0
$\begingroup$

This is a difficult question that I don't understand how to answer but I tried and I hope someone could tell me if my answers are right or wrong. I managed to do a and b only. For c I need some guidance on how to answer it.

The circle with centre (r, 0, r) and radius r in the x-z-plane is rotated through an angle α as shown in the sketch above. We let K denote this rotated circle. circle integration question

a) Find the parametric representation of the original circle in the x-z-plane.

$\begin{pmatrix}r+rcos\left(t\right)\\ 0\\ r+rsin\left(t\right)\end{pmatrix}$

b) Find the parametric representation of the rotated circle K.

$\begin{pmatrix}r+rcos\left(t\right)\cdot cos\left(\theta \right)\\ r+rcos\left(t\right)\cdot sin\left(\theta\right)\\r+rsin\left(t\right)\end{pmatrix}$

c) Compute the path integral$\int _c\vec{F}d\vec{x}\:with\:\vec{F}\left(\vec{x}\right)=\begin{pmatrix}x-rcos\left(\alpha \right)\\ y-rsin\left(\alpha \right)\\ z-r\end{pmatrix}\:$

I understand that I need to find the derivative of x(t)? Is that right? And then I have to insert it into Fx to get F(x(t))and then integrate it? How do I insert x(t) into F(x)?

$\endgroup$
1
$\begingroup$

You did very well for part a. Note that this is not the only parametrization you could have chosen.

You made a small error for part b. From part a, you indicated a set of positions that produce the circle. For part b, you essentially want to rotate these points an angle $\alpha$ about the $z$-axis. You need to have the parametrization of $K$ with position vector $\vec{p}$ as follows:

$ \vec{p}(t)= \begin{pmatrix} x(t) \\ y(t) \\ z(t) \end{pmatrix} = \begin{pmatrix} [r+r\cos(t)]\cos(\alpha) \\ [r+r\cos(t)]\sin(\alpha) \\ r+r\sin(t) \end{pmatrix} = \begin{pmatrix} r\cos(\alpha)+r\cos(t)\cos(\alpha) \\ r\sin(\alpha)+r\cos(t)\sin(\alpha) \\ r+r\sin(t) \end{pmatrix} $

This comes directly from the multiplication of the original parametrization of the position vector with a CCW rotation matrix about the $z$-axis. For part c, I think there might be an error because $x$ isn't a vector in the context of the problem. I think you meant $d\vec{x}$ as a differential path element which is $d\vec{p}$. Similarly, it should be $\vec{F}(\vec{p})$.

You need to establish the function in terms of your parameter $t$ by substituting in for $x$, $y$, and $z$ from your parametrization of $K$. Then you need to account for $d\vec{p}$ by recognizing that $d\vec{p} = \frac{d\vec{p}}{dt}dt$.

The function vector w.r.t. the parametrization is as follows:

$\vec{F}(\vec{p}) = \begin{pmatrix} x - r\cos(\alpha) \\ y - r\sin(\alpha) \\ z - r \end{pmatrix} = \begin{pmatrix} r\cos(\alpha)+r\cos(t)\cos(\alpha) - r\cos(\alpha) \\ r\sin(\alpha)+r\cos(t)\sin(\alpha) - r\sin(\alpha) \\ r+r\sin(t) - r \end{pmatrix} = \begin{pmatrix} r\cos(t)\cos(\alpha) \\ r\cos(t)\sin(\alpha) \\ r\sin(t) \end{pmatrix} $

The differential path vector is as follows:

$d\vec{p} = \frac{d\vec{p}}{dt}dt = \begin{pmatrix} -r\sin(t)\cos(\alpha) \\ -r\sin(t)\sin(\alpha) \\ r\cos(t) \end{pmatrix}dt$

Therefore, the following holds:

$\vec{F}(\vec{p}) \cdot \frac{d\vec{p}}{dt} = \begin{pmatrix} r\cos(t)\cos(\alpha) \\ r\cos(t)\sin(\alpha) \\ r\sin(t) \end{pmatrix} \cdot \begin{pmatrix} -r\sin(t)\cos(\alpha) \\ -r\sin(t)\sin(\alpha) \\ r\cos(t) \end{pmatrix} \\ = -r^2\sin(t)\cos(t)\cos^2(\alpha)-r^2\sin(t)\cos(t)\sin^2(\alpha) + r^2\sin(t)\cos(t)\\ = -r^2\sin(t)\cos(t) + r^2\sin(t)\cos(t) = 0 $

The function vector is orthogonal to the path for all $t$, so the path integral is simply $0$.

$\endgroup$

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.