Finding the distance traveled over a length of time (of an object)

asked 2014-07-02 07:44:06 -0600

Kyouko
231 ●4 ●12

updated 2014-07-02 14:05:58 -0600

Justin
1009 ●2 ●20 ●38

Questions 1c and 2 on the review test as us to find an integral that could be solved in order to find the distance traveled by an object over a parametric plot. I can visualize this process, however, I am not quite sure how to go about setting up an integral for this using both \vec{x} = t + cos(4t) and \vec{y} = -sin(4t).

Here question 1c is asking us to find the integral described above for the following vector

$$ \vec{p}(t) = \langle t + cos(4t), -sin(4t) \rangle $$ and $$ \vec{p'}(t) = \langle 1 - 4sin(4t), -4cos(4t) \rangle $$

I believe setting up the integral would set up using the derivative of the position function integrated over the time interval giving the integral

$$ \int_0^{2\pi} \ <1 - 4sin(4t), -4cos(4t)> \mathrm{d}t $$

Any help would be appreciated

edit retag flag offensive close delete

answered 2014-07-02 08:11:12 -0600

Kyouko
231 ●4 ●12

updated 2014-07-02 08:13:32 -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

$$ \vec{p}(t) = \langle t + cos(4t), -sin(4t) \rangle $$

$$ \vec{p'}(t) = \langle 1 - 4sin(4t), -4cos(4t) \rangle $$

$$ \int_0^{2\pi}\ \sqrt{(1 - 4sin(4t))^2 + 16cos^2(4t)} \mathrm{d}t $$

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)

edit flag offensive delete publish link

Finding the distance traveled over a length of time (of an object)

1 Answer

Question tools

Stats

Related questions