PS1Solutions

# PS1Solutions - MAT235-CALCULUS II SUMMER 2009 ASSIGNMENT#1...

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MAT235-CALCULUS II SUMMER 2009 ASSIGNMENT #1, DUE ON MAY 26 SOLUTIONS Problem 1 Parametric curves. a) Consider the parametric curve x = t 2 + t, y = e t . Find all t such that the tangent line of the curve at ( x ( t ) , y ( t )) intersects the x -axis at (4 , 0). b) Determine the values of t for which the curve x = 2 1 + t, y = R t 2 3 ( u - 1) p 1 + udu , t 0 is concave upward and those for which it is concave downward. Solution: a) We have dy dx ( t ) = dy dt dx dt = e t 2 t +1 . Therefore, the equation of the tangent line at ( x ( t ) , y ( t )) is: y - y ( t ) = dy dx ( t )( x - x ( t )) (1) i.e y - e t = e t 2 t + 1 ( x - t 2 - t ) . (2) To ﬁnd t such that the tangent line at ( x ( t ) , y ( t )) passes through (4 , 0), we replace y with 0 and x with 4 in the last equation to obtain: - e t = e t 2 t + 1 (4 - t 2 - t ) . (3) After canceling the exponential we have: - 2 t - 1 = 4 - t - t 2 . (4) Equivalently, t 2 - t - 5 = 0 . (5) Using the quadratic equation to solve for t yields: t = 1 ± 21 2 . (6) 1

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2 b)We have: dx dt = 1 1 + t . (7) Applying the fundamental theorem of calculus to y , we obtain: dy dt = 2 t ( t - 1) 1 + t. (8) Therefore: dy dx ( t ) = dy dt dx dt = 2 t ( t - 1)( t + 1) = 2 t 3 - 2 t. (9) We also have: d 2 y dx 2 ( t ) = d dt ( dy dx ) dx dt = 6 t 2 - 2 1 1+ t . (10) Since 1 + t is always positive, the sign of d 2 y dx 2 is determined by the sign of 6 t 2 - 2. This implies that d 2 y dx 2 < ( t )0, if 6 t 2 - 2 < 0, which holds if and only if 0 t < 1 3 (remember here that t is positive!!!), and d 2 y dx 2 ( t ) > 0, if 6 t 2 - 2 > 0 which holds if and only if t > 1 3 . We conclude that the curve is concave upward when t > 1 3 , and downward when 0 t < 1 3 . Problem 2 More parametric curves a) Find the length of the curve x = 2 3 t 9 2 + 1, y = t 3 - 5, 0 t 1. b) Consider the curve x = - t 2 , y = t 3 3 - t +1 , t [0 , 1] . Find the area of the surface obtained by revolving the curve about the line y = x . Solution: a) We have dx dt = 3 t 7 2 , dy dt = 3 t 2 . We know that the length is: Z 1 0 s ± dx dt ² 2 + ± dy dt ² 2 dt = Z 1 0 3 t 7 + t 4 dt = Z 1 0 3 t 2 t 3 + 1 dt. (11) Substituting v = t 3 + 1, we have dv = 3 t 2 dt and the integral becomes: Z 2 1 udu = 2 3 u 3 2 | 2 1 = 2 3 (2 3 2 - 1) . (12) b) To answer this question, we need a formula for the area of the surface obtained by revolv- ing the parametric curve ( x ( t ) , y ( t )) , t [ a, b ] about a line. We can modify the derivation in the textbook (which treats the case when the line is the x -axis) to obtain: A = Z b a 2 πD ( t ) s ± dx dt ² 2 + ± dy dt ² 2 dt, (13)
3 where D ( t ) is the distance from the point ( x ( t ) , y ( t )) to the line y = x . For

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## This note was uploaded on 07/18/2010 for the course MATH MAT235 taught by Professor Recio during the Summer '08 term at University of Toronto.

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PS1Solutions - MAT235-CALCULUS II SUMMER 2009 ASSIGNMENT#1...

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