Question

Find the direction in which the function f= x^2-y^2+2xy decreases more rapidly at the point (1,1)

Answer 1

Answer on Question # 41662 – Math – Vector Calculus

Find the direction in which the function $f = x^2 - y^2 + 2xy$ decreases more rapidly at the point (1,1).

Solution.

Partial derivatives of the function: $\frac{\partial f}{\partial x} = 2x + 2y$, $\frac{\partial f}{\partial y} = -2y + 2x$.

The values of the partial derivatives of the function at the point $A(1;1)$:

$$\left. \frac {\partial f}{\partial x} \right| _ {A} = 2 + 2 = 4, \quad \left. \frac {\partial f}{\partial y} \right| _ {A} = - 2 + 2 = 0.$$

The value of the gradient of the function in the direction of the unit vector $\vec{n}(n_x; n_y)$ at the point $A(1;1)$:

$$\left. \frac {\partial f}{\partial x} \right| _ {A} \cdot n _ {x} + \left. \frac {\partial f}{\partial y} \right| _ {A} \cdot n _ {y} = 4 \cdot n _ {x} + 0 \cdot n _ {y} = 4 n _ {x}.$$

As $n_x^2 + n_y^2 = 1$ for a unit vector, then the minimal value of (1) reaches if $n_x = -1$.

So, the function decreases more rapidly at the given point at the negative direction of $X$-axis.

Answer: in the direction of the vector $(-1;0)$.

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