Q. 10 - Calculus

Question

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

($a$) Use the definition of the directional derivative to find $D_{u}f(0, \pi )$.

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

$\underset{h\to 0}{\lim }\frac{e^{k\alpha h}\sin (\pi +k\beta h)}{h}$.

(Hint:$\sin (A + B) = \sin A \cos B + \sin B \cos A$.)

($c$) 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

($a$) By using definition of directional derivative, $D_{u}f(0,\pi )=-\beta$.

($b$)by evaluate the function, $\underset{h\to 0}{\lim }\frac{e^{k\alpha h}\sin (\pi +k\beta h)}{h}$ is $-k\beta$.

($c$) Reason for using unit vector in definition of directional derivative, for every value of $k$, the solution in part $b$ partially is different

Step1: Use definition of directional derivative (part a )

The slope of a line that's tangent to the curve at a particular location is defined because the derivative. A further derivative definition is that the limit of the function's instantaneous rate of change because the time between observations approaches zero.

Then,

$f(0+\alpha h,\pi +\beta h)=e^{(0+\alpha h)}\times \sin (\pi +\beta h)$

$=e^{\alpha h}(\sin \pi \times \cos (\beta h)+\cos \pi \times \sin (\beta h\left)\right)$

and,

$f\left(x_0,y_0\right)=f(0,\pi )=e^0\sin \pi$

$f(0,\pi )=0$                $\left(\sin \pi \right)=0$

Substitute the values,

$\therefore D_{u}f(0,\pi )={Lim}_{h\to 0}\frac{e^{\alpha h}(\sin \pi \times \cos (\beta h)+\cos \pi \times \sin (\beta h\left)\right)-0}{h}$

$\therefore D_{u}f(0,\pi )={Lim}_{h\to 0}\frac{e^{\alpha h}\times \sin \pi \times \cos \left(\beta h\right)+e^{\alpha h}\times \cos \pi \times \sin \left(\beta h\right)}{h}$

$\therefore D_{u}f(0,\pi )={Lim}_{h\to 0}\frac{e^{\alpha h}\times 0\times \cos \left(\beta h\right)+e^{\alpha h}\times (-1)\times \sin \left(\beta h\right)}{h}$             $(\because \sin \pi =0;\cos \pi =-1)$

$\therefore D_{u}f(0,\pi )={Lim}_{h\to 0}\frac{-e^{\alpha h}\times \sin \left(\beta h\right)}{h}$

$\therefore D_{u}f(0,\pi )={Lim}_{h\to 0}\frac{-e^{\alpha h}\times \beta \times \sin \left(h\right)}{h}$

$\therefore D_{u}f(0,\pi )={Lim}_{h\to 0}-e^{\alpha h}\times \beta$                                  $\frac{\sin \left(h\right)}{h}=1$

$\therefore D_{u}f(0,\pi )=-e^{\alpha 0}·\beta$

$\therefore D_{u}f(0,\pi )=-\beta$

Step2: Evaluate the given function (part b ).

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

Then find the directional derivatives:

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

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

Then,

$f(0+k\alpha h,\pi +k\beta h)=e^{(0+k\alpha h)}\times \sin (\pi +k\beta h)$

$=e^{k\alpha h}(\sin \pi \times \cos (k\beta h)+\cos \pi \times \sin (k\beta h\left)\right)$

and,

$f\left(x_0,y_0\right)=f(0,\pi )=e^0\sin \pi$

$f(0,\pi )=0 \left(\sin \pi \right)=0$

Substitute thr values in equation,

$\therefore D_{u}f(0,\pi )={Lim}_{h\to 0}\frac{e^{k\alpha h}(\sin \pi \times \cos (k\beta h)+\cos \pi \times \sin (k\beta h\left)\right)-0}{h}$

$\therefore D_{u}f(0,\pi )={Lim}_{h\to 0}\frac{e^{k\alpha h}\times \sin \pi \times \cos \left(k\beta h\right)+e^{k\alpha h}\times \cos \pi \times \sin \left(k\beta h\right)}{h}$

$\therefore D_{u}f(0,\pi )={Lim}_{h\to 0}\frac{e^{k\alpha h}\times 0\times \cos \left(k\beta h\right)+e^{k\alpha h}\times (-1)\times \sin \left(k\beta h\right)}{h}$  $(\sin \pi =0, \cos \pi =-1)$

$\therefore D_{u}f(0,\pi )={Lim}_{h\to 0}\frac{-e^{k\alpha h}\times \sin \left(k\beta h\right)}{h}$

$\therefore D_{u}f(0,\pi )={Lim}_{h\to 0}\frac{-e^{k\alpha h}\times k\beta \times \sin \left(h\right)}{h}$

$\therefore D_{u}f(0,\pi )={Lim}_{h\to 0}-e^{\alpha h}\times k\beta$

$\therefore D_{u}f(0,\pi )={Lim}_{h\to 0}-e^{\alpha 0}\times k\beta$

$\therefore D_{u}f(0,\pi )=-e^{\alpha 0}\times k\beta$

$\therefore D_{u}f(0,\pi )=-k\beta$                                         $(e^0=1)$

Step3: Necessity to use a unit vector partly a and part b (part c ).

Because each function's directional derivative is set by the direction of $u$ instead of its magnitude. For every value of $k$, the solution $b$ is partially different.