# C once again we need to manipulate r 1 cos θ a bit

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Chapter 3 / Exercise 17
Single Variable Calculus: Early Transcendentals
Stewart
Expert Verified
(c) Once again, we need to manipulate r = 1 - cos( θ ) a bit before using the conversion formulas given in Theorem 2.7 . We could square both sides of this equation like we did in part 2a above to obtain an r 2 on the left hand side, but that does nothing helpful for the right hand side. Instead, we multiply both sides by r to obtain r 2 = r - r cos( θ ). We now have an r 2 and an r cos( θ ) in the equation, which we can easily handle, but we also have another r to deal with. Rewriting the equation as r = r 2 + r cos( θ ) and squaring both sides yields r 2 = ( r 2 + r cos( θ ) ) 2 . Substituting r 2 = x 2 + y 2 and r cos( θ ) = x gives x 2 + y 2 = ( x 2 + y 2 + x ) 2 . Once again, we have performed some algebraic maneuvers which may have altered the set of points described by the original 8 Here, ‘equivalent’ means they represent the same point in the plane. As ordered pairs , (3 , 0) and ( - 3 , τ 2 ) are different, but when interpreted as polar coordinates, they correspond to the same point in the plane. Mathematically speaking, relations are sets of ordered pairs, so the equations r 2 = 9 and r = - 3 represent different relations since they correspond to different sets of ordered pairs. Since polar coordinates were defined geometrically to describe the location of points in the plane, however, we concern ourselves only with ensuring that the sets of points in the plane generated by two equations are the same. This was not an issue, by the way, when we first defined relations as sets of points in the plane. Back then, a point in the plane was identified with a unique ordered pair given by its Cartesian coordinates. 9 In addition to taking the tangent of both sides of an equation (There are infinitely many solutions to tan( θ ) = 3, and θ = 2 τ 3 is only one of them!), we also went from y x = 3, in which x cannot be 0, to y = x 3 in which we assume x can be 0.
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Chapter 3 / Exercise 17
Single Variable Calculus: Early Transcendentals
Stewart
Expert Verified
240 Applications of Trigonometry equation. First, we multiplied both sides by r . This means that now r = 0 is a viable solution to the equation. In the original equation, r = 1 - cos( θ ), we see that θ = 0 gives r = 0, so the multiplication by r doesn’t introduce any new points. The squaring of both sides of this equation is also a reason to pause. Are there points with coordinates ( r, θ ) which satisfy r 2 = ( r 2 + r cos( θ ) ) 2 but do not satisfy r = r 2 + r cos( θ )? Suppose ( r 0 , θ 0 ) satisfies r 2 = ( r 2 + r cos( θ ) ) 2 . Then r 0 = ± ( ( r 0 ) 2 + r 0 cos( θ 0 ) ) . If we have that r 0 = ( r 0 ) 2 + r 0 cos( θ 0 ), we are done. What if r 0 = - ( ( r 0 ) 2 + r 0 cos( θ 0 ) ) = - ( r 0 ) 2 - r 0 cos( θ 0 )? We claim that the coordinates ( - r 0 , θ 0 + τ 2 ), which determine the same point as ( r 0 , θ 0 ), satisfy r = r 2 + r cos( θ ). We substitute r = - r 0 and θ = θ 0 + τ 2 into r = r 2 + r cos( θ ) to see if we get a true statement. - r 0 ? = ( - r 0 ) 2 + ( - r 0 cos( θ 0 + τ 2 ) ) - ( - ( r 0 ) 2 - r 0 cos( θ 0 ) ) ? = ( r 0 ) 2 - r 0 cos( θ 0 + τ 2 ) Since r 0 = - ( r 0 ) 2 - r 0 cos( θ 0 ) ( r 0 ) 2 + r 0 cos( θ 0 ) ? = ( r 0 ) 2 - r 0 ( - cos( θ 0 )) Since cos( θ 0 + τ 2 ) = - cos( θ 0 ) ( r 0 ) 2 + r 0 cos( θ 0 ) X = ( r 0 ) 2 + r 0 cos( θ 0 ) Since both sides worked out to be equal, ( - r 0 , θ 0 + τ 2 ) satisfies r = r 2 + r cos( θ ) which means that any point ( r, θ ) which satisfies r 2 = ( r 2 + r cos( θ ) ) 2 has a representation which satisfies r = r 2 + r cos( θ ), and we are done.