Compute the gradient of the function at the given point.
-$$f ( x , y , z ) = \tan ^ { - 1 } \frac { 7 x } { 9 y + 3 z } , ( 9,0,0 )$$
A)${\nabla} f = \frac { 1 } { 7 } \mathrm { j } - \frac { 1 } { 21 } \mathbf { k }$
B) ${\nabla} f = - \frac { 1 } { 21 } j + \frac { 1 } { 7 } k$
C) ${\nabla} f = - \frac { 1 } { 7 } \mathbf { j } - \frac { 1 } { 21 } \mathbf { k }$
D) ${\nabla} f = - \frac { 1 } { 21 } j \mathbf { j } - \frac { 1 } { 7 } k$

Question

Compute the gradient of the function at the given point.
-$$f ( x , y , z ) = \tan ^ { - 1 } \frac { 7 x } { 9 y + 3 z } , ( 9,0,0 )$$
A)${\nabla} f  = \frac { 1 } { 7 } \mathrm { j } - \frac { 1 } { 21 } \mathbf { k }$
B) ${\nabla} f = - \frac { 1 } { 21 } j + \frac { 1 } { 7 } k$
C) ${\nabla} f  = - \frac { 1 } { 7 } \mathbf { j } - \frac { 1 } { 21 } \mathbf { k }$
D) ${\nabla} f  = - \frac { 1 } { 21 } j \mathbf { j } - \frac { 1 } { 7 } k$

Quizplus · Accepted Answer

The gradient of $f$ at a point is given by $\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)$. For the given function, the partial derivatives with respect to $y$ and $z$ are negative because increasing $y$ or $z$ in the denominator of the fraction inside the arctan function decreases the value of $f$, hence the negative signs in the gradient components corresponding to $y$ and $z$. The absence of $x$ in the options implies its derivative at the point does not contribute to the gradient in the context of the provided choices.

Compute the Gradient of the Function at the Given Point f(x,y,z)=tan⁡−17x9y+3z,(9,0,0)f ( x , y , z ) = \tan ^ { - 1 } \frac { 7 x } { 9 y + 3 z } , ( 9,0,0 )f(x,y,z)=tan−19y+3z7x​,(9,0,0)

Compute the Gradient of the Function at the Given Point $f ( x , y , z ) = \tan ^ { - 1 } \frac { 7 x } { 9 y + 3 z } , ( 9,0,0 )$