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Page 1 6/28/2017 Name (Print Clearly): Student Number: MTH132 Section 1 & 12, Quiz 3 Feb 20, 2009 Instructor: Dr. W. Wu Instructions: Answer the following questions in the space provided. There is more than adequate space provided to answer each question. The total time allowed for this quiz is 15 minutes. 1 [3 pts each]. Find dy / dx for following functions (Do Not Simplify) (a) y 3 4x y (4 x)1 / 3 (apply power chain rule) So y ' 1 d 4 (4 x) 2 / 3 (4 x) (4 x) 2 / 3 3 dx 3 (b) y 4 tan 3 x 3 (chain rule) 4 d y ' sec 2 (3x) (3x) 4 sec 2 (3 x) 3 dx (c) y cos x cos x 3 2 y cos( x3 ) (cos x)2 (use chain rule, the inside function of the first term is x 3 , and the inside function of the second term is cos x ). So y ' sin( x ) 3 d 3 d ( x ) 2 cos x (cos x) sin( x 3 ) (3x 2 ) 2 cos x ( sin x) dx dx (d) x y xy 1 (use implicit differentiation) 2 2 consider y as a function of x (implicitly) apply d to this implicit equation: dx 2 xy x 2 y ' y 2 x 2 y y ' y 2 2 xy so ( x 2 xy) y' ( y 2 xy) y' 2 x 2 xy 2 2 d (1) 0 dx Page 2 6/28/2017 2. [4 pts]. Show that (2,1) lies on the curve xy2 2 y 3 4 . Then find an equation for the tangent line to the curve at this point. Substitute x 2, y 1 into the left hand side 2(1) 2 2(1)3 2 2 4 . So it is on the curve. Consider y as a function of x (implicitly), and use implicit differentiation y 2 x (2 yy ' ) 6 y 2 y ' d (4) 0 dx y2 12 1 So y ' . At (2,1) , the slope is y' 1/ 10 2 2 2 xy 6 y 2 (2) 1 6 1 10 Use point-slope equation: the tangent line at (2,1) is y 1 1 ( x 2) 10 3. Suppose the parametric equations of a curve are x 2 cos t , y 3 sin t , 0 t 2 . (a) [2 pts]. Find dy / dx by using parametric formula. dy dy / dt 3 cos t dx dx / dt 2 sin t (b) [2 pts]. Let p be the point when t / 6 . Find the slope of the tangent line to this curve at p . 3 cos( / 6) 3 3/2 3 2 sin( / 6) 2 1/ 2 2 The slope is the derivative at t / 6 , so it is (c) [bonus 2 pts]. Find a Cartesian equation for this curve, then find x 2 cos t , y 3 sin t cos t dy at p by using implicit differentiation. dx x y 2 2 , sin t , by trigonometric identity cos t sin t 1 2 3 2 2 A Cartesian equation for this ellipse is x y 1 . (Ex 3.5 #69) 4 3 Using 1 2y 2x y' 0 4 3 implicit differentiation: y' x/2 3x 2y /3 4y x 2 cos( / 6) 3 , y 3 sin( / 6) 3 / 2 . So the slope is y ' . When t /6 , 3 3 3 3 3 / 2 . 4 3 / 2 2 3