# Distance/speed trig question

1. Aug 20, 2008

### avr10

1. The problem statement, all variables and given/known data

You are 150 feet away from a road. Looking down the road, you notice a car heading in your direction such that the angle formed by your line of vision to the car and the road is 45 degrees, and this angle is increasing at a rate of 10 degrees per second. How fast is the car traveling?

2. Relevant equations

3. The attempt at a solution

So I drew the diagram such that I'm standing on the positive side of the x-axis, and such that the car is heading down the y-axis (the road) toward the origin. I'm given $$\frac {d\theta}{dt} = 10$$, and I want to find $$\frac {dy}{dt}$$.

What I did was set $$\theta = \arctan { \frac {150}{150-\frac{dy}{dt}*t}}$$ then tried to take the derivative of that...but the $$\frac {dy}{dt}$$ term is leaving me confused as to how to derive such a thing. Ideas?

Last edited: Aug 20, 2008
2. Aug 20, 2008

### HallsofIvy

Staff Emeritus
I think it would be simpler to just say $tan(\theta)= y/150$ (I don't know where you got that "120") and use implicit differentiation:
$$sec^2(\theta)\frac{d\theta}{dt}= \frac{1}{150}\frac{dy}{dt}$$

3. Aug 20, 2008

### avr10

Thanks; the 120 was a typo which I now fixed. When I do that (I used $$\tan{\theta} = 150/y$$), I end up getting something like 3000 ft/s, which I think there is something wrong with... why am I getting such a big number?

4. Aug 21, 2008

### HallsofIvy

Staff Emeritus
Remember that the derivative of tan(x) is sec2(x) when x is measured in radians. sec(45)= $\sqrt{2}$ and you are given that $d\theta/dt[itex]= 10 degrees= [itex](\pi/180)*10$ radians so you should get $dy/dt= (150)(2)(\pi/18)$ which is about 52.2