# Math Help - Find The Critical Points

1. ## Find The Critical Points

Using derivatives

for the function where mgu is a constant and if u = 0.2

$\frac{mgu}{sin\theta+ucos\theta}$

2. Originally Posted by sk8erboyla2004
Using derivatives

for the function where mgu is a constant and if u = 0.2

$\frac{mgu}{sin\theta+ucos\theta}$
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 $\theta$, to find the other critical points (what is your dependent variable?)

to start you off, write as $mgu( \sin \theta + u \cos \theta )^{-1}$.

now use the chain rule (remember how to deal with constants)

3. would I consider

$ucos\theta$

A derivative I would need to use the product rule for?

Ok So I get

$\frac{mgu(cos\theta-usin\theta)}{(sin\theta+ucos\theta)^2}$

So I solve the denom. for 0?

So if that is correct then

$sin\theta+ucos\theta=0$
$ucos\theta=-sin\theta$
$-u=\frac{sin\theta}{cos\theta}$

Then it stats u = 0.2 so

$-0.2=tan\theta$
$arctan(-0.2)=-.197$

which is the critical pt.?

4. Originally Posted by sk8erboyla2004
would I consider

$ucos\theta$

A derivative I would need to use the product rule for?

Ok So I get

$\frac{mgu(cos\theta-usin\theta)}{(sin\theta+ucos\theta)^2}$
.
.
.
it should be $\frac{mgu( u \sin \theta - \cos\theta)}{(\sin\theta+u \cos\theta)^2}$

then, set BOTH the numerator and denominator equal to zero to find all critical points.

5. Originally Posted by Jhevon
it should be $\frac{mgu( u \sin \theta - \cos\theta)}{(\sin\theta+u \cos\theta)^2}$

then, set BOTH the numerator and denominator equal to zero to find all critical points.

Shouldnt this $\frac{mgu( u \sin \theta - \cos\theta)}{(\sin\theta+u \cos\theta)^2}$

be

$-\frac{mgu( u \sin \theta - \cos\theta)}{(\sin\theta+u \cos\theta)^2}$

Just wanted to make sure?

6. Originally Posted by sk8erboyla2004
Shouldnt this $\frac{mgu( u \sin \theta - \cos\theta)}{(\sin\theta+u \cos\theta)^2}$

be

$-\frac{mgu( u \sin \theta - \cos\theta)}{(\sin\theta+u \cos\theta)^2}$

Just wanted to make sure?
$\frac d{d \theta} mgu (\sin \theta + u \cos \theta)^{-1} = -mgu (\sin \theta + u \cos \theta)^{-2}(\cos \theta - u \sin \theta)$ $= mgu (\sin \theta + u \cos \theta)^{-2}(- \cos \theta + u \sin \theta) = \frac {mgu(u \sin \theta - \cos \theta)}{(\sin \theta + u \cos \theta)^2}$

7. so you really just distrbuted the negative in the numer. at rewrote it
Ok well for my critical points I get

x = 1.373
x = -.197

8. Originally Posted by sk8erboyla2004
so you really just distrbuted the negative in the numer. at rewrote it
yes

Ok well for my critical points I get

x = 1.373
x = -.197
yes

just finding two solutions is enough?

9. Well It said find all critcal points and I assume 2 would be good enough?? sorry for being confusing lol

10. Originally Posted by sk8erboyla2004
Well It said find all critcal points and I assume 2 would be good enough?? sorry for being confusing lol
the thing is, you are dealing with trig functions. there are infinitely many solutions. are you sure they want only two?