264_James Stewart Calculus 5 Edition Answers

264_James Stewart Calculus 5 Edition Answers - Stewart...

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Unformatted text preview: Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation 1 , and 3 at x=1 . The three 2 horizontal tangents along the top of the wagon are hard to find, but by limiting the y range of the graph (to [1.6,1.7] , for example) they are distinguishable. (b) There are 9 points with horizontal tangents: 3 at x=0 , 3 at x= / 2 2 35. From Exercise 29, a tangent to the lemniscate will be horizontal if y =0 25x 4x (x + y )=0 22 2 2 25 x [25 4 (x + y )]=0 x + y = ( 1 ). (Note that when x is 0 , y is also 0 , and there is no horizontal 4 25 22 tangent at the origin.) Substituting for x + y in the equation of the lemniscate, 4 2 22 22 2 2 25 2 75 2 25 2 (x + y ) =25 (x y ) , we get x y = ( 2 ). Solving ( 1 ) and ( 2 ), we have x = and y = , so 8 16 16 53 5 the four points are , . 4 4 2 x 36. 2 2 y + 2 a 2x =1 2 b / 2 yy + =0 2 a 2 / y= bx b 2 bx y y= 0 0 xx 0 the ellipse, we have 2 2 37. 2 2 + a 2 yy 2 b 2 a 0 2 x = b 2x =1 2 yy a y y y= 0 0 2 ay 2 b / =0 0 2 0 2 2 xx 0 = 2 x + a b 0 2 . Since (x ,y ) lies on 00 a 2 + a 0 2 =1 . b / y= 2 bx 2 an equation of the tangent line at (x ,y ) is 00 ay (x x ) . Multiplying both sides by 0 2 b yy y 0 0 gives 2 y y 2 bx yy b 0 2 0 (x x ) . Multiplying both sides by ay x 00 y 0 2 an equation of the tangent line at (x ,y ) is 2 ay 0 2 b gives 0 2 b 2 xx y 0 2 b = 0 2 a 2 x 0 2 a . Since (x ,y ) lies on 00 the hyperbola, we have 8 ...
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