The vector field V = (xy^2 + y - 2x, x^2y + x) is a potential vector field. The potential...
Question:
The vector field {eq}V = (xy^{2} + y - 2x, x^{2}y + x) {/eq} is a potential vector field.
The potential function can be written in the form {eq}\phi = Ax^{2}y^{2} + Bxy + Cx^2 + 2020 {/eq}.
Find A (decimal), B (integer) and C (integer).
Let {eq}u= (u_{1}, u_{2}) {/eq} be a unit vector that gives the direction in which {eq}\phi {/eq} does not change at point (1, 1). Assuming that {eq}u_{1}>0 {/eq}, determine {eq}u_{1} {/eq} = (integer) and {eq}u_{2} {/eq} (integer).
Directional Derivatives:
Let {eq}f(x,y) {/eq} be a scalar function and {eq}\mathbf{u} {/eq} be a unit vector. The directional derivative of {eq}f(x,y) {/eq} along the direction {eq}\mathbf{u} {/eq} is defined to be $$D_{\mathbf{u}}f(x,y) = \nabla f(x,y)\cdot \mathbf{u} $$ where {eq}\nabla f = \left\langle f_x, f_y\right\rangle {/eq} is the gradient of {eq}f(x,y). {/eq} If the given direction {eq}\mathbf{u} {/eq} is not a unit vector, then we need to take the vector {eq}\frac{\mathbf{u}}{||\mathbf{u}||} {/eq} in the above computations.
Answer and Explanation: 1
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View this answer(1) Since $$\phi = Ax^2y^2+Bxy+Cx^2+2020 $$ is a potential function of $$\mathbf{V} = \left\langle xy^2+y-2x, x^2y+x \right\rangle $$ then ...
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Chapter 14 / Lesson 6In this lesson, learn about directional derivatives, gradients, and maximum and minimum critical points. Moreover, learn to use the directional derivative formula to calculate slopes at given points.