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`r=2csctheta+3` Find the points of horizontal tangency (if any) to the polar curve.

`r=2csc theta+3`


To solve, express the polar equation in parametric form. To convert it to parametric equation, apply the formula



`x = rcos theta`


`y=r sin theta`



Plugging in  `r=2csctheta +3` , the formula becomes:


`x = (2csctheta +3)cos theta`


`x= 2cot theta + 3cos theta`


`y =(2csc theta+3)sin theta` 


`y=2+3sintheta`


So the equivalent parametric equation of `r=2csctheta +3` is:


`x=2cot theta +3cos theta`


`y=2+3sin theta`


Then, take the derivative of x and y with respect to theta.


`dx/(d theta ) = 2*(-csc^2 theta) + 3*(-sin theta)`


`dx/(d theta)=-2csc^2 theta -3sin theta`


`dy/(d theta)=3costheta`


Take note that the slope of the tangent is equal to dy/dx.


`m= (dy)/(dx)`


To get the dy/dx of a parametric equation, apply the formula:


`dy/dx = (dy/(d theta))/(dx/(d theta))`


When the tangent line is horizontal, the slope of the tangent is zero.


`0 = (dy/(d theta)) / (dx/(d theta))`


This implies that the polar curve will have a horizontal tangent when numerator is zero. So set the derivative of y equal to zero.


`dy/(d theta) = 0`


`3cos theta =0`


`cos theta=0`


`theta =pi/2, (3pi)/2`


So the polar curve have a horizontal tangents at:


`theta_1 = pi/2 + 2pin`


`theta_2 = (3pi)/2 + 2pi n`


where n is any integer.


To determine the points `(r, theta)` , plug-in the values of theta to the polar equation.


`r=2csc theta + 3`


`theta=pi/2+2pin` ,


 `r=2csc(pi/2+2pin)+3= 2csc(pi/2)+3=2*1+3=5`


`theta=(3pi)/2+2pin` ,


`r=2csc((3pi)/2+2pin)+3= 2csc((3pi)/2)+3=2*(-1)+3=1`


Therefore, the polar curve has horizontal tangents at points `(5, pi/2+2pin)` and `(1, (3pi)/2+2pin)` .

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