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Chapter 13: Problem 16

Verify that the vector field is conservative. \(\mathbf{F}(x, y)=\frac{1}{x^{2}}(y \mathbf{i}-x \mathbf{j})\)

Short Answer

Expert verified
The curl of the vector field F equals zero, therefore, the vector field is conservative.

Step by step solution

01

Find Fi and Fj

In order to calculate the curl, the function F needs to be broken down into its i (x) and j (y) components. From F, it can be seen that \(F_i = \frac{y}{x^{2}}\) and \(F_j = -\frac{x}{x^{2}} = -\frac{1}{x}\). These will be substituted into the curl formula.
02

Calculate Partial Derivatives

Now, the partial derivative with respect to x of Fj, denoted as \(\frac{鈭侳_j}{鈭倄}\), and the partial derivative with respect to y of Fi, denoted as \(\frac{鈭侳_i}{鈭倅}\), should be calculated. Using the power rule for differentiation, \(\frac{鈭侳_j}{鈭倄} = \frac{1}{x^2}\) and \(\frac{鈭侳_i}{鈭倅} = \frac{1}{x^{2}}\).
03

Calculate the Curl

The curl is now calculated by substituting the partial derivatives into the formula \(Curl(F) = -\frac{鈭侳_j}{鈭倄} + \frac{鈭侳_i}{鈭倅}\). After substitution, \(Curl(F) = -(\frac{1}{x^{2}}) + \frac{1}{x^{2}} = 0\).

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