# Thread: Derivative problem that has me stumped.

1. ## 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.

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

Originally Posted by Shupp2
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 \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 \displaystyle \begin{align*} x = \arcsin{\left( \frac{2}{\pi} \right)} \end{align*}, and so the angles in the third and fourth quadrants are \displaystyle \displaystyle \begin{align*} \pi + \arcsin{\left( \frac{2}{\pi} \right)} \end{align*} and \displaystyle \displaystyle \begin{align*} 2\pi - \arcsin{ \left( \frac{2}{\pi} \right) } \end{align*}.

Then we need to add multiples of \displaystyle \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 \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*}.