Question

if f(x,y,z)=zi-xk then the corresponding potential function phi(x,y,z)

Answer 1

Answer:

what is your que sorry I did not understand

Answer 2

Step-by-step explanation:

The vector field we'll analyze is

F(x,y,z)=(2xyz3+yexy,x2z3+xexy,3x2yz2+cosz).

We first check if it is conservative by calculating its curl, which in terms of the components of F, is

curlF=(∂F3∂y−∂F2∂z,∂F1∂z−∂F3∂x,∂F2∂x−∂F1∂y).

Since

∂F3∂y∂F1∂z∂F2∂x=∂F2∂z=3x2z2=∂F3∂x=6xyz2=∂F1∂y=2xz3+(1+xy)exy,

the curl of F is zero. The vector field is defined in all R3, which is simply connected, so F is conservative.

We need to find a potential function f(x,y,z) that satisfies ∇f=F, i.e., the three conditions

∂f∂x(x,y,z)∂f∂y(x,y,z)∂f∂z(x,y,z)=2xyz3+yexy=x2z3+xexy=3x2yz2+cosz.

We start with the first condition involving ∂f∂x. We integrate with respect to x, viewing y and z as constants, to obtain that f(x,y,z) must satisfy

f(x,y,z)=x2yz3+exy+g(y,z).

Since we viewed y and z as constants, the constant of integration g(y,z) can be an arbitrary function of y and z. You can verify that f(x,y,z) does satisfy the first condition.

Now, we simply need to determine what g(y,z) must be for f to satisfy the remaining two conditions involving derivatives with respect to y and z. Let's differentiate our new expression for f with respect to y, obtaining

∂f∂y(x,y,z)=x2z3+xexy+∂g∂y(y,z)

We need to f to satisfy the second condition, above, involving ∂f∂y. For this to be true, we require

∂g∂y(y,z)=0.

Since F is conservative, this equation for ∂g∂y must be a function of y and z alone (and not involve x). If x appeared, we would know we had made a mistake somewhere. Since x is absent, we can keep going.

In this case, since we need ∂g∂y(y,z)=0, we conclude that g(y,z) cannot depend on y. It must be a function of z alone, which we'll call h(z). Our expression for f(x,y,z) simplifies to

f(x,y,z)=x2yz3+exy+h(z)

We have one more condition to satisfy, the one involving ∂f∂z. We differentiate our new expression for f with respect to z:

∂f∂z=3x2yz2+dhdz(z).

For f(x,y,z) to satisfy the third condition for f, the function h(z) must satisfy

dhdz(z)=cosz.

This is good news, as dhdz does not depend on x or y. If it had, we would know something went wrong

We can easily integrate to obtain an expression for h,

h(z)=sinz+k

for an arbitrary constant k. A potential function can only be determined up to an arbitrary constant, since we only have conditions on its derivatives. But, line integrals of F depend only on differences among the values of f(x,y,z). The constant k will always cancel out, so we can just set k=0.

Therefore, our potential function for

F(x,y,z)=(2xyz3+yexy,x2z3+xexy,3x2yz2+cosz).

is

f(x,y,z)=x2yz3+exy+sinz.

For any curve C from point p to point q, the integral of F is

∫CF⋅ds=f(q)−f(p)

independent of the path taken by C. Although we had to do a lot of work to calculate f, the last step of computing the integral is simple.

If C is the arc of a helix parametrized by c(t)=(cost,sint,t) for 0≤t≤π/2, the line integral of F is simply

∫CF⋅ds=f(c(π/2))−f(c(0))=f(0,1,π/2)−f(1,0,0)=0+e0+sinπ2−0−e0−sin0=1+1−1=1.