Hey linalg123.
Are the trig functions in radians or degrees?
Mt attempts are in bold. not sure if they are correct
The path of motion of the sun across the sky during the day can be modelled as a circular orbit on an inclined plane called the ecliptic. On the equinox days the sun rises due east and sets due west, taking exactly 12 hours to cross from horizon to horizon.
a) What is the angular speed, w, in degrees/minute, of the sun across the sky.
180deg / 720minutes = 0.25 deg/minute?
b) if the position of the sun on the ecliptic is modelled by a parametric circular arc, find parametric functions r1(t) and r2(t) such that r(t) = r1(t)u + r2(t)v where r(t) is the vector from the local observor at oridion 0, the the sun P.
Assume t=0 corresponds to sunrise. u and v are unit vectors.
r1(t) = r*cos(wt)
r2(t)= r*sin(wt)
c) If the directions E= east, N=north, U=up are aligned to the basis {i,j,k} and the orthonormal basis for the ecliptic plane is {u,v} where u = i and v= 0i + sin( )j + cos( )k. Find the vector p(t) = x(t)i + y(t)j + z(t)k descibing the displacement of P about 0 with respect to the basis {i,j,k}, as a function of , w ,r and t
r(t) = rcos(wt)u + r(sin(wt)v
r(t) = rcos(wt) i + rsin(wt)(0,sin( ),cos( ))
r(t) = rcos(wt)i + rsin(wt)sin( )j + rsin(wt)cos( )k
Since the angles only appear inside the trig functions why would that be of any consequence?
Now if you want to do a claculation using these equations on some form of caluclator you will need to ensure that you and the calculator are using the same angular mode, but again it makes no difference to the formulae.
.
It makes a difference in terms of what t corresponds to for a specific location. If its in degrees then its usually the case where you introduce a scaling constant because trig functions are conventionally used for radian measures.