Using derivatives
for the function where mgu is a constant and if u = 0.2
$\displaystyle \frac{mgu}{sin\theta+ucos\theta}$
critical points occur where the derivative is zero or undefined.
so first find your derivative, then you can see where it is undefined. that will be a critical point. then set it equal to zero and solve for $\displaystyle \theta$, to find the other critical points (what is your dependent variable?)
to start you off, write as $\displaystyle mgu( \sin \theta + u \cos \theta )^{-1}$.
now use the chain rule (remember how to deal with constants)
would I consider
$\displaystyle ucos\theta$
A derivative I would need to use the product rule for?
Ok So I get
$\displaystyle \frac{mgu(cos\theta-usin\theta)}{(sin\theta+ucos\theta)^2}$
So I solve the denom. for 0?
So if that is correct then
$\displaystyle sin\theta+ucos\theta=0$
$\displaystyle ucos\theta=-sin\theta$
$\displaystyle -u=\frac{sin\theta}{cos\theta}$
Then it stats u = 0.2 so
$\displaystyle -0.2=tan\theta$
$\displaystyle arctan(-0.2)=-.197$
which is the critical pt.?
$\displaystyle \frac d{d \theta} mgu (\sin \theta + u \cos \theta)^{-1} = -mgu (\sin \theta + u \cos \theta)^{-2}(\cos \theta - u \sin \theta) $ $\displaystyle = mgu (\sin \theta + u \cos \theta)^{-2}(- \cos \theta + u \sin \theta) = \frac {mgu(u \sin \theta - \cos \theta)}{(\sin \theta + u \cos \theta)^2}$