Let $f ( x , y ) = \ln \sqrt { x ^ { 2 } + y ^ { 2 } }$ . Then the directional derivative of f at (3,4) in the direction of the unit vector $\mathbf { U }$ which is parallel to $\mathbf { v } = 5 \mathbf { i } + 12 \mathbf { j }$ is

A) $\frac { \sqrt { 3 } + 1 } { 2 }$
B) $\sqrt { 15 }$
C) $- \frac { 7 } { 13 }$
D) $- \frac { 5 } { 13 }$
E) $\frac { 63 } { 325 }$

Question

Let

f ( x , y ) = \ln \sqrt { x ^ { 2 } + y ^ { 2 } }

. Then the directional derivative of f at (3,4) in the direction of the unit vector

\mathbf { U }

which is parallel to

\mathbf { v } = 5 \mathbf { i } + 12 \mathbf { j }

is

A) $\frac { \sqrt { 3 } + 1 } { 2 }$
B) $\sqrt { 15 }$
C) $- \frac { 7 } { 13 }$
D) $- \frac { 5 } { 13 }$
E) $\frac { 63 } { 325 }$

Quizplus · Accepted Answer

The directional derivative of $f$ at a point $(x_0, y_0)$ in the direction of a unit vector $\mathbf{U}$ is given by $\nabla f(x_0, y_0) \cdot \mathbf{U}$, where $\nabla f$ is the gradient of $f$. For $f(x, y) = \ln\sqrt{x^2 + y^2}$, we first find the gradient $\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)$. The partial derivatives are $\frac{\partial f}{\partial x} = \frac{x}{x^2 + y^2}$ and $\frac{\partial f}{\partial y} = \frac{y}{x^2 + y^2}$. At $(3,4)$, $\nabla f(3,4) = \left(\frac{3}{25}, \frac{4}{25}\right)$.The unit vector $\mathbf{U}$ in the direction of $\mathbf{v} = 5\mathbf{i} + 12\mathbf{j}$ is $\mathbf{U} = \frac{1}{\sqrt{5^2 + 12^2}}(5\mathbf{i} + 12\mathbf{j}) = \frac{1}{13}(5\mathbf{i} + 12\mathbf{j})$.The directional derivative is then $\nabla f(3,4) \cdot \mathbf{U} = \left(\frac{3}{25}, \frac{4}{25}\right) \cdot \left(\frac{5}{13}, \frac{12}{13}\right) = \frac{3 \cdot 5 + 4 \cdot 12}{25 \cdot 13} = \frac{15 + 48}{325} = \frac{63}{325}$.

Let f(x,y)=ln⁡x2+y2f ( x , y ) = \ln \sqrt { x ^ { 2 } + y ^ { 2 } }f(x,y)=lnx2+y2​

Let $f ( x , y ) = \ln \sqrt { x ^ { 2 } + y ^ { 2 } }$