# r=1-sintheta Find the points of horizontal and vertical tangency (if any) to the polar curve.

r=1-sin theta

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=1-sin theta , the formula becomes:

x=(1-sin theta)cos theta=cos theta -sin theta cos theta

y = (1-sin theta)sin theta=sin theta -sin^2...

r=1-sin theta

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=1-sin theta , the formula becomes:

x=(1-sin theta)cos theta=cos theta -sin theta cos theta

y = (1-sin theta)sin theta=sin theta -sin^2 theta

So the equivalent parametric equation of r= 1-sin theta is:

x=cos theta -sin theta cos theta

y=sin theta -sin^2 theta

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

dx/(d theta) = -sintheta - (sintheta*(-sintheta) + costheta*costheta)

dx/(d theta)=-sintheta+sin^2theta-cos^2theta

dy/(d theta) = costheta - 2sinthetacostheta

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 dy/(d theta)=0 and dx/(d theta) !=0.

Setting the derivative of y yields:

dy/(d theta) = 0

costheta - 2sinthetacostheta=0

costheta(1 - 2sintheta) =0

costheta = 0

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

1-2sintheta=0

-2sintheta=-1

sintheta=1/2

theta=pi/6,(5pi)/6

Take note that at theta=pi/2 , the value of dx/(d theta) is zero. Since both dy/(d theta) and dx/(d theta)  are zero, the slope at this value of theta is indeterminate.

m=0/0   (indeterminate)

So the polar curve has horizontal tangents at:

theta_1 = pi/6 + 2pin

theta_2= (5pi)/6+2pin

theta_3= (3pi)/2+2pin

where n is any integer.

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

r=1-sin theta

theta_1 = pi/6 + 2pin

r_1=1-sin(pi/6 + 2pin)=1-sin(pi/6) = 1-1/2=1/2

theta_2= (5pi)/6+2pin

r_2=1-sin((5pi)/6+2pin)=1-sin((5pi)/6)= 1 -1/2=1/2

theta_3= (3pi)/2+2pin

r_3=1-sin((3pi)/2+2pin)=1-sin((3pi)/2)=1-(-1)=2

Therefore, the polar curve has horizontal tangent at points

(1/2, pi/6+2pin) ,   (1/2, (5pi)/6+2pin) ,  and  (2, (3pi)/2+2pin) .

Moreover, when the tangent line is vertical, the slope is undefined.

u n d e f i n e d =(dy/(d theta)) / (dx/(d theta))

This implies that the polar curve will have vertical tangent when dx/(d theta)=0 and dy/(d theta)!=0 .

Setting the derivative of x equal to zero yields:

dx/(d theta) = 0

-sintheta+sin^2theta-cos^2theta=0

-sin theta + sin^2 theta-(1-sin^2 theta) = 0

2sin^2 theta -sin theta -1=0

(2sin theta +1)(sin theta -1) = 0

2sin theta + 1=0

sin theta=-1/2

theta = (7pi)/6,(11pi)/6

sin theta -1=0

sin theta=1

theta=pi/2

Take note that at theta =pi/2 , both dy/(d theta ) and dx/(d theta) are zero. So the slope is indeterminate at this value of theta.

m=0/0  (indeterminate)

So the polar curve has vertical tangents at:

theta_1 =(7pi)/6+2pin

theta_2=(11pi)/6+2pin

where n is any integer.

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

r=1-sin theta

theta_1=(7pi)/6+2pin

r_1=1-sin((7pi)/6+2pin)=1-sin((7pi)/6)=1-(-1/2)=3/2

theta_2=(11pi)/6+2pin

r_2=1-sin((11pi)/6+2pin)=1-sin((11pi)/6)=1-(-1/2)=3/2

Therefore, the polar curve has vertical tangent at points (3/2, (7pi)/6+2pin) and (3/2, (11pi)/6+2pin)` .

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