Calculus problem from hell.

[bob000555]

Two pieces of wood, both one meter long, are attached with a hinge; the angle formed by the two pieces of wood is termed theta. When theta is pi over two radians a string is stretched attached to the ends of both sticks such that its length is exactly square root of two. A block is used to hold the pieces of wood such that theta remains constant at ninety degrees then the apparatus is hung from the ceiling such that when the block is released the pieces of wood will close together and the string will fall in a perfect arc. The block is removed. As the pieces of wood fall together we imagine the line AB completes the triangle formed by the two sticks and the line CD extends perpendicularly from the midpoint of AB to the apex of the arc formed by the string. Find the equation of the derivative of the length of CD with respect to theta.

 

I have no clue what to do.

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So far I've done this, using the law of sines

 

[math]

\frac{\sin(\theta)}{f(CD)} = \frac{sin( \frac{2\pi-\theta}{2})}{1}

[/math]

Where [math]f(x)[/math] is a function that takes CD as an input and outputs AB. Does that sound at all right? If so does anyone know what [math] f(x) [/math] is?

[zule]

1- Firstly, you can use the Sine Theorem to relate AB to theta. So, you would have the equation of AB in function of theta.

 

2- You know the length of the string, which is constant, square root of 2. The string would be all the time half an ellipse, being one diameter AB and the other one CD. You can get an equation of CB in function of AB.

 

3- Now, you get together 1 and 2 for getting how much CD is depending on theta.

[bob000555]

1- Firstly, you can use the Sine Theorem to relate AB to theta. So, you would have the equation of AB in function of theta.

 

2- You know the length of the string, which is constant, square root of 2. The string would be all the time half an ellipse, being one diameter AB and the other one CD. You can get an equation of CB in function of AB.

 

3- Now, you get together 1 and 2 for getting how much CD is depending on theta.

 

Did you actually do that and find the solution? If you sit down and try it steps 2 and 3 are nearly imposable. I don't think the law of sines is the way to go. Mooey was trying to solve it with trig for the better part of an hour last night to no avail, and she is a physics major.

[mooeypoo]

Did you actually do that and find the solution? If you sit down and try it steps 2 and 3 are nearly imposable. I don't think the law of sines is the way to go. Mooey was trying to solve it with trig for the better part of an hour last night to no avail, and she is a physics major.

Mooey was trying to get *YOU* to do the work instead of solving it for you, at 11pm at night after a whole day of school and work.

 

The method works, bob, if you just sit down and follow it through.

[bob000555]

Mooey was trying to get *YOU* to do the work instead of solving it for you, at 11pm at night after a whole day of school and work.

 

The method works, bob, if you just sit down and follow it through.

What method works? Did you actually solve it?

[zule]

I am not a physics or mathematics major, but you can solve this problem with basic knowledge. You only have to visualize it.

 

1-You say you have problems with 2 and 3. So, I suppose you don’t have problems with 1 and you have been able to know AB respect to theta.

 

2-The string forms have half an ellipse. You have to use the simpler formula for the ellipse because with a more complex one, I don’t see how to work out the value of CD respect to AB.

I realize I made a mistake: CD wouldn’t be a diameter, but a radius.

Substituting in the formula of the ellipse 2*square root of 2=2pi*square root of {[(AB/2)^2 +CD^2]/2} You can work out the value of CD respect AB from here.

 

 

3- You have AB respect to theta and CD respect to AB. You only have to substitute.

[mooeypoo]

What method works? Did you actually solve it?

Yes, I think I did, and I continued the method we started.

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2-The string forms have half an ellipse. You have to use the simpler formula for the ellipse because with a more complex one, I don’t see how to work out the value of CD respect to AB.

I realize I made a mistake: CD wouldn’t be a diameter, but a radius.

Substituting in the formula of the ellipse 2*square root of 2=2pi*square root of {[(AB/2)^2 +CD^2]/2} You can work out the value of CD respect AB from here.

 

Can you write the full formula? I thought I remembered soemthing like that and didn't find it the other day, I'm just curious. Thanks

 

~moo

[zule]

Can you write the full formula? I thought I remembered soemthing like that and didn't find it the other day, I'm just curious. Thanks

 

~moo

 

You can’t calculate exactly the perimeter of a ellipse, but I have used this approximation:

 

p= 2*pi*square root of[(r^2+s^2)/2]

where p is perimeter and r and s the radii of the ellipse.

 

There is a more exact approximation, but I think is not possible here to work out CD respect to AB (r respect to s). I have found it here (it's the second one, the first one is the one I had used):

http://www.mathsisfun.com/geometry/ellipse-perimeter.html

 

In that same web there are formulas for an exact solution, but they are an addition of infinite terms, so, in the practice, you never get the exact solution. The more term you add, the more exact approximation you get. What I don’t know is if those late formulas could be integrated and so you could work out the derivate of CD respect to AB. But I haven’t done an integral or derivate in the last 20 years, I don’t remember almost anything about them. So, it is only an idea I don’t know if it is possible.

[mooeypoo]

You can’t calculate exactly the perimeter of a ellipse, but I have used this approximation:

 

p= 2*pi*square root of[(r^2+s^2)/2]

where p is perimeter and r and s the radii of the ellipse.

 

There is a more exact approximation, but I think is not possible here to work out CD respect to AB (r respect to s). I have found it here (it's the second one, the first one is the one I had used):

http://www.mathsisfun.com/geometry/ellipse-perimeter.html

 

In that same web there are formulas for an exact solution, but they are an addition of infinite terms, so, in the practice, you never get the exact solution. The more term you add, the more exact approximation you get. What I don’t know is if those late formulas could be integrated and so you could work out the derivate of CD respect to AB. But I haven’t done an integral or derivate in the last 20 years, I don’t remember almost anything about them. So, it is only an idea I don’t know if it is possible.

 

You're right, calculating the perimeter of the ellipse is a pain - which is why that's not what I did. I divided the "half ellipse" into triangles. CD is the common 'leg' of 2 triangles trapped inside the half ellipse. Then, trig relationships can show you how to get CD in terms of theta.

 

I'm at work at the moment, but when I get back home and with a few spare moments I will try to put up an image.

[zule]

Yes, I had though about the triangles and it supposes only a little less of work than using the formula of the ellipse I have used. And I though the result was more exact by considering the ellipse. However, I now read on the page I have previously linked that the simplest ellipse equation has little exactitude when one of the radius is more than three times bigger than the other.

 

So, perhaps it is more exact to use the approximation of triangles when r > 3s and when r < s/3? I don’t know. In any case, I think it would be better to continue using the formula of ellipse when s/3 < r <3s

 

If I haven’t had any mistake in the calculus, CD would be 1.57 times bigger if you calculate it by the approximation of triangles than if you calculate it by the formula of ellipse.