# Rate of change in Vector Calc

Looking for some help with this vector calc review question:

In what direction at the point $(2,0)$ does the function $f(x,y) = xy$ have rate of change $-1$? Anad are there directions in which the rate is $-2$

Not quite sure how to approach this problem. Do I need to take the partial derivatives of $f(x,y)$ which would be $fx = y$ and $fy = x$ and then plug my values in at the point $(2,0)$ just looking for some help

## 1 Answer

Note that it asks "in what direction" hinting at a directional derivative. For that we need the gradient of $f$ which is $$\nabla f=<y,x>$$ Then we are looking for a unit vector such that \begin{align} \nabla f(2,0)\cdot\hat u &= <0,2>\cdot\hat u \\ &= -1 \end{align} Since $\hat u$ is a unit vector it is of the form $$\hat u=\frac{<x,y>}{\sqrt{x^2+y^2}}$$ Plugging this into the above equation we find that the components of $\hat u$ must satisfy $$y=-\frac{1}{2}\sqrt{x^2+y^2}\qquad (1)$$ Here $x$ can be thought of as a free variable, so we solve for $y$ in terms of it. Doing this we get $$y=\frac{x}{\sqrt{3}}\qquad x<0$$ Or we could say that $y=-|x|/\sqrt{3}$, this is because $y$ must be negative. Plugging in this definition of $y$ into $\hat u$ we find that \begin{align} \hat u &=\frac{<x,-|x|/\sqrt{3}>}{\sqrt{x^2/3+x^2}} \\ &=\frac{<x,-|x|/\sqrt{3}>}{2|x|(1/\sqrt{3})} \\ &=\left<\frac{\sqrt{3}}{2}\cdot\frac{x}{|x|},-\frac{1}{2}\right> \\ &= \frac{1}{2}\left<\pm\sqrt{3},-1\right> \end{align} Are the directions in which the rate of change of $f$ at $(2,0)$ are $-1$. A similar process will show that the rate of change is $-2$ in the direction $$\hat v=<0,-1>$$

• why do i use v = <0,-1> for the rate of change being -2 – mp12345 Jul 24 '16 at 17:04
• Follow the same steps. You are looking for a unit vector $\hat u$ so that $$<0,2>\cdot\hat u=-2$$ This means equation $(1)$ becomes $y=-\sqrt{x^2+y^2}$ which implies $x=0$ and $y=-|y|$. Then $\hat u$ is of the form $$\hat u=\frac{<0,-|y|>}{\sqrt{y^2}}=\frac{<0,-|y|>}{|y|}$$ – Will Fisher Jul 24 '16 at 17:15
• okay awesome so what do i plug in for y then? – mp12345 Jul 24 '16 at 17:21
• You can have any $y\in\mathbb{R}$ but notice how no matter what that simplifies to $\hat u=<0,-1>$ – Will Fisher Jul 24 '16 at 17:23
• okay so the answer wouls just be \$ u = <0,-1> for the rate of changing being -2 – mp12345 Jul 24 '16 at 18:04