Vector field
the vector field $F(x,y)= [ -2 \frac{x}{(x^{2} +y^{2}+1)^2 },-2\frac{y}{(x^{2}+y^{2}+1)^2 } ] $ is conservative throughout the planet. Determine a potential U to F and calculate $\int_{\gamma} \mathbf F \cdot d\,\mathbf r$ where γ is an arbitrary curve in the plane from (3,−4) to (4,1).
calculate U and $\int_{\gamma} \mathbf F \cdot d\,\mathbf r$
Answer
A potential is:
$$U(x,y) = \frac1{x^2+y^2+1}$$
as you may see via the chain rule:
$$\partial_x U(x,y) = -2\frac x{(x^2+y^2+1)^2},\qquad \partial_y U(x,y) = -2\frac y{(x^2+y^2+1)^2}$$
So for any curve $\gamma$ from $(3,-4)$ to $(1,4)$ we have:
$$\int_{\gamma} F\cdot d\gamma = \int_\gamma \mathrm{grad}(U)\cdot d\gamma = U(4,1)-U(3,-4) = \frac1{18}-\frac1{26}$$
by the fundamental theorem of calculus / Stokes' theorem, whatever you are using.
- answered
- 1268 views
- $3.20
Related Questions
- Reduction formulae
- Bivariate Normality questions
- Calculus problems on improper integrals
- Trying to solve this system of simultaneous equations. A solution with work shown would be appreciated.
- True-False real analysis questions
- Find slope intercept equation
- Find the derivative of the function $f(x)=\sqrt{\sin^2x+e^x+1}$
- Limits : $x^{-1} \sin(x) $ as x -> 0 and $\tfrac{\ln(x)}{1-x}$ as x-> 0