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Thread: Trig Derivatives

  1. #1
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    Question Trig Derivatives

    I'm looking for some help on these problems:

    1. A ladder 10 ft long rests against a vertical wall. Let theta be the angle between the top of the ladder and the wall and let x be the distance from the bottom of the ladder to the wall. If the bottom of the ladder slides away from the wall, how fast does x change with respect to theta when theta = pi/3

    2. lim t -> 0 tan6t / sin 2t

    3. lim x -> 0 sin(cosx) / secx

    Thanks a lot!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by coolio View Post
    I'm looking for some help on these problems:

    1. A ladder 10 ft long rests against a vertical wall. Let theta be the angle between the top of the ladder and the wall and let x be the distance from the bottom of the ladder to the wall. If the bottom of the ladder slides away from the wall, how fast does x change with respect to theta when theta = pi/3
    $\displaystyle x = L~sin(\theta)$
    where L is the length of the ladder.

    So
    $\displaystyle \frac{dx}{d \theta} = L~cos(\theta)$

    So when $\displaystyle \theta = \frac{\pi}{3}$,
    $\displaystyle \frac{dx}{d \theta} = L~cos \left ( \frac{\pi}{3} \right ) = L \cdot \frac{1}{2}$

    Note that the units for $\displaystyle \frac{dx}{d \theta}$ are going to be "$\displaystyle ft/rad$."

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by coolio View Post
    2. lim t -> 0 tan6t / sin 2t
    I'll borrow a trick from one of the other members here:
    $\displaystyle \lim_{t \to 0} \frac{tan(6t)}{sin(2t)}$

    $\displaystyle = \lim_{t \to 0} \frac{\frac{6t}{6t} \cdot tan(6t)}{\frac{2t}{2t} \cdot sin(2t)}$

    $\displaystyle = \lim_{t \to 0} \frac{6t}{2t} \cdot \frac{\frac{tan(6t)}{6t}}{\frac{sin(2t)}{2t}}$

    $\displaystyle = 3 \cdot \lim_{t \to 0} \frac{\frac{tan(6t)}{6t}}{\frac{sin(2t)}{2t}}$

    Now,
    $\displaystyle \lim_{t \to 0}\frac{tan(6t)}{6t} = 1$
    and
    $\displaystyle \lim_{t \to 0}\frac{sin(2t)}{2t} = 1$

    So
    $\displaystyle \lim_{t \to 0} \frac{tan(6t)}{sin(2t)} = 3 \cdot \lim_{t \to 0} \frac{\frac{tan(6t)}{6t}}{\frac{sin(2t)}{2t}} = 3 \cdot \frac{1}{1} = 3$

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by coolio View Post
    3. lim x -> 0 sin(cosx) / secx
    Why is this a problem?
    $\displaystyle \lim_{x \to 0} \frac{sin(cos(x))}{sec(x)} = \frac{sin(cos(0))}{sec(0)} = \frac{1}{1} = 1$

    -Dan
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