`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`
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`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)` .