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Exam 2 - Version 034 Exam02 Gilbert(56380 This print-out...

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Version 034 – Exam02 – Gilbert – (56380) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine a Cartesian equation for the curve given in parametric form by x ( t ) = 3 ln(16 t ) , y ( t ) = t . 1. y = 1 3 e x/ 4 2. y = 1 4 e 6 /x 3. y = 1 3 e x/ 8 4. y = 1 3 e 8 /x 5. y = 1 4 e x/ 3 6. y = 1 4 e x/ 6 correct Explanation: We have to eliminate the parameter t from the equations for x and y . Now from the equation for x it follows that t = 1 16 e x/ 3 . But then y = parenleftBig 1 16 e x/ 3 parenrightBig 1 / 2 = 1 4 e x/ 6 . 002 10.0 points Let Q, R be the points where the ray of angle θ intersects circles centered at the origin as shown in 2 4 6 8 2 4 6 8 2 4 6 8 2 4 6 8 θ P R Q and let P be the point of intersection of the horizontal line through Q and the vertical line through R . As θ varies, P traces out a curve. Write this curve in parametric form ( x ( θ ) , y ( θ )) , 0 θ 2 π . (Hint: find the coordinates of Q and R .) 1. (7 sin θ, 4 cos θ ) 2. (4 cos θ, 7 sin θ ) 3. (4 tan θ, 7 sec θ ) 4. (7 sec θ, 4 tan θ ) 5. (7 cos θ, 4 sin θ ) correct 6. (4 sec θ, 7 tan θ ) Explanation: From the graph it follows that the inner circle has radius 4, while the outer circle has radius 7. On the other hand, any circle of radius r centered at the origin can be written in parametric form as (cos θ, sin θ ) , 0 θ 2 π , Thus Q and R are the points Q (4 cos θ, 4 sin θ ) , R (7 cos θ, 7 sin θ ) . But P, Q have the same y -coordinate, and P, R have the same x -coordinate.
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Version 034 – Exam02 – Gilbert – (56380) 2 Consequently, the curve traced out by P has the parametric form (7 cos θ, 4 sin θ ) for 0 θ 2 π . Eliminating θ , we see that P traces out the ellipse x 2 49 + y 2 16 = 1 whose relation to the circles is shown in 2 4 6 8 2 4 6 8 2 4 6 8 2 4 6 8 P 003 10.0 points Find an equation for the tangent line to the curve given parametrically by x ( t ) = e 2 t , y ( t ) = 3 t 2 + t + 2 at the point P (1 , 2). 1. y = 1 2 x + 5 2 2. y = 1 2 x + 3 2 3. y = x + 3 2 4. y = 1 2 x + 3 2 correct 5. y = x + 5 2 6. y = 1 2 x + 5 2 Explanation: Notice first that P (1 , 2) is the point corre- sponding to the choice t = 0. We can thus use the point slope formula with t = 0 to find an
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