Q. 7 - Calculus

Question

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

(a) Lee the definition of the directional derivative to find $D_{u}f(1,2)$.

(b) Explain why $⟨ka,k\beta )$ is a unit vector-only when $\left|k\right|=1$

(c) Assume that $\left|k\right|\ne 1$ and evaluate the limit

$\underset{h\to 0}{\lim }\frac{(1+(k\alpha \left)h\right)+(2+(k\beta \left)h\right)-(1+2)}{h}$

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

Step-by-Step Solution

Answer

Part ($c$) has a distinct answer for each value of $k$.

Step1: Find the alpha.

Since;

$f(1+\alpha h,2+\beta h)=(1+\alpha h)+(2+\beta h)$
$=1+\alpha h+2+\beta h$

$=3+h(\alpha +\beta )$                              $\left(2\right)$

And

$f\left(x_0,y_0\right)=f(1,2)=1+2$

$f(1,2)=3$

Substituting equation ($2$) and ($3$) in equation ($1$), get

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

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

$\therefore D_{u}f(1,2)=\alpha +\beta$

Step2: Unit vector.

$\left(b\right)$Given $u=(\alpha ,\beta )$ is a unit vector means,

$\left|\hat{u}\right|=\sqrt{{\alpha }^2+{\beta }^2}=1$

When $u=(k\alpha ,k\beta )$ then $u$ will be the unit vector-only when $\left|k\right|=1$ because

$\left|\hat{u}\right|=\sqrt{k^2{\alpha }^2+k^2{\beta }^2}=\sqrt{1^2{\alpha }^2+1^2{\beta }^2}$

$\left|\hat{u}\right|=\sqrt{{\alpha }^2+{\beta }^2}=1$

Step3: Find the directional derivative.

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

Let us find the directional derivative as;

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

$\therefore D_{u}f(1,2)={Lim}_{h\to 0}\frac{f(1+k\alpha h,2+k\beta h)-f(1,2)}{h}$                         $\left(4\right)$

Since;

$f(1+k\alpha h,2+k\beta h)=(1+k\alpha h)+(2+k\beta h)$

$=1+k\alpha h+2+k\beta h$

$=3+kh(\alpha +\beta )$                                                                                             $\left(5\right)$

And

$f\left(x_0,y_0\right)=f(1,2)=1+2$

$f(1,2)=3$                                                                                                          $\left(6\right)$

Step4: Equation of substituting.

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

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

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

$\therefore D_{u}f(1,2)=k(\alpha +\beta )$

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