1. ## 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!

2. Originally Posted by coolio
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
$x = L~sin(\theta)$
where L is the length of the ladder.

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

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

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

-Dan

3. Originally Posted by coolio
2. lim t -> 0 tan6t / sin 2t
I'll borrow a trick from one of the other members here:
$\lim_{t \to 0} \frac{tan(6t)}{sin(2t)}$

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

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

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

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

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

-Dan

4. Originally Posted by coolio
3. lim x -> 0 sin(cosx) / secx
Why is this a problem?
$\lim_{x \to 0} \frac{sin(cos(x))}{sec(x)} = \frac{sin(cos(0))}{sec(0)} = \frac{1}{1} = 1$

-Dan