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?
On
$$ Gradient f = <-8x,-2y>$$
$$ \frac {dx}{dt} = 8x$$ $$ \frac {dy}{dt} = 2y$$
$$ \frac {dy}{dx} = \dfrac{\frac {dy}{dt}}{\frac {dx}{dt}}$$
$$\frac {dy}{2y} = \frac {dx}{8x}$$ $$\frac {dy}{y} = \frac {dx}{4x}$$ 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 $$ln(y) = \frac {1}{4}ln(x) + ln(2)$$
$$ln(y) = ln(2.x^{\frac{1}{4}})$$
$$ y = 2.x^{\frac{1}{4}}$$
Note that any directional partial derivative can be expressed as: $$ \frac{\partial f}{\partial l}=f_x\cos \alpha +f_y\sin \alpha $$ where $\alpha$ is the directional angle of $l$.
Based on this, we apply Cauchy-Schwarz's inequality:
$$ |\langle x,y\rangle|\le ||x||\ ||y|| $$ Plug [$f_x$, $f_y$] and [$\cos \alpha$, $\sin\alpha$] into x and y, respectively.
the = sign holds iff x and y are linearly dependent. More specifically, if x=ky,where k>=0,
$$ \langle x,y\rangle=||x||\ ||y||=||x|| $$
if k<0; then
$$ \langle x,y\rangle=- ||x||\ ||y||=-||x|| $$ note that $||y||=\sqrt{\cos^2\alpha+\sin^2\alpha}=1$.
the first case is the gradient case(steepest increasing), the latter is the opposite gradient case(steepest decreasing).