Question

Determine the direction in which the scalar field
φ( x, y) = xy^2 + x^3y increases the
fastest at the point (1, 2).

Answer 1

Answer on Question #75245-Math-Other

Determine the direction in which the scalar field $\phi(x, y) = xy^2 + x^3y$ increases the fastest at the point $(1, 2)$ .

Solution

$$\boldsymbol {\nabla} \varphi = \frac {\partial \varphi}{\partial x} \boldsymbol {i} + \frac {\partial \varphi}{\partial y} \boldsymbol {j}$$ $$\boldsymbol {\nabla} \varphi (\mathrm {x}, \mathrm {y}) = (\mathrm {y} ^ {2} + 3 \mathrm {x} ^ {2} \mathrm {y}) \boldsymbol {i} + (2 \mathrm {x y} + \mathrm {x} ^ {3}) \boldsymbol {j}$$ $$\boldsymbol {\nabla} \varphi (1, 2) = (2 ^ {2} + 3 (1 ^ {2}) 2) \boldsymbol {i} + (2 (1) 2 + 1 ^ {3}) \boldsymbol {j} = 1 0 \boldsymbol {i} + 5 \boldsymbol {j}$$ $$| \boldsymbol {\nabla} \varphi | = \sqrt {1 0 ^ {2} + 5 ^ {2}} = 5 \sqrt {5}$$

The direction in which the scalar field $\phi(x, y)$ increases the fastest at the point $(1, 2)$ is

$$\frac {\boldsymbol {\nabla} \varphi}{| \boldsymbol {\nabla} \varphi |} = \frac {1 0 \boldsymbol {i} + 5 \boldsymbol {j}}{5 \sqrt {5}} = \frac {2 \boldsymbol {i} + \boldsymbol {j}}{\sqrt {5}} = \frac {2}{\sqrt {5}} \boldsymbol {i} + \frac {1}{\sqrt {5}} \boldsymbol {j}$$

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