In Exercises 17–24, find a potential function for the given vector field.

F(x,y)=(eysec2x,eytanx)

Short Answer

Expert verified

A potential function for the given vector field is f(x,y)=eytanx.

Step by step solution

01

Step 1. Given Information

In given exercises we have to find a potential function for the given vector field.

F(x,y)=(eysec2x,eytanx)

02

Step 2. Since F(x, y)=(eysec2x, eytanx)

f(x,y)=(eysec2x)dx+Bf(x,y)=eysec2xdx+Bf(x,y)=eytanx+α+B

where α is an arbitrary constant and B is the integral with respect to y of the terms in F2(x,y) in which the factor x does not appear.

03

Step 3. In this case, that is all of F2(x,y), so

B=(eytanx)dyB=tanxeydyB=tanx·ey+βB=eytanx+β

whereβ is an arbitrary constant.

04

Step 4. Setting the constants equal to zero since they do not affect the gradient of f(x,y)

We have,

f(x,y)=eytanx

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