UNIVERSITY OF TORONTO DEPARTMENT OF MATHEMATICS MAT 235 Y - CALCULUS II TEST #1. OCTOBER 30, 2008.PROBLEMS AND SOLUTIONS.1. Let Cbe the curve defined by the parametric equations x= 4t– t2and y= 2t3. a) (10 marks) Find the coordinates of each of the points on the curve Cwhere the tangent line has slope 3 . Solution:We just have to find the points on Cwhere 3dydx=. Computing the derivates, we obtain 2/6/42dydy dttdxdx dtt==−. So, we need 6t2= 12 – 6tor t2+ t– 2 = 0 . That is: t= – 2 or t= 1 . The points on the curve Cwhere the tangent has slope 3 are ( – 12 , – 16 ) and ( 3 , 2 ) . b) (10 marks) Find the values of tfor which the curve Cis concave upward. Solution:We just have to find the values of tfor which 220dyd x>. Again, computing the corresponding derivatives, we obtain 2233 (4)2(2)ddydyttdtdxdxd xtdt⎛⎞⎜⎟−⎝⎠==−. So, we need 33 (4)02(2)ttt−>−and the values of tfor which this condition holds are 0 < t< 2 or t> 4 .
has intentionally blurred sections.
Sign up to view the full version.