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Math Help - related rate angle so close...

  1. #1
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    related rate angle so close...

    A truck 20 feet high drives away from a video camera located on the ground at 10 feet per second. How fast is the angle between the road and the top of the truck shrinking from the point of view of the camera when the truck is 20 feet away?

    So far what I've gotten is that I can use tan and I can find the derivative...I'll use @ for theta.

    tan@(t) = 20/d(t) or 20 * d(t) ^-1

    derivative: sec^2(t)@'(t) = -20(d(t))^-2(d'(t))

    I know my d(t) = 20
    and d'(t) = 10
    so I plug those in and get .5 for an answer. I know the ultimate answer is actually -1/4 radians but I don't know where I went wrong or how to get that.

    Thanks so much for any help!!
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  2. #2
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    \tan{\theta} = \frac{20}{x}

    \sec^2{\theta} \cdot \frac{d\theta}{dt} = -\frac{20}{x^2} \cdot \frac{dx}{dt}

    \frac{d\theta}{dt} = -\frac{20}{x^2} \cdot \frac{dx}{dt} \cdot \cos^2{\theta}

    \frac{d\theta}{dt} = -\frac{20}{20^2} \cdot 10 \cdot \left(\frac{\sqrt{2}}{2}\right)^2 = -\frac{1}{4} \, \, rad/s<br />
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    where does the cos^2 come from??

    and how do I know that cos^2 of theta = sqrt(2)/2 ? sorry if this is obvious to everyone but me...
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  4. #4
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    Quote Originally Posted by littlejodo View Post
    where does the cos^2 come from??
    \frac{1}{\sec^2{\theta}} = \cos^2{\theta}

    and how do I know that cos^2 of theta = sqrt(2)/2 ?
    look at the triangle formed ... you did sketch a picture, right?

    x = 20 , height = 20 ... an isosceles right triangle, correct? ... \theta = \frac{\pi}{4}

    \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}
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