# The position of an object moving along a line is given by p(t) = 3t - tcos(( pi )/4t) . What is the speed of the object at t = 7 ?

Nov 29, 2017

$3 - \frac{\sqrt{2}}{2} - \frac{7 \sqrt{2} \pi}{8}$

#### Explanation:

You're looking for the velocity of the object. You can find the velocity $v \left(t\right)$ like this:
$v \left(t\right) = p ' \left(t\right)$
Basically, we have to find $v \left(7\right)$ or $p ' \left(7\right)$.
Finding the derivative of $p \left(t\right)$, we have:
$p ' \left(t\right) = v \left(t\right) = 3 - \cos \left(\frac{\pi}{4} t\right) + \frac{\pi}{4} t \sin \left(\frac{\pi}{4} t\right)$ (if you don't know how I did this, I used power rule and product rule)
Now that we know $v \left(t\right) = 3 - \cos \left(\frac{\pi}{4} t\right) + \frac{\pi}{4} t \sin \left(\frac{\pi}{4} t\right)$, let's find $v \left(7\right)$.
$v \left(7\right) = 3 - \cos \left(\frac{\pi}{4} \cdot 7\right) + \frac{\pi}{4} \cdot 7 \sin \left(\frac{\pi}{4} \cdot 7\right)$
$= 3 - \cos \left(\frac{7 \pi}{4}\right) + \frac{7 \pi}{4} \cdot \sin \left(\frac{7 \pi}{4}\right)$
$= 3 - \frac{\sqrt{2}}{2} - \frac{7 \pi}{4} \cdot \frac{\sqrt{2}}{2}$
$v \left(7\right) = 3 - \frac{\sqrt{2}}{2} - \frac{7 \sqrt{2} \pi}{8}$