# Related rates, finding an angle.

• Mar 27th 2011, 06:14 PM
david13579
Related rates, finding an angle.
I was trying to solve this problem:

A ladder 13ft long is leaning against a wall. It suddenly starts to slide away at the base of the ladder. When it is 12ft away from the wall it is sliding away at 5ft/s

if it is 12 feet away then at that moment the other side is 5ft high
I found that at that moment the side on the wall is coming down at -12ft/s

What I can't find is how much the angle between the base and the ground is changing at that moment. I know it is -1 rad/s but I can't get there.
• Mar 27th 2011, 07:01 PM
skeeter
Quote:

Originally Posted by david13579
I was trying to solve this problem:

A ladder 13ft long is leaning against a wall. It suddenly starts to slide away at the base of the ladder. When it is 12ft away from the wall it is sliding away at 5ft/s

if it is 12 feet away then at that moment the other side is 5ft high
I found that at that moment the side on the wall is coming down at -12ft/s

What I can't find is how much the angle between the base and the ground is changing at that moment. I know it is -1 rad/s but I can't get there.

I assume you mean the angle between the ladder and the ground ... the rate of change is not -1 rad/s , it's -13/12 rad/s
• Mar 27th 2011, 07:05 PM
david13579
Quote:

Originally Posted by skeeter
I assume you mean the angle between the ladder and the ground ... the rate of change is not -1 rad/s , it's -13/12 rad/s

-1 is the answer my books gives. And yes, I meant the base of the ladder and the ground.
• Mar 28th 2011, 05:29 AM
HallsofIvy
If $\displaystyle \theta$ is the angle between the ladder and the ground then $\displaystyle tan(\theta)= \frac{h}{x}$ where h is the height of the ladder on the wall and x is the distance from the base of the ladder to the wall.

$\displaystyle \frac{1}{1+ \theta^2}\frac{d\theta}{dt}= \frac{1}{x}\frac{dh}{dt}- \frac{h}{x^2}\frac{dx}{dt}$.

We are told that, at this point, x= 12, h= 5, $\displaystyle \frac{dx}{dt}= 5$, and you have calculated that $\displaystyle \frac{dh}{dt}= -12$. Also $\displaystyle tan(\theta)= \frac{5}{12}$ so $\displaystyle \theta= 0.3948$ radians.

Putting all of that into the formula:
$\displaystyle \frac{1}{1+ 0.1559}\frac{d\theta}{dt}= \frac{1}{12}(-12)- \frac{5}{144}(5}= \frac{12+ 25}{144}= \frac{169}{144}$
$\displaystyle = -1.0187$ which is very close to -1.
• Mar 28th 2011, 11:32 AM
skeeter
well dip me in dog ...

... sure does work out to be -1

$\displaystyle \theta = \arctan\left(\dfrac{h}{x}\right)$

$\displaystyle \dfrac{d\theta}{dt} = \dfrac{x \cdot \frac{dh}{dt} - h \cdot \frac{dx}{dt}}{x^2} \cdot \dfrac{1}{1 + \left(\frac{h}{x}\right)^2}$

$\displaystyle \dfrac{d\theta}{dt} = \dfrac{12 \cdot (-12) - 5 \cdot 5}{12^2} \cdot \dfrac{1}{1 + \left(\frac{5}{12}\right)^2}$

$\displaystyle \dfrac{d\theta}{dt} = \dfrac{-169}{12^2} \cdot \dfrac{1}{1 + \left(\frac{5}{12}\right)^2}$

$\displaystyle \dfrac{d\theta}{dt} = \dfrac{-169}{12^2\left[1 + \left(\frac{5^2}{12^2}\right)\right]} = -1$

need to watch my arithmetic.
• Mar 28th 2011, 03:17 PM
david13579
Quote:

Originally Posted by skeeter
well dip me in dog ...

... sure does work out to be -1

$\displaystyle \theta = \arctan\left(\dfrac{h}{x}\right)$

$\displaystyle \dfrac{d\theta}{dt} = \dfrac{x \cdot \frac{dh}{dt} - h \cdot \frac{dx}{dt}}{x^2} \cdot \dfrac{1}{1 + \left(\frac{h}{x}\right)^2}$

$\displaystyle \dfrac{d\theta}{dt} = \dfrac{12 \cdot (-12) - 5 \cdot 5}{12^2} \cdot \dfrac{1}{1 + \left(\frac{5}{12}\right)^2}$

$\displaystyle \dfrac{d\theta}{dt} = \dfrac{-169}{12^2} \cdot \dfrac{1}{1 + \left(\frac{5}{12}\right)^2}$

$\displaystyle \dfrac{d\theta}{dt} = \dfrac{-169}{12^2\left[1 + \left(\frac{5^2}{12^2}\right)\right]} = -1$

need to watch my arithmetic.

Thanks :). Though I had already worked it out with the help of my professor.