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Pavel314

Joined: 29 Apr 2005
Posts: 78 Posted: Thu Jul 06, 2006 1:39 am    Post subject: Trig Derivatives Problem 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 Frederick Williams

Joined: 19 Nov 2005
Posts: 97 Posted: Thu Jul 06, 2006 2:08 am    Post subject: Re: Trig Derivatives Problem 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).

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Joined: 31 May 2005
Posts: 180 Posted: Thu Jul 06, 2006 5:54 pm    Post subject: Re: Trig Derivatives Problem "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.  Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year Oldest FirstNewest First
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