anonymous
  • anonymous
A particle moves through the force field F(x,y,z)=<4,3,1> along the twisted cubic TC r(t)= as t goes from 0 to 2. What is the work?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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IrishBoy123
  • IrishBoy123
\(W = \int \vec F \bullet d \vec r\)
anonymous
  • anonymous
Assuming you know about conservative vector fields, you can try determining whether \({\textbf{F}}\) is conservative and find \(f(x,y,z)\) such that \(\nabla f={\textbf{F}}\). A vector field \({\textbf{F}}\) is conservative if \(\mathrm{curl}\,{\textbf{F}}=0\) You have \[\begin{align*}\nabla\times{\textbf{F}}&=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\[1ex]\dfrac{\partial}{\partial x}&\dfrac{\partial}{\partial y}&\dfrac{\partial}{\partial z}\\[1ex]4&3&1 \end{vmatrix}\\[1ex] &=\left(\frac{\partial}{\partial y}1-\frac{\partial}{\partial z}3\right)\vec{i}-\left(\frac{\partial}{\partial x}1-\frac{\partial}{\partial z}4\right)\vec{j}+\left(\frac{\partial}{\partial x}3-\frac{\partial}{\partial y}4\right)\vec{k}\\[1ex] &=0\vec{i}+0\vec{j}+0\vec{k}\end{align*}\]hence \({\textbf{F}}\) is conservative. This means you can apply fundamental theorem for line integrals, ie. \[\int_C{\textbf{F}}\bullet\mathrm{d}\vec{r}=\int_C\nabla f\bullet\mathrm{d}\vec{r}=f(\vec{b})-f(\vec{a})\]where \(\vec{a}\) is the starting point on the path described by \(\vec{r}(t)\) and \(\vec{b}\) is the terminal point.
Kainui
  • Kainui
Just for variety, I just decided to show whichever method @SithsAndGiggles didn't do. $$\int \vec F \cdot d \vec r = \int_0^2 \vec F \cdot \frac{d \vec r}{dt} dt$$ The heuristic way of thinking about it is just to imagine multiplying \(\frac{dt}{dt}=1\) into the integral. $$\int_0^2 \langle 4, 3, 1 \rangle \cdot \langle 1, 2t, 3 t^2 \rangle dt = \int_0^2 4+6t+3t^2 dt$$ Since F doesn't really depend on the x,y,z coordinates it makes it pretty smooth to just plug it in since normally you'd have to plug in the components of the path you take. --- There's also an alternate route we could have done it starting from here, although this is a pretty tricky case that doesn't work in general, only works because the force components don't depend on the other variables. $$\int \vec F \cdot d \vec r = \int F_xdx+F_ydy+F_zdz$$ Now we split up the integral into the individual parts and since all the force components are constants they come out of the integrals: $$ F_x\int dx+F_y \int dy + F_z \int dz $$ Now to determine the limits of integration, we just plug in \(r(0)\) and \(r(2)\) and plug those in on top and bottom for the corresponding ones: $$ 4\int_0^2 dx+3 \int_0^{2^2} dy + 1 \int_0^{2^3} dz $$

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