Q. 9 - Calculus

Question

Let $f(x, y)=xy$ and $\mathbf{u}=⟨\alpha ,\beta ⟩$ be a unit vector.

($a$) Use the definition of the directional derivative to find$D_{\mathrm{u}}f(-1,3)$

($b$) Assume that $\left|k\right|\ne 1$ and evaluate

$\underset{h\to 0}{\lim }\frac{(-1+(k\alpha \left)h\right)(3+(k\beta \left)h\right)-(-1·3)}{h}.$

($c$) Use your results from parts ($a$) and ($b$) to explain why it is necessary to use a unit vector in the definition of the directional derivative.

Step-by-Step Solution

Answer

Part ($b$) has an alternative answer for each value of $k$

Step1: Find the values.

Since,

$f(-1+\alpha h,3+\beta h)=(-1+\alpha h)·(3+\beta h)$

$=-1·3-\beta h+3·\alpha h+\alpha \beta h^2$

$=-3-\beta h+3·\alpha h+\alpha \beta h^2$                              ($2$)

And

$f\left(x_0,y_0\right)=f(-1,3)=-1·3$

$f(-1,3)=-3 \left(3\right)$

Step2: Find the equation.

By substituting equation ($2$) and ($3$) in equation ($1$), get

$\therefore D_{u}f(-1,3)={Lim}_{h\to 0}\frac{-3-\beta h+3·\alpha h+\alpha \beta h^2+3}{h}$

$\therefore D_{u}f(-1,3)={Lim}_{h\to 0}\frac{-\beta h+3·\alpha h+\alpha \beta h^2}{h}$

$\therefore D_{u}f(-1,3)={Lim}_{h\to 0}\frac{-\beta h}{h}+\frac{3·\alpha h}{h}+\frac{\alpha \beta h^2}{h}$

$\therefore D_{u}f(-1,3)={Lim}_{h\to 0}-\beta +3·\alpha +\alpha \beta h$

$\therefore D_{u}f(-1,3)=3\alpha -\beta$

$\left(b\right)$ Let $\left|k\right|\ne 1$ Assume that $u=(k\alpha ,k\beta )$

Let us find the directional derivative as;

$\therefore D_{u}f\left(x_0,y_0\right)={Lim}_{h\to 0}\frac{f\left(x_0+k\alpha h,y_0+k\beta h\right)-f\left(x_0,y_0\right)}{h}$

$\therefore D_{u}f(-1,3)={Lim}_{h\to 0}\frac{f(-1+k\alpha h,3+k\beta h)-f(-1,3)}{h} \left(4\right)$

Step3: Find the value.

Since$f(-1+k\alpha h,3+k\beta h)=(-1+k\alpha h)·(3+k\beta h)$

$=-1·3-k\beta h+3·k\alpha h+\alpha \beta k^2{h}^2$

$=-3-k\beta h+3k\alpha h+\alpha \beta k^2{h}^2 \left(5\right)$

And

$f\left(x_0,y_0\right)=f(-1,3)=-1.3$

$f(-1,3)=-3 \left(6\right)$

Step4: substituting equation.

By substituting equation ($5$) and ($6$) in equation ($4$), get

$\therefore D_{u}f(-1,3)={Lim}_{h\to 0}\frac{-3-k\beta h+3k\alpha h+\alpha \beta k^2{h}^2+3}{h}$

$\therefore D_{u}f(-1,3)={Lim}_{h\to 0}\frac{-k\beta h+3k\alpha h+\alpha \beta k^2{h}^2}{h}$

$\therefore D_{u}f(-1,3)={Lim}_{h\to 0}\frac{-k\beta h}{h}+\frac{3k\alpha h}{h}+\frac{\alpha \beta k^2{h}^2}{h}$

$\therefore D_{u}f(-1,3)={Lim}_{h\to 0}-k\beta +3k\alpha +\alpha \beta k^{2}h$

$\therefore D_{u}f(-1,3)=3k\alpha -k\beta$

$\left(C\right)$Because, the directional derivative for any function, depends upon the direction of $u$, not its magnitude. The answer in part ($b$) is different for each value of $k$