waltgrace83
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I feel like I am missing crucial elements to this question. Thoughts? This really doesn't seem hard, but I CANNOT figure it out (work enclosed).
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HallsofIvy
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"h". in your picture, is the height above the center of the circle. You want want the height above the ground which is 8+h to be 12 feet. Then h= 12- 8= 4.
You say you have \(\displaystyle \frac{dh}{dt}= \frac{2\pi}{3min}\). I don't know what the "min" in the denominator means. But the fact that the rate of rotation is constant does not mean that the rate of change of h is a constant. If the rate of rotation is \(\displaystyle 2\pi/3\) radians per minute then \(\displaystyle h= 8(sin(2\pi t/3)+ 1)\). That will be 12 when \(\displaystyle 12= 8(sin(2\pi t/3)+ 1)\). \(\displaystyle sin(2\pi t/3)+ 1= 12/8= 3/2\). \(\displaystyle sin(2\pi t/3)= 1/2\). \(\displaystyle 2\pi t/3= 30\), \(\displaystyle t= 45/\pi\).
Evaluate the derivative of h, \(\displaystyle dh/dt= (16\pi/3)cos(2\pi t/3)\), at \(\displaystyle t= 45/\pi\).
You say you have \(\displaystyle \frac{dh}{dt}= \frac{2\pi}{3min}\). I don't know what the "min" in the denominator means. But the fact that the rate of rotation is constant does not mean that the rate of change of h is a constant. If the rate of rotation is \(\displaystyle 2\pi/3\) radians per minute then \(\displaystyle h= 8(sin(2\pi t/3)+ 1)\). That will be 12 when \(\displaystyle 12= 8(sin(2\pi t/3)+ 1)\). \(\displaystyle sin(2\pi t/3)+ 1= 12/8= 3/2\). \(\displaystyle sin(2\pi t/3)= 1/2\). \(\displaystyle 2\pi t/3= 30\), \(\displaystyle t= 45/\pi\).
Evaluate the derivative of h, \(\displaystyle dh/dt= (16\pi/3)cos(2\pi t/3)\), at \(\displaystyle t= 45/\pi\).
Dr.Peterson
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I'm guessing you meant to say that \(\frac{d\theta}{dt} = \frac{2\pi}{3}\) radians per minute.
From the diagram:
Rising speed is:
[math]v_{up}=\dfrac{dh}{dt}[/math]
[math]h=R\sin{\theta}[/math]
deriving it with respect to time:
[math]\dfrac{dh}{dt}=R\cos{\left(\theta\right)}\dfrac{d\theta}{dt}=R\omega\cos{\left(\theta\right)}=v_{up}[/math] (Since [math]\omega=\dfrac{d\theta}{dt}[/math])
[math]\sin{\theta}=\dfrac{h}{R}\rightarrow \cos{\theta}=\sqrt{1-\sin^2\theta}=\dfrac{1}{R}\sqrt{R^2-h^2}[/math]
Then:
[math]\boxed{v_{up}=\omega\sqrt{R^2-h^2}}[/math]
NOTE:
As @HallsofIvy said, here [math]h=Height-R[/math] where Height is the given one in the problem.
Rising speed is:
[math]v_{up}=\dfrac{dh}{dt}[/math]
[math]h=R\sin{\theta}[/math]
deriving it with respect to time:
[math]\dfrac{dh}{dt}=R\cos{\left(\theta\right)}\dfrac{d\theta}{dt}=R\omega\cos{\left(\theta\right)}=v_{up}[/math] (Since [math]\omega=\dfrac{d\theta}{dt}[/math])
[math]\sin{\theta}=\dfrac{h}{R}\rightarrow \cos{\theta}=\sqrt{1-\sin^2\theta}=\dfrac{1}{R}\sqrt{R^2-h^2}[/math]
Then:
[math]\boxed{v_{up}=\omega\sqrt{R^2-h^2}}[/math]
NOTE:
As @HallsofIvy said, here [math]h=Height-R[/math] where Height is the given one in the problem.
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