# Thread: Related Rates - Angle rate

1. ## Related Rates - Angle rate

I'm not sure weather this is correct or not, but I gave it my best shot and it looks correct... but I'm not sure if it is.

A 10 ft plank is leaning against a wall. If at a certain instant the bottom of the plank is 2 ft from the wall and is pushed toward the wall at a rate of 6 in/sec, how fast is the acute angle that the plank makes with ground increasing?
So here is my try:

cos(theta) = 2/10 = 2/10
-sin(theta) d(theta)/dt = 2/10 dx/dt
-sin(theta) d(theta)/dt = 2/10 * .5ft/sec
-sin(theta) d(theta)/dt = 1/10
d(theta)/dt = 1/10 / -sin(theta)
d(theta)/dt = (1/10) / (sqrt(96) / 10)

I'm not sure weather or not to convert inches to feet. But when I don't convert then I get 12/sqrt(96) as my answer. :| Am I on the right track here?

2. Originally Posted by nautica17
I'm not sure weather this is correct or not, but I gave it my best shot and it looks correct... but I'm not sure if it is.

So here is my try:

cos(theta) = 2/10 = 2/10
-sin(theta) d(theta)/dt = 2/10 dx/dt
-sin(theta) d(theta)/dt = 2/10 * .5ft/sec
-sin(theta) d(theta)/dt = 1/10
d(theta)/dt = 1/10 / -sin(theta)
d(theta)/dt = (1/10) / (sqrt(96) / 10)

I'm not sure weather or not to convert inches to feet. But when I don't convert then I get 12/sqrt(96) as my answer. :| Am I on the right track here?

$\cos{\theta} = \frac{x}{10}$

take the time derivative ...

$-\sin{\theta} \cdot \frac{d\theta}{dt} = \frac{1}{10} \cdot \frac{dx}{dt}$

note that $\frac{dx}{dt} = -\frac{1}{2} \, ft/sec$ because the variable distance x is decreasing.

$-\frac{\sqrt{96}}{10} \cdot \frac{d\theta}{dt} = \frac{1}{10} \cdot \left(-\frac{1}{2}\right)$

$\frac{d\theta}{dt} = \frac{1}{10} \cdot \left(-\frac{1}{2}\right) \cdot \left(-\frac{10}{\sqrt{96}}\right)$

$\frac{d\theta}{dt} = \frac{1}{8\sqrt{6}} \, rad/sec$

### a 10 ft plank is leaning against a wall

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