practice test 2 answers

practice test 2 answers - MAT 111 Practice Test 2 Solutions...

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Unformatted text preview: MAT 111 Practice Test 2 Solutions Spring 2010 1 1. (10 points) Find the equation of the tangent line to 2 x + 2 y = 1+ x 2 y 2 at the point (1 , 1). The equation is y- y = dy dx ( x- x ) So all we need is dy/dx . Differentiating the above equation implicitly, we get 1 2 2 x + 2 y (2 + 2 dy dx ) = 2 xy 2 + 2 y dy dx x 2 So, substituting x = 1 and y = 1, we get 1 4 (2 + 2 dy dx ) = 2 + 2 dy dx solving, we get dy dx =- 1. So the equation of the tangent is y- 1 =- 1( x- 1) OR y =- x + 2 2 2. (30 points) Evaluate the following: (a) lim x 2 x 4- 2 x- 12 x- 2 Solution: Notice that if we let f ( x ) = x 4- 2 x then f (2) = 12 and so the above limit is actually equal to lim x 2 f ( x )- f (2) x- 2 which is precisely f (2). Thus, all we need to do is evaluate f ( x ) and substitute 2. f ( x ) = 4 x 3- 2 and so lim x 2 f ( x )- f (2) x- 2 = f (2) = 4 2 3- 2 = 30 . (For the PE class students, the same solution with f ( x ) = x 4 + 2 x and f ( x ) = 4 x 3 + 2 and limit = 34). (b) Find f ( x ) if f ( x ) = sec 3 x 1+sin x Solution: By Quotient Rule, f ( x ) = (1 + sin x ) d dx (sec3 x )- sec3 x d dx (1 + sin x ) (1 + sin x ) 2 By Chain Rule d dx (sec(3 x )) = sec(3 x ) tan(3 x ) d dx (3 x ) = 3sec(3 x ) tan(3 x ). So the final answer is: 3sec(3 x ) tan(3 x )(1 + sin x )- cos ( x ) sec (3 x ) ( i + sin x ) 2 (c) Find f ( x ) if f ( x ) = sin 2 x x 3 +1 cos x +2 . Solution: This is just a repeated application of chain rule, quotient rule and product rule: Outer most functions is square, next is sin and then there is a quotient to be differentiated. f ( x ) = 2sin x x 3 + 1 cos x + 2 ! d dx " sin x x 3 + 1 cos x + 2 !# 3 = 2sin x x 3 + 1 cos x + 2 !" cos x x 3 + 1 cos x + 2 !# d dx x x 3 + 1 cos x + 2 = 2sin x x 3 + 1 cos x + 2 !" cos x x 3 + 1 cos x + 2 !# (cos x + 2) d dx ( x x 3 + 2)- x x 3 + 2 d dx (cos x + 2) (cos x + 2) 2 = 2sin...
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practice test 2 answers - MAT 111 Practice Test 2 Solutions...

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