For the function $f ( x , y ) = x ^ { 2 } y ^ { 3 }$ find a unit vector in the direction of the steepest increase at the point (a, b)= (1, 1).
A) $\vec { u } = \frac { 2 } { \sqrt { 13 } } \vec { i } - \frac { 3 } { \sqrt { 13 } } \vec { j }$
B) $\vec { u } = 2 \vec { i } + 3 \vec { j }$
C) $\vec { u } = \frac { 2 } { \sqrt { 13 } } \vec { i } + \frac { 3 } { \sqrt { 13 } } \vec { j }$
D) $\vec { u } = \frac { 2 } { \sqrt { 13 } } \vec { i } + \frac { 6 } { \sqrt { 13 } } \vec { j }$

Question

Quizplus · Accepted Answer

The direction of the steepest increase of a function $f(x, y)$ at a point is given by the gradient vector $\nabla f$, and the unit vector in that direction can be found by dividing the gradient by its magnitude. For $f(x, y) = x^2y^3$, the gradient is $\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (2xy^3, 3x^2y^2)$. At the point (1, 1), this becomes $(2, 3)$. The magnitude of this vector is $\sqrt{2^2 + 3^2} = \sqrt{13}$, so the unit vector in the direction of the steepest increase is $\left(\frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}}\right)$.

For the Function f(x,y)=x2y3f ( x , y ) = x ^ { 2 } y ^ { 3 }f(x,y)=x2y3

For the Function $f ( x , y ) = x ^ { 2 } y ^ { 3 }$