# Critical Number Trig Problem

• Oct 17th 2013, 12:13 PM
Jason76
Critical Number Trig Problem
Find critical numbers:

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

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

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

$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta$ http://www.freemathhelp.com/forum/im...s/confused.png How do we solve for $\displaystyle \theta$ at this point?
• Oct 17th 2013, 12:23 PM
SlipEternal
Re: Critical Number Trig Problem
How do you normally find critical numbers? Set it equal to zero and solve for $\displaystyle \theta$.

$\displaystyle -18\sin \theta + 18\sin \theta \cos \theta = 0$.

Factor out $\displaystyle 18\sin \theta$:

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

Hence, either $\displaystyle 18\sin \theta = 0$ or $\displaystyle \cos \theta -1 = 0$.
• Oct 17th 2013, 12:38 PM
Jason76
Re: Critical Number Trig Problem
Find critical numbers:

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

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

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

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

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

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

$\displaystyle f '(\theta) = 18 (\sin \theta) \cos \theta = 18 \sin \theta$
• Oct 17th 2013, 12:51 PM
SlipEternal
Re: Critical Number Trig Problem
I stand by my initial response.
• Oct 17th 2013, 05:58 PM
Jason76
Re: Critical Number Trig Problem
Quote:

I stand by my initial response.
I agree with you my approach to the problem was all wrong. The way to go is finding things to factor out.

Again:

Find critical numbers:

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

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

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

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

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

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

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

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

$\displaystyle f '(\theta) = (\sin \theta) -18 + \cos \theta = 0$ ?? Next move?
• Oct 17th 2013, 06:17 PM
Prove It
Re: Critical Number Trig Problem
Did you not read SlipEternal's post at all? He left you with ONE step...
• Oct 17th 2013, 06:32 PM
Jason76
Re: Critical Number Trig Problem
Find critical numbers:

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

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

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

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

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

$\displaystyle f '(\theta) = - 18 (\sin \theta)[ 1 + \cos \theta]$ ?? Now?
• Oct 17th 2013, 06:38 PM
SlipEternal
Re: Critical Number Trig Problem
You seem to have tremendous difficulty with algebra. If I multiply out what you wrote:

$\displaystyle -18(\sin \theta)[1+\cos \theta] = -18\sin\theta - 18 \sin\theta \cos\theta$. That is not equal to what you started with. As Prove It said, it appears you either did not read or did not understand my post at all.
• Oct 17th 2013, 06:46 PM
Jason76
Re: Critical Number Trig Problem
Find critical numbers:

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

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

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

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

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

$\displaystyle 18 (\sin \theta)[cos \theta - 1] = 0$ ??
• Oct 17th 2013, 06:49 PM
SlipEternal
Re: Critical Number Trig Problem
Let's multiply out again:

$\displaystyle 18(\sin\theta)[\cos\theta-1] = 18\sin\theta \cos\theta - 18\sin\theta$ Yes, this is correct. Next step, set the derivative equal to zero.
• Oct 17th 2013, 06:55 PM
Jason76
Re: Critical Number Trig Problem
Find critical numbers:

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

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

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

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

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

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

$\displaystyle 18 (\sin \theta)(cos \theta) - 18 (\sin \theta) = 0$ ??
• Oct 17th 2013, 06:58 PM
SlipEternal
Re: Critical Number Trig Problem
Quote:

Originally Posted by Jason76
Find critical numbers:

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

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

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

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

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

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

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

From line 6, you can find solutions. You have the product of two expressions is equal to 0. The expressions are $\displaystyle [18\sin\theta]$ times $\displaystyle [\cos\theta -1]$ equals 0. This means $\displaystyle 18\sin\theta=0$ or $\displaystyle \cos\theta-1=0$.

So, when does $\displaystyle 18\sin\theta=0$? You can divide both sides by 18, so really, I am asking when is $\displaystyle \sin\theta=0$?

Next, when is $\displaystyle \cos\theta -1=0$?