Fall Exam 4: 8&11-11/14/13 - Write all responses on separate paper. Show your work for credit.

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1 Math Fall - Exam : 8& - // - Write all responses on separate paper. Show your work for credit. Name (Print):. Convert the rectangular equation to polar coordinates and solve for r. (a) x + (y ) = 6 Solution: Expanding we have x + y 8y = r = r sin θ r = 8 sin θ (b) (x + y + y) = (x + y ) (r + y) = r r + yr + y r = Substituting y = r sin θ and collecting like terms r + sin θr + (sin θ )r =. Now divide through by r (assuming it s not zero) and setting up to complete the square: r + sin θr = (r + sin θ) = so r = sin θ ± Here s a graph of this (note that the origin is also part of the graph since r = is a solution to the equation:. Convert the polar equation to rectangular coordinates and solve for y. (a) r = sin θ + cos θ Solution: Multiply both sides by the denominator and you get r sin θ + r cos θ = y = x (b) r = sec θ(tan θ ) Solution: Multiply both sides by cos θ to get r cos θ = tan θ then substitute from the Rosetta stone to get x = y x y = x + x. Consider the polar function r = sin θ (a) Test the function for symmetry. What do you find? f( θ) = f(θ) so there is y-axis symmetry. (b) Write the function as a conic section in standard rectangular form. Multiply both sides by the denominator solve for r and equate squares to get r = (y + ) x + y = y + y + (y + ) = x a parabola opening upwards from a vertex at ( ) with focus at () directrix along y = and with a focal diameter of.

2 Math Exam : 8& - Page of // (c) Complete the table below for r x and y for the given θ 7 θ r x - y (d) construct a graph for the function. -. Find all solutions to each equation including the complex solutions. Hint: first convert the number to polar form and use DeMovire s theorem. (a) x = ( ) ( ) (k + ) (k + ) Solution: x = cos(k + ) + i sin(k + ) x = cos + i sin for k =. Since these expressions are solvable by radicals (see your class notes) we can write these as z = + ( ) + i z = ( + ) + i z = z = ( + ) i z = + ( ) i z z R(z) (b) x 6 = 8 + i Solution: Let θ = arctan ( ) 8. Then x 6 = 7 (cos (θ + k) + i sin (θ + k)) x = 6 ( ( ) ( )) θ + k θ + k 7 cos + i sin for k =. I don t think there s any 6 6 particular insight to be gained by simplifying further. z I(z) z z

3 Math Exam : 8& - Page of // (x ). Consider the ellipse described by + y 9 = (a) Find the center x-intercepts y-intercepts and the coordinates of the foci. Solution: The center is ( ) and since a = and b = c = 9 =. Thus the x- intercepts are ( ) and (9 ). The y-intercepts can be found by setting x = and solving for y = ± 6 = ±9. Finally the foci are at ( ) and (8 ). (b) Sketch a graph showing these features. Solution: It s easier for a graphing device to graph the parametric form so I ll convert first: { x(t) = + cos(t) y(t) = sin(t) (c) What is the eccentricity e = c a? e = c a = (d) What is the polar form? Hint: it s in the r = Solution: Substituting c = and multiplying by ed e cos θ form we have r = d cos θ Now f() = d and f() = d 9 need to match up with the x-intercepts (9 ) and ( ) which the will if we set d = 9. 9 Thus r = cos θ 6. Consider the hyperbola describe described by r = (a) Find the eccentricity. Solution: e =. (b) Complete the table: sin θ

4 Math Exam : 8& - Page of // θ r ( arcsin ) 6 x y 6 ( arcsin ) 6 6 (c) Given that the vertices of the hyperbola are the y-intercepts what are the coordinates of the center? Solution: Halfway between the vertices at ( ) and ( ) is the center at ( 6) (d) ( points) Sketch a graph (see attached graph paper). (e) ( points) What is the rectangular form? Solution: Inspecting the graph we see that a = c = 6 so that b = 6 6 = whence (y + 6) x = is the rectangular form Find parametric equations for each given conic. (a) x + (y ) 9 = (b) (x ) y = x = cos(t) y = + sin(t)

5 Math Exam : 8& - Page of // (c) (y ) = (x ) x = + sec(t) y = tan(t) x = + t y = + t 8. Make a table of values and sketch a graph for the given parametric equations. x = cos(t) () y = sin (t) () t ± 6 ± ± ± ± ± ± 6 ± x Note that it s easy to eliminate the parameter: y = x but that the graph oscillates only on the part of the parabola where y y