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Distance/speed trig question

  1. Aug 20, 2008 #1
    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 [tex] \frac {d\theta}{dt} = 10 [/tex], and I want to find [tex] \frac {dy}{dt} [/tex].

    What I did was set [tex] \theta = \arctan { \frac {150}{150-\frac{dy}{dt}*t}} [/tex] then tried to take the derivative of that...but the [tex]\frac {dy}{dt}[/tex] term is leaving me confused as to how to derive such a thing. Ideas?
     
    Last edited: Aug 20, 2008
  2. jcsd
  3. Aug 20, 2008 #2

    HallsofIvy

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    I think it would be simpler to just say [itex]tan(\theta)= y/150[/itex] (I don't know where you got that "120") and use implicit differentiation:
    [tex]sec^2(\theta)\frac{d\theta}{dt}= \frac{1}{150}\frac{dy}{dt}[/tex]
     
  4. Aug 20, 2008 #3
    Thanks; the 120 was a typo which I now fixed. When I do that (I used [tex]\tan{\theta} = 150/y [/tex]), 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?
     
  5. Aug 21, 2008 #4

    HallsofIvy

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