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Math Help - Critical Number Trig Problem

  1. #1
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    Critical Number Trig Problem

    Find critical numbers:

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

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

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

    f '(\theta) = 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta How do we solve for \theta at this point?
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  2. #2
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    Re: Critical Number Trig Problem

    How do you normally find critical numbers? Set it equal to zero and solve for \theta.

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

    Factor out 18\sin \theta:

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

    Hence, either 18\sin \theta = 0 or \cos \theta -1 = 0.
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  3. #3
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    Re: Critical Number Trig Problem

    Find critical numbers:

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

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

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

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

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

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

    f '(\theta) = 18 (\sin \theta) \cos \theta = 18 \sin \theta
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  4. #4
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    Re: Critical Number Trig Problem

    I stand by my initial response.
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    Re: Critical Number Trig Problem

    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:

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

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

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

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

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

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

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

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

    f '(\theta) = (\sin \theta) -18 + \cos \theta = 0 ?? Next move?
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  6. #6
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    Re: Critical Number Trig Problem

    Did you not read SlipEternal's post at all? He left you with ONE step...
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    Re: Critical Number Trig Problem

    Find critical numbers:

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

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

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

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

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

    f '(\theta) = - 18 (\sin \theta)[ 1 + \cos \theta] ?? Now?
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    Re: Critical Number Trig Problem

    You seem to have tremendous difficulty with algebra. If I multiply out what you wrote:

    -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.
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    Re: Critical Number Trig Problem

    Find critical numbers:

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

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

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

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

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

    18 (\sin \theta)[cos \theta - 1] = 0 ??
    Last edited by Jason76; October 17th 2013 at 07:52 PM.
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    Re: Critical Number Trig Problem

    Let's multiply out again:

    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.
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  11. #11
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    Re: Critical Number Trig Problem

    Find critical numbers:

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

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

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

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

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

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

    18 (\sin \theta)(cos \theta) - 18 (\sin \theta) = 0 ??
    Last edited by Jason76; October 17th 2013 at 07:59 PM.
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  12. #12
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    Re: Critical Number Trig Problem

    Quote Originally Posted by Jason76 View Post
    Find critical numbers:

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

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

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

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

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

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

    7. 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 [18\sin\theta] times [\cos\theta -1] equals 0. This means 18\sin\theta=0 or \cos\theta-1=0.

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

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