MT1 w soln

# MT1 w soln - Midterm 1 Problem 1 Sketch the curve given in...

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Midterm 1 Problem 1 Sketch the curve given in Cartesian coordinates x = t - 2 sin( t ) and y = 1 - 2 cos( t ) t [ - 3 π, 3 π ]. Find an equation of the tangent to the curve. Find the points on the curve where the tangent is vertical. How many points of self-intersection does the curve have? Solution The curve is a trochoid. Problem 34 of Section 10.1 that you had in your homework introduced it to you. Here is the sketch –4 –2 0 2 4 –10 –5 5 10 Figure 1: A trochoid The equation of the tangent to the curve is given by dy dx = 2 sin( t ) 1 - 2 cos( t ) . The tangent is vertical where dy/dx = ±∞ . It follows that it holds for the values of the parameter t such that cos( t ) = 1 / 2. Solving it we obtain t = ± π/ 3 + 2 πk , where k = - 1 , 0 , 1 since t [ - 3 π, 3 π ]. Plugging these values of the parameter into the parametric equations we get x = ± π/ 3 3 + 2 πk and y = 0, where k = - 1 , 0 , 1. Therefore, we have six points on the curve at which the tangent is vertical. From the sketch it’s clear that there are 3 points of self-intersection. Problem 2 Sketch the curve given in polar coordinates r 2 = sin(2 θ ). Find the area it encloses and the area of the surface obtained by rotating the curve about the y -axis. Solution The curve is called lemniscate of Bernoulli. We already met it in one of our discussion section. It lies in the first and the third quadrants. The sketch is given in Fig. 2. Using the symmetry of the curve we obtain the area it encloses A = 2 · 1 2 Z π/ 2 0 r 2 = Z π/ 2 0 sin(2 θ ) = 1 . 1

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–0.8 –0.6 –0.4 –0.2 0 0.2 0.4 0.6 0.8 –0.8 –0.6 –0.4 –0.2 0.2 0.4 0.6 0.8 Figure 2: lemniscate of Bernoulli Again employing the symmetry of the curve for the area of the surface obtained by rotating the curve about the y -axis we get A y = 2 · 2 π Z π/ 2 0 r ( θ ) cos( θ ) r r 2 + dr 2 dθ.
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