SOLUTIONS TO EXAM II, MATH f(x)dx where a table of values for the function f(x) is given below.

Transcription

1 SOLUTIONS TO EXAM II, MATH 56 Use Simpson s rule with n = 6 to approximate the integral f(x)dx where a table of values for the function f(x) is given below x f(x) - - x f(x) We can compute δx = 6 = / therefore applying Simpson s rule we have f(x) dx δx f() f(5) f() f(5) f() f(5) f() = ( ) ( ) 6 = 5 Evaluate the following improper integral 5 π tan x dx The problem point is x = π/ Therefor we need to break up this improper integral into two component proper integrals π tan(x) dx = b π/ b tan(x) dx a π/ π Let s compute these separately beginning with the first term b π/ b tan(x) dx = a tan(x) dx ln cos(b) ln cos() b π/ = ln cos(b) b π/ = Now that we know one of the components diverges, we know the entire thing diverges Diverges Which of the following is the correct expression for the arc length of the curve between the points (, ln() and (, ln(5))? y = ln( x )

2 SOLUTIONS TO EXAM II We would like to use the formula s = To do this we need to compute dy dx t t dy dx = ( ) dy dx dx x x Furthermore we need to find t and t By inspection we see y() = ln() and y() = ln(5) Therefore t = and t = Therefore s = x ( x ) dx s = x ( x ) dx Use Euler s method with step size to estimate y(6) where y(x) is the solution to the initial value problem y = x y, y() = Recall the recursions for Euler s method: x n = x n x and y n = y n x dy dx (xn,y n) In our case, we start with x =, y =, and step size x =, so the first iteration gives an approximation of y(): x = = y = = The second iteration brings x to = 6, which is the estimate we need: x = 6 y = = = y(6) 5 Which of the five alternatives below is the differential equation whose direction field is shown in the figure?

3 SOLUTIONS TO EXAM II y = y x y = x y y = x y y = x y y = y x Notice that whenever x = y or x = y, we have a slope of zero in the direction field This tells us that the sign of x and y is irrelevant, so either the slope field is given by y x, or by x y To figure out which one is correct, observe that if we move along a vertical line, the slope increases with the absolute value of y Similarly, if we move along a horizontal line, the slope decreases with the absolute value of x Hence y should have a positive sign, and x should have a negative sign Thus the corresponding differential equation is y = y x 6 The solution to the initial value problem satisfies the implicit equation y = cos y x y() = π tan y = arctan(x) π y sin(y) = arctan(x) π tan y = arctan(x) arctan y = tan(x) π arctan y = tan(x) This is a separable differential equation, so we separate the variables as sec (y)dy = dx x

4 SOLUTIONS TO EXAM II which we integrate to get tan y = arctan x C To determine the value of C, use our initial condition and set x = and y = π/: tan π/ = arctan C = π/ C C = π = π so our solution is tan y = arctan x π 7 Find ( ) n n 9 n 9 n Since ( ) n n 9 n = n = and 9 n = 9 n = 9 both of them are absolutely convergent Hence we have ( ) n n 9 n 9 n = = 9 = 9 ( ) n n 9 n ( ) n n = 9 n 9 ( ) n 9 = = 9 = 9 9 = n 8 Find the sum of the following series e (n) n e (n) n

5 SOLUTIONS TO EXAM II 5 = e (n) n N N = N = N (e e e (n) n e (n) n e e (N) N e (n) n ) ( e e 5 = e ) ( e (N) N e (N) ) ( e (N) N N e (N) ) N 9 Which are the following sequences converge? { (I) ( ) n n } { n } n (II) n (III) I) Since n n n = n n = n the alternating sequence diverges II) Set e f(x) = x x then f(n) = n Apply the L Hospital rule to f(x), we get n Hence III) Since we have f(x) = x x x x ln = x 6x x (ln ) = x n n n = f(x) = x ln 5 n ln(5/n ) = n n = n 5/n = eln(5/n) == e n ln 5 n = n I) diverges, II) and III) converge The sequence given by a n = cos(/n) arctan(n) a n = n cos() n arctan(n) = π/ = π {5 /n} 6 x (ln ) =

6 6 SOLUTIONS TO EXAM II π ( Find) the arc length of the curve y = f(x) from the point (, /) to the point, e e, where f(x) = ex e x Recall the general formula for the length of the curve y = f(x) from the point a to the point b: b length = (f (x)) dx In our case the endpoints are and The derivative is That gives us and Now the length is To sum it up, the length is f (x) = ex e x a = ex e x (f (x)) = e x e x (f (x)) = e x e x ( e x e x ) = e x e x Solve the initial value problem e x e x dx = e e Note that we can separate the variables: x y y =, y() = x y = y dy dx = y x y = x y dy = x dx = e e

7 SOLUTIONS TO EXAM II 7 We integrate both LHS and RHS to get ln y = x C We find the constant C using the initial conditions y() = ln 5 = C, hence C = ln 5 Now ln y = x ln 5 ln y = x ln 5 And the solution is y = 5e x y = 5 e x Find the family of orthogonal trajectories to the family of curves given by y = k x The coefficient k is k = y x We differentiate y = k x with respect to x: y = k y x = x x = y x In other words, we have a family of curves given by y = y satisfies the following differential equation y = y, or dy x dx = x y Let us try to solve this equation Separate the variables and integrate ydy = xdx ydy = xdx y = x C Hence the family of orthogonal trajectories to y = k x is given by y = x C x The orthogonal family