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Math Help - Derivative problem that has me stumped.

  1. #1
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    Derivative problem that has me stumped.

    Hello,

    This is my first time using this site, so if I'm not clear with the equations or formulas I present please let me know.

    The problem is:
    Find the value of x for F(x) =2cos(x)+1 where the tangent line is parallel to the secant line that passes through x=pi and x= 3pi/2, answers for x must me in the interval 0<=x<=2pi.


    I started out plugging in x = pi and x= 3pi/2 gettng the points (pi, -1) and (3pi/2, 1)

    using the slope formula y2-y1 \ x2-x1, I got the answer 4/pi for the slope of the secant line.

    then I got the derivative using the constant = 0 rule to turn the 1 into a zero , the derivative of cos(x) = -sin(x) , and the constant multiplying a function rule to end up with F'(x) = -2sin(x)

    Then I set 4/pi equal to -2sin(x) : 4/pi = -2sin(x) , 4/-2pi = sin(x) , -2/pi = sin(x)
    and I'm stuck at this point.

    I'm not sure how to isolate the x at this point. I'm pretty sure I have the work correct up to this point but I could be wrong, any help would be deeply appreciated.

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  2. #2
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    Re: Derivative problem that has me stumped.

    Quote Originally Posted by Shupp2 View Post
    Hello,

    This is my first time using this site, so if I'm not clear with the equations or formulas I present please let me know.

    The problem is:
    Find the value of x for F(x) =2cos(x)+1 where the tangent line is parallel to the secant line that passes through x=pi and x= 3pi/2, answers for x must me in the interval 0<=x<=2pi.


    I started out plugging in x = pi and x= 3pi/2 gettng the points (pi, -1) and (3pi/2, 1)

    using the slope formula y2-y1 \ x2-x1, I got the answer 4/pi for the slope of the secant line.

    then I got the derivative using the constant = 0 rule to turn the 1 into a zero , the derivative of cos(x) = -sin(x) , and the constant multiplying a function rule to end up with F'(x) = -2sin(x)

    Then I set 4/pi equal to -2sin(x) : 4/pi = -2sin(x) , 4/-2pi = sin(x) , -2/pi = sin(x)
    and I'm stuck at this point.

    I'm not sure how to isolate the x at this point. I'm pretty sure I have the work correct up to this point but I could be wrong, any help would be deeply appreciated.

    Since \displaystyle \begin{align*} \sin{(x)} = -\frac{2}{\pi} \end{align*}, we have the sine values being negative, which means it is in the third and fourth quadrants.

    The focus angle is the angle in the first quadrant: \displaystyle \begin{align*} x = \arcsin{\left( \frac{2}{\pi} \right)} \end{align*}, and so the angles in the third and fourth quadrants are \displaystyle \begin{align*} \pi + \arcsin{\left( \frac{2}{\pi} \right)} \end{align*} and \displaystyle \begin{align*} 2\pi - \arcsin{ \left( \frac{2}{\pi} \right) } \end{align*}.

    Then we need to add multiples of \displaystyle \begin{align*} 2\pi \end{align*} to signify going around the unit circle any number of times in either direction.

    So your solution is \displaystyle \begin{align*} x = \left\{ \pi + \arcsin{ \left( \frac{2}{\pi} \right)} , 2\pi - \arcsin{ \left( \frac{2}{\pi} \right) } \right\} + 2\pi \, n \textrm{ where } n \in \mathbf{Z} \end{align*}.
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