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Math Help - Critical Numbers

  1. #1
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    Critical Numbers

    I need some help with these problems I have to find the critical numbers fo the problems.

    First I need to differentiate the problem and then I need to set it equal to 0 and solve for the variable.

    The first one I just need help differentiating:

    f(theta)=2cos(theta)+sin^2(theta)

    This second one I differentiated but I don't know how to solve for theta:

    g(theta)=4(theta)-tan(theta)
    and
    g'(theta)=4-sec^2(theta)

    I appreciate any help.
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  2. #2
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    Solution

    Well, for the first one, you take the derivative of each term
    Keep in mind:
    The derivative of cosx = -sinx
    The derivative of sinx = cosx

    The first term's derivative would be 2 * -sin(theta) = -2sin(theta)
    The second term's derivative involves the chain rule
    It is gotten by first using the power rule, and then differentiating the trig function inside the parenthesis:
    So, it is 2sin(theta)*cos(theta)

    So, f'(theta) = -2sin(theta) +2sin(theta)cos(theta)

    I'll leave the factoring and solving for theta to you.
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  3. #3
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    Second Question

    g'(theta)=4-sec^2(theta)

    So, you set the right side equal to zero:
    4 - sec^2(theta) = 0

    Using a little algebra:
    sec^2(theta) = 4

    Take the square root of both sides to get:
    sec(theta) = 2

    Now remember, sec(theta) = 1/cos(theta)
    So, 1/cos(theta) = 2

    Take the inverse of both sides to get:
    cos(theta) = 1/2

    I'm sure you know how to do the rest. I hope this helps.
    Also, for the first problem, do some review of the chain rule. It will help you understand how to get the derivatives a lot better.
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  4. #4
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    Quote Originally Posted by ajj86 View Post
    g'(theta)=4-sec^2(theta)

    So, you set the right side equal to zero:
    4 - sec^2(theta) = 0

    Using a little algebra:
    sec^2(theta) = 4

    Take the square root of both sides to get:
    sec(theta) = 2

    Now remember, sec(theta) = 1/cos(theta)
    So, 1/cos(theta) = 2

    Take the inverse of both sides to get:
    cos(theta) = 1/2

    I'm sure you know how to do the rest. I hope this helps.
    Also, for the first problem, do some review of the chain rule. It will help you understand how to get the derivatives a lot better.
    Acutally I don't know what to do next. That's where I'm stuck.
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  5. #5
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    Additional

    Sorry for not replying, but I haven't been on the forum since thursday or so.
    The last part of the second problem involves taking the arccos of both sides to yield:

    theta = arccos(1/2)

    Now, we think what values of theta will cause cos(theta) to equal 1/2.

    well, we know cos(Pi/3) will yield a value of 1/2. Also, since the 4th quadrant is also positive for cosine, the theta value of 5Pi/3 will yield 1/2. Now, the trick is writing these so that it covers all values that will make cos(theta) equal to 1/2.

    theta = n*Pi/3 for n = 1,5,7,11,13,17,...
    Last edited by ajj86; November 2nd 2008 at 05:41 PM. Reason: mistake
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