Revision history [back]
 | 1 | initial version | posted 2014-07-02 08:11:12 -0600  |
Professor Mcclure answered this in class.
Using the general formula for the distance traveled by an object over a parametrically described motion we get the integral:
$$ \int_0^{2\pi} \ ||\vec{p'}(t)|| \mathrm{d}t $$
By plugging into this formula we get the following integral
$$ \vec{p}(t) = <t +="" cos(4t),="" -sin(4t)>="" $$<="" p="">
$$ \vec{p'}(t) = <1 - 4sin(4t), -4cos(4t)> $$
$$ \int_0^{2\pi}\ \sqrt{(1 - 4sin(4t))^2 + 16cos^2(4t)} \mathrm{d}t $$
| | 2 | No.2 Revision |
Professor Mcclure answered this in class.
Using the general formula for the distance traveled by an object over a parametrically described motion we get the integral:
$$ \int_0^{2\pi} \ ||\vec{p'}(t)|| \mathrm{d}t $$
By plugging into this formula we get the following integral
$$ \vec{p}(t) = <t +="" cos(4t),="" -sin(4t)>="" $$<="" p="">
\langle t + cos(4t), -sin(4t) \rangle $$
$$ \vec{p'}(t) = <1 \langle 1 - 4sin(4t), -4cos(4t)> -4cos(4t) \rangle $$
$$ \int_0^{2\pi}\ \sqrt{(1 - 4sin(4t))^2 + 16cos^2(4t)} \mathrm{d}t $$
And because were not asked to solve this integral, the line above is the final answer (in which you can expand the quantity square under the root sign)
