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 $\int_{\gamma} \mathbf F \cdot d\,\mathbf r$
Asara
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.
Dynkin
- answered
- 1938 views
- $3.20
Related Questions
- Need help with finding equation of the plane containing the line and point. Given the symmetric equation.
- Solve $abc=2(a-2)(b-2)(c-2)$ where $a,b $ and $c$ are integers
- Evaluate the integral $\int_{-\infty}^{+\infty}e^{-x^2}dx$
- A function satifying $|f(x)-f(y)|\leq |x-y|^2$ must be constanct.
- Integral of $\arctan x$
- College Algebra Help
- Multivariable Calc: Vectors, Equations of Lines, Shapes of Curves
- Compounding interest of principal P, where a compounding withdrawal amount W get withdrawn from P before each compounding of P.
