MHB Angle $\theta$ Changing Rate of Ladder $13ft$ Long

[karush]

A Ladder $13ft$ long is leaning against the side of a building.
If the foot of the ladder is pulled away from the building
at a constant rate of $\displaystyle \frac{2 in}{sec}$
how fast is the angle formed by the ladder and the ground changing in $\displaystyle\frac {rad}{sec}$
at the instant when the top of the ladder is $12 ft$ above the ground.

View attachment 1511

I started with
$\displaystyle\theta = cos^{-1}\left(\frac{\sqrt{169-x^2}}{13}\right)$

$\displaystyle\frac{d\theta}{dt}=\frac{dx}{dt} cos^{-1}\left(\frac{\sqrt{169-x^2}}{13}\right)$

I didn't know how to use the $\displaystyle \frac{2 in}{sec}$ after this

 

[Ackbach]

I would probably go with
$$ \cos( \theta)= \frac{ \sqrt{13^{2}-x^{2}}}{13}.$$
While we're at it, utilize implicit differentiation to the full:
$$ \cos^{2}( \theta)= \frac{13^{2}-x^{2}}{13^{2}}.$$
Next, you must differentiate this equation w.r.t. $t$:
$$-2 \cos( \theta) \sin( \theta) \frac{d \theta}{dt}= \frac{d}{dt} \frac{13^{2}-x^{2}}{13^{2}} = - \frac{2x}{13^{2}} \frac{dx}{dt}.$$
Now solve for $d \theta/dt$ and plug in everything you know.

 

[karush]

$- \frac{2x}{13^{2}} \frac{dx}{dt}$
Now solve for $d \theta/dt$ and plug in everything you know.

so if $\displaystyle\frac{dx}{dt}$ when x=12 is
$\displaystyle\frac{60in}{30sec}$ or 2

then $\displaystyle\frac{2\cdot 12 \cdot 2}{169} \approx \frac {0.28 rad}{sec}$

or is this completely off...

 

[Ackbach]

You need to solve for $d \theta/dt$ first. Never forget that you're going after your Target Variable (which I call the T. V.). You've plugged things in, but not into an expression for $d \theta/dt$.

 

[karush]

$$-2 \cos( \theta) \sin( \theta) \frac{d \theta}{dt}= \frac{d}{dt} \frac{13^{2}-x^{2}}{13^{2}} = - \frac{2x}{13^{2}} \frac{dx}{dt}.$$

so from this to isolate
$$\frac{d \theta}{dt}=
\frac{1}{-2 \cos( \theta)\sin( \theta)}
\cdot\left( - \frac{2x}{13^{2}} \frac{dx}{dt}\right)$$

so if

$$\frac{dx}{dt}=2$$
$$x=12$$
$$\theta \approx 1.17 rad$$

then plug..

 

[karush]

I got $$-\frac{1}{72} \frac {\text {rad}}{\text {sec}}$$

for ans

 

[Ackbach]

I think I see a problem: how did you define your coordinate system? Is $x$ positive to the right? Or up? If $x$ is positive to the right, then $dx/dt=2$, and $x=5$. Also, we may not have set the problem up correctly.

You always need to define your coordinate system early in solving any problem of this sort.