Find the directional derivative of the function $f ( x , y ) = x ^ { 2 } + y ^ { 2 }$ at the point $( 1,1 )$ in the direction $\theta = \frac { \pi } { 4 }$ .
A) $\sqrt { 2 }$
B) $2 \sqrt { 2 }$
C) $4 \sqrt { 2 }$
D) $\frac { 1 } { \sqrt { 2 } }$
E)1
F)2
G)4
H)
$\frac { 1 } { 2 }$

Question

Find the directional derivative of the function $f ( x , y ) = x ^ { 2 } + y ^ { 2 }$ at the point $( 1,1 )$ in the direction $\theta = \frac { \pi } { 4 }$ .
A) $\sqrt { 2 }$ 
B) $2 \sqrt { 2 }$
C) $4 \sqrt { 2 }$
D) $\frac { 1 } { \sqrt { 2 } }$
E)1
F)2
G)4
H)
$\frac { 1 } { 2 }$

Quizplus · Accepted Answer

The directional derivative of a function $f(x, y)$ at a point $(x_0, y_0)$ in the direction of a unit vector $\mathbf{u} = \langle \cos \theta, \sin \theta \rangle$ is given by $D_{\mathbf{u}}f = f_x(x_0, y_0) \cos \theta + f_y(x_0, y_0) \sin \theta$, where $f_x$ and $f_y$ are the partial derivatives of $f$ with respect to $x$ and $y$, respectively. For $f(x, y) = x^2 + y^2$, we have $f_x = 2x$ and $f_y = 2y$. At the point $(1,1)$ and direction $\theta = \frac{\pi}{4}$, we have $\cos \theta = \sin \theta = \frac{1}{\sqrt{2}}$. Thus, $D_{\mathbf{u}}f = 2(1) \cdot \frac{1}{\sqrt{2}} + 2(1) \cdot \frac{1}{\sqrt{2}} = 2\sqrt{2}$.

Find the Directional Derivative of the Function f(x,y)=x2+y2f ( x , y ) = x ^ { 2 } + y ^ { 2 }f(x,y)=x2+y2

Find the Directional Derivative of the Function $f ( x , y ) = x ^ { 2 } + y ^ { 2 }$