a k 0, then k + 1 = 2 lim 1 + 1

Size: px

Start display at page:

Download "a k 0, then k + 1 = 2 lim 1 + 1"

Clifton Jennings
5 years ago
Views:

1 Math 7 - Midterm - Form A - Page From the desk of C. Davis Buenger. Problem a) [3 pts] If lim a k = then a k converges. False: The divergence test states that if lim a k, then b) [3 pts] k = False: After a partial fraction decomposition, c) [3 pts] a k diverges. k = k ( k + = lim + n n ) = 3 n + x dx = (x + 3) 5/ tan θ sec 3 θ dθ where x = 3 tan θ. True: If x = 3 tan θ, then dx = 3 sec θ dθ, and x dx = (x + 3) 5/ ( 3 tan θ) 5 3 tan ( ( 3 tan θ) + 3 ) 5/ 3 sec θ sec θ tan θ θ dθ = 5 dθ = 3 sec5 θ sec 3 θ dθ d) [3 pts] The parametric equations x = 3 + sin(3t), y = + cos(3t), for t < π, generate a circle centered at (3, ) and trace the circle exactly once. False: The circle will be centered at the point ( 3, ) and will trace the circle 3 times. e) [3 pts] k= k 5 k+ = 3 False: Since 5 <, k= k 5 = k+ k+ 5 = k+ 5 ( ) k = = 5 f) [3 pts] Given a = and a n+ 3 a n for n, the first four terms of the sequence {a n } n= are nonnegative False: a 3 a 3 = a 3 3 a 3 a 3 a =.

2 Math 7 - Midterm - Form A - Page 3 Problem [ pts] Determine whether the following improper integral converges or diverges. If it converges, find its value. x + 6 x(x + x 3) dx. x+6 Solution: First let us perform a partial fraction decomposition for. Notice the x(x +x 3) denominator factors into x(x + 3)(x ), so there exists constants A, B, C such that x + 6 x(x + x 3) = A x + B x C x. After multiplying through by the denominator of the lefthand side we find At x =, we have that x + 6 = A(x + 3)(x ) + Bx(x ) + Cx(x + 3). + 6 = A( + 3)( ) + B ( ) + C ( + 3) = 3A. So A =. At x = 3, we have that = A(3 + 3)(3 ) + B ( 3)( 3 ) + C ( 3)( 3 + 3) = B. So B =. At x =, we have that So C =. Thus x + 6 x(x + x 3) + 6 = A( + 3)( ) + B ( ) + C ( + 3) = C. b dx = lim b x + x x dx = lim ln x + ln x ln x b b ( ) (x + 3)(x ) b = lim ln b x ( ) ( ) (b + 3)(b ) ( + 3)( ) = lim ln ln b b ( ) ( ) (b + 3)(b ) 5 = ln lim ln b b ( ) 5 = ln () ln = ln 5.

3 Math 7 - Midterm - Form A - Page Problem 3 a) [ pts] Find the radius and interval of convergence for the power series. k (x + 5) k 3 k Solution: Fix x and let us perform the ratio test on the resulting series a k+ lim a k = lim (k+) (x+5) k+ 3 (k+) k (x+5) k 3 k (k + ) x + 5 k+ 3 k = lim 3 (k+) k x + 5 k (k + ) = lim k ) = lim = x ( + k x + 5 k+ x + 5 k x k 3 k+ Thus the series converges if x+5 3 <. Equivalently, the series converges if x + 5 < 9, 9 < x + 5 < 9, < x <. Thus the radius of convergence is 9 and the interval of convergence is (, ). b) [6 pts] Multiple Choice. Choose the function that is represented by the series. k x k+ k +. or Solution: Recall that the power series for ln x is x k k= k Thus k x k+ k + = k+ x k+ (k + ) = (x) k+ k + = = x k+ k+ ln x. a If you forgot this, then you can derive this equation by integrating the series for x. for x <.a

4 Math 7 - Midterm - Form A - Page 5 Problem [ pts] Taylor Series. Find the first four nonzero terms of the Taylor series for the given function centered at a. a) [ pts] f(x) = ( + x), a = Solution: By taking derivatives and evaluating these derivatives at, we have that f(x) = f (x) = ( + x) 3, ( + x), f() = ( + ) =, f () = () 3 =, f (x) ( + x), f () () = 3 8, f (x) = ( + x) 5, and f () = () 5 = 3. Thus the 3rd order polynomial is P 3 (x) = + (x ) + 3 (x ) 8! b) [ pts] g(x) = x 3 5x +, a = + 3 (x ) 3. 3! Solution: By taking derivatives and evaluating these derivatives at, we have that Thus the 3rd order polynomial is f(x) = x 3 5x +, f( ), f (x) = 3x 5, f ( ) =, f (x) x, f ( ) = 6, f (x), and f ( ). (x + ) (x + )3 P 3 (x) (x + ) ! 3!

5 Math 7 - Midterm - Form A - Page 6 Problem 5 a) [ pts] Find an equation of the line tangent to the curve at the point corresponding to the given value of t. x = t 3 + t, y = t t; t = Solution:Recall that for parametric equations, dy dy/dt (t) =. In our case, dy/dt = dx dx/dt t3, and dx/dt = 3t +. Thus dy dx (t) = t3 3t +. At the time t = dy dx () = = 3. At t = x = 3 + = and y = =. Thus in point slope form our line is and in standard form the equation is y = 3 (x ), y = 3 x 3. b) [ pts] Find the area of the region bounded by the polar curve r = + sin θ. Solution: The equation is of the form r = A + B sin θ with A > B. Hence the graph of the function is a cardioid and one must integrate from to π to find its area. Area = = = π π π π ( + sin θ) dθ + sin θ + sin θ dθ + sin θ + cos(θ) = 9 cos(θ) + sin θ + = [ 9 sin(θ) x cos θ = = 9π. [( 9 ] π dθ dθ (π) cos(π) sin((π)) ) ( )] 9 sin( ) cos

y = x 3 and y = 2x 2 x. 2x 2 x = x 3 x 3 2x 2 + x = 0 x(x 2 2x + 1) = 0 x(x 1) 2 = 0 x = 0 and x = (x 3 (2x 2 x)) dx

y = x 3 and y = 2x 2 x. 2x 2 x = x 3 x 3 2x 2 + x = 0 x(x 2 2x + 1) = 0 x(x 1) 2 = 0 x = 0 and x = (x 3 (2x 2 x)) dx Millersville University Name Answer Key Mathematics Department MATH 2, Calculus II, Final Examination May 4, 2, 8:AM-:AM Please answer the following questions. Your answers will be evaluated on their correctness,