0
$\begingroup$

Let $F(x,y) = (3x^2,4y^3)$. Determine the value of $\int_c F(x,y)\cdot \mathrm dr$, where $c$ is the path from $(0,1)$ to $(\pi,-1)$ along graph of $y=\cos x$.

  1. Is it good to always check for path independence first? Here i can do $\frac{\mathrm dp}{\mathrm dy}=0$, $\frac{\mathrm dq}{\mathrm dx} = 0$ so path independent, $\frac{\mathrm dp}{\mathrm dx}= x^3$, $\frac{\mathrm dp}{\mathrm dy}= y^4$.

    Therefore I get, $f(x,y)=x^3+y^4$, $$f(b)-f(a) = f(\pi,-1)-f(0,1) = (\pi^3)-(1)^4-(0^3+1^4) = \pi^3.$$

  2. Or do $x = t$, $y = \cos t$, $r = (t, \cos t)$, $\mathrm dr = (1,-\sin t)$?

    $F(x,y)\cdot\mathrm dr = (3t^2,4\cos t^3)\cdot(1, -\sin t)$ dot product but how do I find here the $t=?$ to $t=?$

    And is this the way to do it if the path independece does not equal?

$\endgroup$
1
$\begingroup$

For 2.: We have $x(t)=t$ and $y(t)= \cos t$ for $t \in [0, \pi].$

The integral then $= \int_0^{\pi} <3t^2,4 \cos^3 t> \cdot<1, -\sin t> dt.$

$\endgroup$
  • $\begingroup$ thanks, is the 1. correct? should i always be checking for path integral and solving it before trying to do 2.? $\endgroup$ – MasterYoshi Mar 15 at 8:41
  • 1
    $\begingroup$ We have $ f(\pi,-1)-f(0,1) = (\pi^3)+(-1)^4-(0^3+1^4) = \pi^3.$ ! Checking of independence is always a good idea ! $\endgroup$ – Fred Mar 15 at 8:45
  • $\begingroup$ Please don't use < and > for anything but comparisons. The spacing is wrong. $\LaTeX$ provides \langle and \rangle for angular brackets. $\endgroup$ – Christoph Mar 15 at 9:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.