# Calcccc R.Rates

• Dec 7th 2008, 01:14 PM
Jason2JZ
Calcccc R.Rates
Two Cyclists depart at the same time from a starting point aloing routes that make an angle 60 degrees with each other. the first cyclist is travelling at 15km/h while the second is moving at 20km/h. how fast are the cyclist moving apart after 2 hours?(Surprised)
• Dec 7th 2008, 01:44 PM
galactus
For this one we do not have a right triangle. We have an angle of $\displaystyle {\theta}=\frac{\pi}{3}$

We can use the law of cosines.

Draw the diagram. It'll help.

Let $\displaystyle \frac{da}{dt}=15$ be the rate of one biker and

$\displaystyle \frac{db}{dt}=20$ be the rate of the other.

In two hours, one biker will have traveled 15*2=30 km

and the other will have traveled 20*2=40 km

So, we have a whole bunch of info to enter into the law of cosines formula after it is differentiated.

We need dc/dt.

Use the law of cosines to find c with a=30, b=40, theta=Pi/3

$\displaystyle c^{2}=a^{2}+b^{2}-2ab\cdot cos({\theta})$

Remember, the bikers maintain a constant angle, so $\displaystyle \frac{d{\theta}}{dt}=0$. That makes it easier.

Now, go for it. You have all you need. If you get hung up, let me know.