I am having some difficulties doing some line integrals exercises, could someone please check my working and solution?
The question is as follows :
A particle moves from point $A = (0,0,0)$ to point $B=(2\pi,0,2\pi)$, under the action of the force $\mathbf F=x\ \mathbf i + y\ \mathbf j -z\ \mathbf k$ where $\mathbf i,\mathbf j,\mathbf k$ are the direction vectors.
- Calculate the work done by the force $\mathbf F$ on the particle if it moves along the conic-helical curve :$\mathbf r(t)= (t\cos t)\ \mathbf i +(t \sin t)\ \mathbf j+t\ \mathbf k$ with $t$ between $0$ and $2\pi$.
My approach was as follows :
Find the integral $$\int_C\mathbf F(x(t),y(t),z(t))\cdot\mathbf r'(t)\mathrm dt$$
I calculate $\mathbf r'(t)$ to be $(\cos t-\sin t)\ \mathbf i+(t\cos t+\sin t)\ \mathbf j +\mathbf k$
$$\int_0 ^{2\pi}(t\cos t\ \mathbf i + t\sin t\ \mathbf j-\mathbf k)\cdot\mathbf r'(t)\mathrm dt$$
this reduced to $$[t^3 + t]_0^{2\pi} = 2\pi(4\pi^2 +1)$$
- Find the parametric vector equation for the straight line connecting $A$ to $B$, and calculate the work done by the force $\mathbf F$ on the particle as it moves along the straight line.
I think in this question my methods is not correct, but it was as follows :
$$x=t,y=t,z=t\implies\mathbf r(t)= t\ \mathbf i+t\ \mathbf j+t\ \mathbf k$$
I determined the tangent vector to be $\frac{d\mathbf r}{dt}=\mathbf i +\mathbf j -\mathbf k$
then determined the force field of the tracjectory:
$$\int_0 ^{2\pi}(t\ \mathbf i+t\ \mathbf j-t\mathbf k)(\mathbf i+\mathbf j-\mathbf k)\mathrm dt = 2\pi^2$$
- Determine whether it is conservative, which I don't think I will have a problem with once these are checked.
Best Answer
Ok Lets start with part 1:
We want to calculate the work done by a force field on the particle along a path $$ \int \vec{F}(\vec{r}) \cdot \mathrm{d}\vec{r} = \int \vec{F}(\vec{r}(t))\cdot \vec{r}'(t) \mathrm{d}t $$ We are given that the path is a conical helix given by $$\vec{r}(t) = t \cos{t} \;\hat{i} + t \sin{t} \;\hat{j} + t \hat{k} \quad t \in(0,2 \pi)$$ And $$\vec{F}(\vec{r}) = x \; \hat{i} + y\;\hat{j} + z\;\hat{k}$$ Using the product rule we obtain for $\vec{r}'$: $$\vec{r}'(t) = (\cos{t} - t\sin{t})\; \hat{i} + (\sin{t} + t\cos{t})\; \hat{j} + \hat{k}$$ And $$\vec{F}(\vec{r}(t)) = t\cos{t}\;\hat{i} + t\sin{t}\;\hat{j} + t\;\hat{k}$$ We take the dot product: \begin{eqnarray*} \vec{F}(\vec{r}(t)) \cdot\vec{r}'(t) &=& (\cos{t} - t\sin{t})t\cos{t} + (\sin{t} + t\cos{t}) t\sin{t} + t \\ &=& t\cos^2{t} - t^2 \sin{t}\cos{t} + t\sin^2{t} + t^2\cos{t}\sin{t} + t \\ &=& 2t \end{eqnarray*} Thus the resulting integral is: $$\int_0^{2 \pi} 2t \mathrm{d}t = \left. t^2 \right|_0^{2\pi} = 4 \pi^2$$
For part 2: We want to parameterize the straight line from $(0,0,0)$ to $(2\pi,0,2\pi)$ in the variable t, you were close with your parameterization however you didn't need y so the correct line is: $$\vec{r}(t)= t \;\hat{i}+ t\;\hat{k} \quad t\in(0,2\pi)$$ You can likely carry on from there. But if you need more I can continue on