Critical Number Trig Problem

[Jason76]

Find critical numbers:

$\displaystyle f(\theta) = 18 \cos \theta + 9 \sin^{2} \theta$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 9 (u) du$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (u) \cos \theta$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta$ How do we solve for $\displaystyle \theta$ at this point?

 

[Deleted member 4993]

Find critical numbers:

$\displaystyle f(\theta) = 18 \cos \theta + 9 \sin^{2} \theta$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 9 (u) du$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (u) \cos \theta$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta$ How do we solve for $\displaystyle \theta$ at this point?

For what value of f'(Θ)?

 

[HallsofIvy]

Find critical numbers:

$\displaystyle f(\theta) = 18 \cos \theta + 9 \sin^{2} \theta$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 9 (u) du$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (u) \cos \theta$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta$ How do we solve for $\displaystyle \theta$ at this point?

First, you have to have an equation to solve!

If I remember correctly, a "critical number" for a function is a value of the variable at which the derivative is 0 or does not exist. Sine and cosine are defined for all $\displaystyle \theta$ so the question is just "where is
$\displaystyle -18 sin(\theta)+ 18sin(\theta)cos(\theta)= -18 sin(\theta)(1- cos(\theta))= 0$
So $\displaystyle sin(\theta)$ and $\displaystyle cos(\theta)$ must be equal to what?

 

[Jason76]

Sorry guys, this makes no sense to me at all.

 

[Jason76]

Find critical numbers:

$\displaystyle f(\theta) = 18 \cos \theta + 9 \sin^{2} \theta$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 9 (u) du$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (u) \cos \theta$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta = 0$ Sorry left the $\displaystyle 0$ out.

$\displaystyle f '(\theta) = 18 \sin \theta + 18 (\sin \theta) \cos \theta = 0$

$\displaystyle f '(\theta) = 18 (\sin \theta) \cos \theta = 18 \sin \theta$ Next move?

 

[JeffM]

Find critical numbers:

$\displaystyle f(\theta) = 18 \cos \theta + 9 \sin^{2} \theta$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 9 (u) du$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (u) \cos \theta$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta = 0$ Sorry left the $\displaystyle 0$ out.

$\displaystyle f '(\theta) = 18 \sin \theta + 18 (\sin \theta) \cos \theta = 0$

$\displaystyle f '(\theta) = 18 (\sin \theta) \cos \theta = 18 \sin \theta$ Next move?

$\displaystyle 18 \{-sin (\theta )\} + 18 \{sin (\theta ) \cos (\theta )\} = 0 \implies -sin(\theta ) + sin(\theta )cos(\theta) = 0 \implies sin(\theta ) * \{cos(\theta ) - 1\} = 0.$

Now apply the zero product property just as HOI explained above.

 

[Jason76]

Now let's use factoring as our approach to solving this problem.

Find critical numbers:

$\displaystyle f(\theta) = 18 \cos \theta + 9 \sin^{2} \theta$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 9 (u) du$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (u) \cos \theta$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta = 0$

$\displaystyle f '(\theta) = 18[(-\sin \theta) + (\sin \theta) \cos \theta] = 0$

$\displaystyle f '(\theta) = -18[(\sin \theta) + (\sin \theta) \cos \theta] = 0$

$\displaystyle f '(\theta) = (\sin \theta) [-18 + \cos \theta] = 0$

$\displaystyle f '(\theta) = (\sin \theta) -18 + \cos \theta = 0$ :?:

 

[Imum Coeli]

Maybe if you write f'(x) = sin(x)*(18*cos(x)-18) then find the values of x such that f'(x)= 0?

 

[Jason76]

Maybe if you write f'(x) = sin(x)*(18*cos(x)-18) then find the values of x such that f'(x)= 0?

I'm trying to solve for $\displaystyle \theta$ How is this possible in this situation?

 

[Jason76]

How about this?

Find critical numbers:

$\displaystyle f(\theta) = 18 \cos \theta + 9 \sin^{2} \theta$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 9 (u) du$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (u) \cos \theta$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta$

$\displaystyle f '(\theta) = -[18 (\sin \theta) + 18 (\sin \theta) \cos \theta]$

$\displaystyle f '(\theta) = - 18 (\sin \theta)[ 1 + \cos \theta]$ :?:

 

[Imum Coeli]

I'm trying to solve for $\displaystyle \theta$ How is this possible in this situation?

As Halls of Ivy pointed out you just need to find values of x such that f'(x) = 0 (as the function is continuous). By inspection and a knowledge of the unit circle it should be possible to see what values of x satisfy this condition. Those are your critical points.

(Sorry, for my example let theta = x)

 

[Jason76]

Another way to do this:

Find critical numbers:

$\displaystyle f(\theta) = 18 \cos \theta + 9 \sin^{2} \theta$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 9 (u) du$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (u) \cos \theta$

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta$

$\displaystyle 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta = 0$

$\displaystyle 18 (\sin \theta)[cos \theta - 1] = 0$

$\displaystyle 18 (\sin \theta)(cos \theta) - 18 (\sin \theta) = 0$ :?:

 

[HallsofIvy]

You are just going around and around in circles. Before taking Calculus, you should have learned algebra! And one of the things you should have learned in algebra is the "zero product property": if ab= 0 then either a= 0 or b= 0 (or both).
You have (as you wrote way back in your first post) $\displaystyle 18sin(\theta)(cos(\theta)- 1)= 0$. "18" can never be 0, so you must have either $\displaystyle sin(\theta)= 0$ or $\displaystyle cos(\theta)- 1= 0$. The second is, of course, equivalent to $\displaystyle cos(\theta)= 1$.