Question

Calculus 283 Chapter 16​

Answer 1

(a) The vector field $\vec F$ is conservative if it is a gradient field. So we try to find a scalar function $f$ such that

$\nabla f(x,y) = \dfrac{\partial f}{\partial x} \, \vec\imath + \dfrac{\partial f}{\partial y} = \vec F(x,y)$

Given the field

$\vec F(x,y) = y\,\vec\imath+x\,\vec\jmath$

we need to have

$\dfrac{\partial f}{\partial x} = y$

$\dfrac{\partial f}{\partial y} = x$

Integrate both sides of the first equation with respect to $x$.

$\displaystyle \int \frac{\partial f}{\partial x} \, dx = \int y \, dx \implies f(x,y) = xy + g(y)$

Differentiate both sides with respect to $y$.

$\dfrac{\partial f}{\partial y} = x + \dfrac{dg}{dy} = x \implies \dfrac{dg}{dy} = 0 \implies g(y) = C$

So we have

$f(x,y) = xy + C \implies \nabla f = \vec F$

is indeed conservative.

(b) By the gradient theorem, the integral is path-independent and

$\displaystyle \int_C \vec F \cdot d\vec r = f(5,4) - f(0,2) = \boxed{20}$

for any path $C$ that starts at (0, 2) and ends at (5, 4).

Answer 2

Answer:

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