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Determine whether or not f is a conservative vector field. If it is, find a function f such that f = ∇f. If it is not, enter none. F(x, y) = (exsin(y))i (excos(y))j

User JonF
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1 Answer

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Final answer:

After computing partial derivatives of the x-component with respect to y and the y-component with respect to x, and finding them equal, we conclude that F(x, y) is a conservative vector field. Therefore, there exists a scalar potential function f such that F = ∇f.

Step-by-step explanation:

To determine whether F(x, y) = (
e^x sin(y))i + (
e^x cos(y))j is a conservative vector field, we can apply the condition for conservative fields. In a conservative field, the curl of the vector field should be zero; mathematically, this translates to the partial derivative of the x-component with respect to y being equal to the partial derivative of the y-component with respect to x.

First, we compute the partial derivative of the x-component (
e^x sin(y)) with respect to y, which is
e^x cos(y). Next, we compute the partial derivative of the y-component (
e^x cos(y)) with respect to x, which is also
e^x cos(y).

Since the partial derivatives are equal ((dFx/dy) = (dFy/dx)), F(x, y) is a conservative vector field. Hence, there exists a scalar potential function f such that F = ∇f.

User Sweet Suman
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