path of steepest descent I'm given an elevation on a hill by $f(x,y)= 200 - y^2 -4x^2$. Now I have to show that from the point (1,2) the path of steepest descent is $y=2x^{1/4}$ as it travels down the hill.
I attempted it by finding the partial derivatives of x and y.
So I have $df/dx = -8x$ and $df/dy = -2y$.
I also know that the path of steepest descent is in the opposite direction of the gradient, so the signs would be switched.
I have no idea how to prove this from here though.
Any suggestions?

$$ Gradient f = <-8x,-2y>$$
$$ \frac {dx}{dt} = 8x$$
$$ \frac {dy}{dt} = 2y$$
$$ \frac {dy}{dx} = \dfrac{\frac {dy}{dt}}{\frac {dx}{dt}}$$
$$\frac {dy}{2y} = \frac {dx}{8x}$$
$$\frac {dy}{y} = \frac {dx}{4x}$$
Integrating, we get
$$ ln(y) = \frac {1}{4}ln(x) + C$$
Evaluate at (1,2) => ln2 = C
Substituting back
we get the path of steepest descent
$$ln(y) = \frac {1}{4}ln(x) + ln(2)$$
$$ln(y) = ln(2.x^{\frac{1}{4}})$$
$$ y = 2.x^{\frac{1}{4}}$$
