Find a function f such that 𝐅=∇f and (b) use part (a) to evaluate ∫C 𝐅 ·d 𝐫 along the given curv

Find a function f such that 𝐅=∇f and (b) use part (a) to...

Find a function $f$ such that $\mathbf{F}=\nabla f$ and $(b)$ use part (a) to evaluate $\int_{C} \mathbf{F} \cdot d \mathbf{r}$ along the given curve $C.$ $\mathbf{F}(x, y, z)=\left(y^{2} z+2 x z^{2}\right) \mathbf{i}+2 x y z \mathbf{j}+\left(x y^{2}+2 x^{2} z\right) \mathbf{k},$ $C : x=\sqrt{t}, y=t+1, z=t^{2}, \quad 0 \leqslant t \leqslant 1$

Find a function $f$ such that $\mathbf{F}=\nabla f$ and $(b)$ use part (a) to evaluate $\int_{C} \mathbf{F} \cdot d \mathbf{r}$ along the given curve $C.$
$\mathbf{F}(x, y, z)=\left(y^{2} z+2 x z^{2}\right) \mathbf{i}+2 x y z \mathbf{j}+\left(x y^{2}+2 x^{2} z\right) \mathbf{k},$ $C : x=\sqrt{t}, y=t+1, z=t^{2}, \quad 0 \leqslant t \leqslant 1$

Key Concepts

Conservative Vector Fields

A conservative vector field is one that can be expressed as the gradient of a scalar potential function. This implies that the field has no circulation around any closed loop, and the line integral between two points is independent of the chosen path. Recognizing a vector field as conservative is crucial in simplifying the evaluation of line integrals in multivariable calculus.

Potential Function

A potential function is a scalar function whose gradient equals a given vector field. Finding this function allows one to relate the vector field to its underlying field of scalar values, and it simplifies the computation of integrals and the analysis of the field’s properties.

Gradient Operator

The gradient operator acts on a scalar function to produce a vector field composed of its partial derivatives. It is used to determine the direction and rate of fastest increase of the function, and its application is fundamental in linking potential functions to their corresponding conservative vector fields.

Fundamental Theorem for Line Integrals

The fundamental theorem for line integrals states that if a vector field is conservative, then the line integral of the field along a curve depends only on the values of its potential function at the endpoints of the curve. This theorem significantly simplifies the evaluation of integrals over paths in conservative fields.

Transcript

00:01 In this problem, we are given the shown information, and our first step in this might be to make sure that f is a conservative vector field.

00:08 The way that we might do this is to set this component here equal to some function p, this component equal to some function q, and this component equal to some function r, and we're going to make sure that the partial of p with respect to y is equal to the partial of q with respect to x.

00:31 We also want to make sure that the partial of p of p with respect to z is equal to the partial of r with respect to x and lastly we want to make sure that the partial of q with respect to z is equal to the partial of r with respect to y now the partial of p with respect to y is going to be 2y and the partial of q with respect to x is going to be 2y z so that one's good.

01:06 The partial of p with respect to z is going to be y squared plus 2x times 2 z.

01:20 And the partial of p with respect to x is going to be y squared plus 2 times 2x z.

01:30 As you can see those are equal.

01:32 So that's good.

01:38 So that's good.

01:40 And the partial of q with respect to z is going to be 2xy and the partial of r with respect to y is going to be x times 2y so those are equal so that's good so it is a conservative vector field and now we can find our function f and the way that we're going to do this is to write our function p in an integral and evaluated in terms of y so we're going to take p with y squared z plus 2x z squared and write that in terms of dx.

02:20 And when we evaluate this, we get xy squared z plus x squared z squared, plus some function n, y, and z.

02:37 Because the derivative of that would go to zero if we took the partial of our new function in terms of x.

02:42 Okay, and now what we're going to do is take the partial of this function in terms of y.

02:51 And when we do this, we get x times 2y times z plus g prime of y z.

03:13 Now we're going to set that equal to our function q, which is 2x y z.

03:26 And from this we can see that g prime of some y z is equal to zero.

03:35 And we write this in an integral to solve for g of y z...