Consider F and C below. F(x, y, z) = (2xz + y^2) i + 2xy\space j + (x^2 + 6z^2) k C: x = t^2,...


Consider {eq}\vec{F} {/eq} and {eq}C {/eq} below.

{eq}\vec{F}(x, y, z) = (2xz + y^2) \hat{i} + 2xy\space \hat{j} + (x^2 + 6z^2) \hat{k} {/eq}

{eq}C: x = t^2, y = t + 2, z = 2t -1, 0 \leq t \leq 1 {/eq}

(a) Find a function {eq}f {/eq} such that {eq}\vec{F} = \nabla f. f(x, y, z) =\square {/eq}

(b) Use part (a) to evaluate {eq}\displaystyle \int \nabla f \cdot d\vec{r} {/eq} along the given curve {eq}C{/eq}.

Gradient Theorem:

The Gradient Theorem is the Fundamental Theorem of Calculus for Line Integrals. The former name is a bit more revealing, as it implies the restriction that it is valid only for gradient vector fields. Like the original theorem, it works only because of path independence. Also, in order to use it we must find a potential function for the field. The Gradient Theorem says

{eq}\begin{align*} \int_C \vec F \cdot d\vec r &= \int_{(x_1,y_1,z_1)}^{(x_2,y_2,z_2)} \nabla f \cdot d\vec r = f(x_2,y_2,z_2) - f(x_1,y_1,z_1) \end{align*} {/eq}

Answer and Explanation:

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Part A

We want to find the potential function, and since we have the vector field, we have all of its partials:

{eq}\begin{align*} \frac{\partial...

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Learn more about this topic:

Directional Derivatives, Gradient of f and the Min-Max


Chapter 14 / Lesson 6

In 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.

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