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

## Expert Answers 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...

(The entire section contains 529 words.)

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