Final - Version 019 – FINAL – Gilbert – (57495) 1...

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Unformatted text preview: Version 019 – FINAL – Gilbert – (57495) 1 This print-out should have 25 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 ) = 4 e t , y ( t ) = 2 e − 2 t . 1. x 2 y = 16 2. x 2 y = 32 correct 3. x y 2 = 8 4. xy 2 = 8 5. x 2 y = 16 6. x 2 y = 32 Explanation: We have to eliminate the parameter t from the equations for x and y . Now from the equation for x it follows that e t = x 4 , from which in turn it follows that y = 2 parenleftBig 4 x parenrightBig 2 . Consequently, x 2 y = 32 . 002 10.0 points Find dy dx when x ( t ) = 6 t + t 6 , y ( t ) = t ln( t 2 ) . 1. dy dx = 1 6 parenleftBig 1 + 2 ln t 1 + t 5 parenrightBig 2. dy dx = 1 3 parenleftBig 1 + ln t 1 + t 5 parenrightBig correct 3. dy dx = 2 parenleftBig 1 + ln t 1 + 6 t 5 parenrightBig 4. dy dx = 6 parenleftBig 1 + t 5 1 + 2 ln t parenrightBig 5. dy dx = 6 + t 5 1 + 2 ln t 6. dy dx = 3 parenleftBig 1 + t 5 1 + ln t parenrightBig Explanation: First noticed that y ( t ) = 2 t ln t . Thus, after differentiation with respect to x , we see that x ′ ( t ) = 6 + 6 t 5 , y ′ ( t ) = 2(1 + ln t ) . Consequently, dy dx = 1 3 parenleftBig 1 + ln t 1 + t 5 parenrightBig . 003 10.0 points Use the graph in Cartesian coordinates r θ π 2 π of r as a function of θ to determine which one of the following is the graph of the corre- sponding polar function? 1. Version 019 – FINAL – Gilbert – (57495) 2 2. 3. correct 4. 5. 6. Explanation: As the Cartesian graph of r never crosses the θ-axis, there is no value of θ in [0 , 2 π ] at which r = 0; in particular, therefore, the polar graph never passes through the origin. This eliminates 3 of the graphs, leaving only the ‘dimpled’ ones as possibilities. We now need to look in detail at the Carte- sian graph. It shows that (a) r increases on [0 , π/ 2], (b) r decreases on[ π/ 2 , π ]. Thus the polar graph has a local maximum (furthest from origin but maximum because r > 0) at θ = π/ 2. In addition (c) r increases on [ π, 3 π/ 2], (d) r decreases on [3 π/ 2 , 2 π ]. Thus the polar graph has a local minimum (closest to origin but minimum because r > 0), a ‘dimple’, at θ = π , a local maximum at θ = 3 π/ 2, and another local minimum at θ = 0. Hence the polar graph has ‘dimples’ at θ = 0 and π only. Consequently, its graph is keywords: Cartesian graph, polar graph, 004 10.0 points The shaded region in lies inside the polar curve r = 3 sin θ and outside the polar curve r = 2 sin θ . Which of the following integrals gives the area of this region? 1. I = 1 2 integraldisplay π 5 sin 2 θ dθ correct Version 019 – FINAL – Gilbert – (57495) 3 2. I = 1 2 integraldisplay π/ 2 sin θ dθ 3. I = integraldisplay π 5 sin 2 θ dθ 4. I = 1 2 integraldisplay π sin θ dθ 5. I = integraldisplay π sin θ dθ 6. I = 1 2 integraldisplay π/ 2 5 sin 2 θ dθ Explanation:...
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This note was uploaded on 12/17/2009 for the course M 57505 taught by Professor Gilbert during the Fall '09 term at University of Texas.

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Final - Version 019 – FINAL – Gilbert – (57495) 1...

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