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Math Help - Help setting up rate of change of angle

  1. #1
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    Help setting up rate of change of angle

    A camera mounted 132ft above the ground is videotaping a racecar moving at 264ft/sec. How fast will the camera angle be changing when the car is right in front of you? a half second later?

    ">" is the angle

    tan>=y/x and dtan>/dt= x(dy/dt)-y(dx/dt)/x^2

    y=132ft and dx/dt=264ft/sec

    How do I find x. or where do I go from here
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  2. #2
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    I'd do this problem a completely different way. I'm going to call the angle 'a.'

    tan(a) = \frac {x} {132} Notice that 132 is constant, don't need a y variable.

    a = Arctan(\frac{x}{132}) I solve for 'a' so we can take a derivative here and find \frac{da}{dx}.

    Then since we know \frac{dx}{dt}, the rest is straightforward.
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  3. #3
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    what is the value of x?
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  4. #4
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    x is the distance from directly underneath the camera that the car is. And therefore \frac{dx}{dt} is just going to be equal to the speed of the car.
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  5. #5
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    I am so lost. so does x=y and dx/dt=dy/dt? I am a 1st semester high school junior, so I need baby steps.
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  6. #6
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    I think I understand the problem now.

    t is the independant variable, time.

    x is the position of the car.

    a is the angle of the camera.

    You can call them whatever you like, but those are the three quantities we're measuring. We're trying to find \frac{da}{dt}, the angle change per second. They tell us what \frac{dx}{dt} is when they tell us the speed of the car. We then have to find \frac{da}{dx} so we can plug it into the equation below.

    \frac{da}{dt}=\frac{da}{dx}*\frac{dx}{dt}

    That's why you need to find the derivative of the equation in my first reply.
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