Triangle rate problem with derivative and limited operators

[SansMath]

Triangle rate problem with derivative and limited operators

A side-angle-side triangle with angle-A and side-AC are fixed. Side-AB changes with time. I'm attempting to find the change of side-CB and angle-C with time. The catch is the end result can only have mult/div/add/sub operations. If not, then an approximation where side-AC is much larger than side-AB would be appreciated. Thanks.

 

[Ishuda]

The relationship between angle A and the known sides AC an AB can be determined by the Law of Cosines (a generalization of the Pythagorean theorem).

This 'law' also shows that the relationship is not a simple arithmetic one. However, as a hint, an approximation can be made by expanding the square root function.

If you were to tell us what this is about and what you have tried then more appropriate help might be obtained.If you have not read
http://www.freemathhelp.com/forum/threads/41538-Read-Before-Posting!!
it would be a good thing to read.

 

[SansMath]

Thanks for the response.
My problem is to determine the range rate (R’) and azimuth rate (AZ’) with respect to time (t) of a balloon from a given location range old (Ro), azimuth old (AZo), constant wind speed (Ws), and constant wind direction (Wd). My program allows pre-calculating constants with any operators, but can only use mult/div/add/sub when the function is running real time.

I believe the approach is to first determine the function of new range (Rn) with time using the law of cosines:
Rn(t) = sqrt( Ro^2 + (Ws*t)^2 – t*2*Ro*Ws*COS(Wd – AZo) )
Create a constant:
k = 2*Ro*Ws*COS(Wd – AZo)
substitute:
Rn(t) = sqrt( Ro^2 + (Ws*t)^2 – t*k )
Take the derivative with respect to time:
Rn’(t) = 1 / sqrt(Ro^2 + (Ws*t)^2 – t*k) * 2*Ws*t^2 – k)
Is there a way around the sqrt from here? Thanks.