finalsol_v2

# finalsol_v2 - Math 20A Final Examination December 9 2008 1...

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Math 20A Final Examination. December 9, 2008 1. Find the following limits: (a) (2 points) 0 51 lim x x x , By L’Hopital’s Rule, 00 5 1 ln(5)5 0 lim lim ln(5) 1 xx H x →→ −− == . (b) (2 points) 0 lim log x x x + , 0 0 2 1 log lim log lim lim lim 0 11 H x x x x x ++ + + === = = . (c) (2 points) 5 0 lim x x x + . If y = x 5 x , then ln( y ) = 5 x ln( x ), and lim 5 ln( ) 5 lim ln( ) 0 = = , from (b). Since e x is a continuous, one-to-one function, we have that . 50 0 lim 1 x x xe + 2. Compute the following integrals: (a) (3 points) 9 0 xdx 9 99 1/2 3/2 0 22 2 (9) (0) 18 33 3 xdx x dx x ⎡⎤ =−= ⎢⎥ ⎣⎦ ∫∫ . (b) (3 points) 5 2 0 25 x dx (Hint: Use geometry and interpret as an area.) The function 2 25 x represents the top half of a circle with radius 5. The bounds on the integral from 0 to 5 represent the area in the first quadrant. So, the integral equals one-fourth the area of a circle with radius 5. That is given by (1/4) p (5) 2 = (25/4) p . 1

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3. (6 points) Find the point(s) on the ellipse given by x 2 + 4 y 2 = 1 where the tangent line has slope 1. Using implicit differentiation, we have: 2 28 0 84 x x xy y y y y ′′ += = = . Setting 1 y ′ = , we have: 14 4 x y y = =− ⇒ =− . Substituting this into our equation for the ellipse, we have: 22 2 1 (4) 4 1 2 0 1 20 yy y y −+= = = ± . Plugging these y-values into our relationship, x = -4 y , we obtain the two points: 11 4, 20 20 ⎛⎞ ⎜⎟ ⎝⎠ and 20 20 . 4. Let 3 () 3 x fx x + = .
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finalsol_v2 - Math 20A Final Examination December 9 2008 1...

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