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Forum index » Science and Technology » Math » Recreational
Trig Derivatives Problem
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Pavel314
science forum addict


Joined: 29 Apr 2005
Posts: 78

PostPosted: Thu Jul 06, 2006 1:39 am    Post subject: Trig Derivatives Problem Reply with quote

I'm trying to solve a problem which involves trigonometric derivatives,
something I haven't worked with in about 35 years. The problem is to find
the velocity at any point along a path of travel as described at my webpage

http://mywebpages.comcast.net/pavel314/jhocdoor.html

To summarize my dilemma, for constant D and angle A, I end up with the
derivative

d(D*cot(A))/dA = -D*csc(A)^2

But this could also be stated as d(D*(cos(A)/sin(A))/dA = -D according to
the interactive derivative website http://www.webmath.com/diff.html

Any help would be greatly appreciated.

Paul
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Frederick Williams
science forum addict


Joined: 19 Nov 2005
Posts: 97

PostPosted: Thu Jul 06, 2006 2:08 am    Post subject: Re: Trig Derivatives Problem Reply with quote

Pavel314 wrote:
Quote:

I'm trying to solve a problem which involves trigonometric derivatives,
something I haven't worked with in about 35 years. The problem is to find
the velocity at any point along a path of travel as described at my webpage

http://mywebpages.comcast.net/pavel314/jhocdoor.html

To summarize my dilemma, for constant D and angle A, I end up with the
derivative

d(D*cot(A))/dA = -D*csc(A)^2

You're right.

Quote:
But this could also be stated as d(D*(cos(A)/sin(A))/dA = -D according to
the interactive derivative website http://www.webmath.com/diff.html

It's wrong (or you and I don't know how to use it).

--
Remove "antispam" and ".invalid" for e-mail address.
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Greg Neill
science forum Guru Wannabe


Joined: 31 May 2005
Posts: 180

PostPosted: Thu Jul 06, 2006 5:54 pm    Post subject: Re: Trig Derivatives Problem Reply with quote

"Pavel314" <Pavel314@NOSPAM.comcast.net> wrote in message
news:tJWdnQKujsQk9DHZnZ2dnUVZ_u6dnZ2d@comcast.com...
Quote:
I'm trying to solve a problem which involves trigonometric derivatives,
something I haven't worked with in about 35 years. The problem is to find
the velocity at any point along a path of travel as described at my webpage

http://mywebpages.comcast.net/pavel314/jhocdoor.html

To summarize my dilemma, for constant D and angle A, I end up with the
derivative

d(D*cot(A))/dA = -D*csc(A)^2

But this could also be stated as d(D*(cos(A)/sin(A))/dA = -D according to
the interactive derivative website http://www.webmath.com/diff.html

Any help would be greatly appreciated.

Paul


Let r be the instantaneous radius of the section of
door that intersects with the "latitude" of the
desired chord trajectory. Let h be that latitude,
i.e., the height of the chord from the horizontal
diameter of the revolving door. Also let w be the
rotational velocity of the door.

The tangential velocity of the end of the radius vector
r is given by Vt = w*r. Construct a triangle using
this tangential velocity as a hypotenuse and horizontal
vertical projections from ends of it as its legs. The
horizontal leg will be the horizontal component of the
velocity Vt. Call this V.

This triangle will be similar to the one comprising
the door radius r as hypotenuse with vertical leg
of length h being the height of the trajectory chord
from the centerline. So construct the ratios:

h/r = V/Vt

Thus,

V = Vt*h/r

= (w*r)*h/r

= w*h

The velocity is constant and equal to w*h so long
as the moving body is in contact with the door.
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