3 a = 3 b c 2 = a 2 + b 2 = 2 2 = 4 c 2 = 3b 2 + b 2 = 4b 2 = 4 b 2 = 1 b = 1 a = 3b = 3. x 2 3 y2 1 = 1.

Transcription

1 MATH 222 LEC SECOND MIDTERM EXAM THU NOV 8 PROBLEM ( 5 points ) Find the standard-form equation for the hyperbola which has its foci at F ± (±2, ) and whose asymptotes are y ± 3 x The calculations b a 3 a 3 b c 2 a 2 + b c 2 3b 2 + b 2 4b 2 4 b 2 b a 3b 3 yield the standard-form equation x 2 3 y2 ( 5 points ) Find the standard-form equation for the ellipse which has a focus at F (4, ) and whose eccentricity is e 2 3 The calculations c 4 e c a a a a 2 b 2 + c 2 36 b b 2 2 yield the standard-form equation x y2 2

2 MATH 222 LEC SECOND MIDTERM EXAM THU NOV 8 PROBLEM 2 ( 5 points ) Replace the Cartesian equation x 2 + (y 2) 2 4 by the equivalent polar equation and simplify The calculations x 2 + (y 2) 2 4 x 2 + y 2 4y x 2 + y 2 4y r 2 4r sin θ r 4 sin θ or r result in the equivalent polar equation r 4 sin θ Note: Since the point with r is part of the solution set for this equation (when θ ), no point is lost by working only with this polar equation ( 6 points ) Find the slope of the curve r + sin θ at the point on it where θ The function involved in the given polar equation, f(θ) +sin θ has the value f() Its derivative f (θ) df dθ cos θ has the value f () cos dy dx f (θ) sin θ + f(θ) cos θ f (θ) cos θ f(θ) sin θ ( ) dy f () sin + f() cos dx θ f () cos f() sin f() cos f () cos ( )() ()()

3 MATH 222 LEC SECOND MIDTERM EXAM THU NOV 8 PROBLEM 3 ( points ) Find the shared area inside the two curves r 2 cos θ and r 2 sin θ where θ π The shared boundary points are found to be P (r, θ) P (, θ) and Q(r, θ) Q(2, π) 4 The line θ π is an axis of reflection symmetry for the region in question 4 Hence the area can be computed by π/4 Area 2 2 (2 sin(θ))2 dθ π/4 π/4 4 sin 2 (θ) dθ 2( cos(2θ)) dθ 2 π 4 2 sin(2θ) 2 π 2 sin(π 2 ) π 2 π/4

4 MATH 222 LEC SECOND MIDTERM EXAM THU NOV 8 PROBLEM 4 ( ) n 3n + Consider the sequence a n 3n ( points ) Determine if this sequence converges or diverges In case this sequence diverges provide a verification of its divergence In case this sequence converges find its limit ln a n ( ) 3n + n ln 3n n (ln(3n + ) ln(3n )) ln(3n + ) ln(3n ) n The limit of this logarithm when n is of the form L Hôpital s rule yields the computation lim ln a n lim ln(3n + ) ln(3n ) ( n ) 3n+ 3n 3 lim n 2 n 2 ((3n ) (3n + )) 3 lim (3n + )(3n ) 2n 2 3 lim 9n It follows from the continuity of the exponential function that the sequence a n converges to the limit L lim a n e lim ln an e 2/3 3 e 2

5 MATH 222 LEC SECOND MIDTERM EXAM THU NOV 8 PROBLEM 5 ( 5 points ) Express the number as a ratio of two integers n ()(999) (25)(999) n n ( + (ln n) 2 ) ( 4 points ) Verify that the hypotheses of the Integral Test are satisfied by this series The series terms are f(n) (n( + (ln n) 2 )) for the function f(x) (x( + (ln x) 2 )) For x > this function is positive and continuous, since + (ln x) 2 The functions x, ln x, (ln x) 2, + (ln x) 2, x( + (ln x) 2 ) are all increasing Therefore, as the reciprocal of a positive, increasing function, f(x) is decreasing The hypotheses of the integral test are satisfied for the given series and the function f(x) ( 3 points ) Evaluate the integral arising from the integral test applied to the above series in order to determine its convergence or divergence Using the substitution u ln x, du dx x one calculates (x( + (ln x) 2 )) dx ( + u 2 ) du arctan(u) π 2 According to the Integral Test the convergence of this integral is equivalent to the convergence of the given series

6 MATH 222 LEC SECOND MIDTERM EXAM THU NOV 8 PROBLEM 6 n n + n 2 n ( 5 points ) Determine if this series converges or diverges by applying the Limit Comparison Test Replacing the term n + by n in the numerator of the series yields the comparison terms n+ n 2 n n n 2 n n n 2 n n 2 /2 n 3/2, which are the terms of the convergent p-series with p 3/2 Since n + lim lim n ( lim + ) / {, } n the Limit Comparison Test implies the convergence of the series (ln n) n n n/2 n ( 6 points ) Determine if this series converges or diverges by using an appropriate test The terms of the series are positive for n > In order to apply the Root Test one calculates (using L Hôpital s rule for a limit of the form ) n n (ln n) ln n n n/2 n /2 ln n/2 n /2 ln n /2 ρ lim lim n /2 n /2 d dn n/2 d dn n/2 lim n /2 Since ρ, the Root Test guarantees the convergence of this series

7 MATH 222 LEC SECOND MIDTERM EXAM THU NOV 8 PROBLEM 7 ( n ) ( ) n + n n ( 5 points ) Determine if this series is convergent Hint: What is the value of ( n + n ) ( n + + n )? ( n ) ( n ) ( n ) 2 ( ) 2 + n + + n + n (n + ) (n) Writing u n ( n + + n ) the terms of the given series are of the form a n ( ) n u n, with u n > This shows that the given series is alternating Since the sequences n +, n, n + + n are all positive and increasing, the reciprocal of the last one, u n, is decreasing Since lim n++ n the Alternating Series Convergence Test applies and assures the convergence of the alternating series ( 4 points ) Determine if this series is absolutely convergent The series of absolute values is n n + + n u n n The p-series for p /2 is divergent Comparing the given series to this p-series the Limit Comparison Test lim / n /( n + + n) lim lim 2 / {, } n + + n n ) ( + n + shows that the absolute value series is divergent ( 2 points ) Determine if this series is conditionally convergent The given alternating series is convergent but not absolutely convergent By definition, the given series is therefore conditionally convergent