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Unformatted text preview: MATH 271 Test #4T Solutions You do not need to evaluate the integrals in problems (2)–(5); just set them up. (1) (20 points) Consider the curve which is represented by the parametric equations x = 1 + t + t 2 , y = 1 + e t , where t is any real number. Find the equation of the line tangent to this curve at the point (1 , 2). Solution: The slope of the tangent line to a curve is dy/dx . Here, x and y are given in terms of the parameter t , so dy dx = dy/dt dx/dt = e t 1 + 2 t . To find the slope of the line tangent to the specific point (1 , 2) , we need to find the value of t which gives x ( t ) = 1 and y ( t ) = 2 . Hence, t must satisfy 1 + t + t 2 = 1 and 1 + e t = 2 . Only one value of t will work, namely t = 0 . The slope of the tangent line at the point (1 , 2) is e 1 + 2(0) = 1 . Now we have a point and the slope, so we can use the pointslope form of the equation of a line: y 2 = 1( x 1) , or y = x + 1 . Grading: +5 points for writing dy/dx in terms of t , +5 points for finding t , +5 points for the slope of the tangent line, +5 points for the equation of the tangent line. 1 (2) The polar coordinate equation r = 1 + cos θ traces out a cardioid, covering it exactly once, when θ ranges from 0 to 2 π . (a) (5 points) Sketch the graph of this cardioid. Solution: x y ↑ ↓ ← → . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ....
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 Spring '08
 Hurlbert
 Calculus, Equations, Integrals, Parametric Equations, ... ...

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