MATH 1207 R02 FINAL SOLUTION

Size: px

Start display at page:

Download "MATH 1207 R02 FINAL SOLUTION"

Kelly Mitchell
5 years ago
Views:

1 MATH 7 R FINAL SOLUTION SPRING 6 - MOON Write your answer neatly and show steps. Except calculators, any electronic devices including laptops and cell phones are not allowed. () Let f(x) = x cos x. (a) (4 pts) Find f (x). ( ) ( ) d d f (x) = dx x cos x + x dx cos x = cos x + x ( sin x) = cos x x sin x Knowing the product rule: pts. Getting the answer cos x x sin x: 4 pts. (b) (5 pts) Find f(x)dx. u = x, dv = cos x dx du = dx, v = sin x x cos x dx =x sin x sin x dx =x sin x ( cos x) + C = x sin x + cos x + C Finding appropriate u = x and dv = cos x dx: pts. Applying integration by parts and getting x sin x sin x dx: 4 pts. Getting the answer x sin x + cos x + C: 5 pts. Date: May 9, 6.

MATH 7 Final Spring 6 - Moon () Two cars A and B, start side by side and accelerate from rest. The figure shows the graphs of their velocity functions. The unit of time is minute.

2 MATH 7 Final Spring 6 - Moon () Two cars A and B, start side by side and accelerate from rest. The figure shows the graphs of their velocity functions. The unit of time is minute. (a) ( pts) On the graph above, indicate the region whose area is the moving distance of B during first minutes. See the blue region on the graph. (b) ( pts) Which car is ahead after one minute? Explain your answer. Car A is ahead, because the area under the graph, which is the moving distance is larger. (c) ( pts) What is the meaning of the area of the shaded region S? It is the distance between two cars A and B after one minute. Not mentioning the time: - pt. (d) ( pts) Estimate the time at which the cars are again side by side. (You don t need to find the precise time. Give an approximation and explain your answer.) After two minutes, the area under the graph of A and that of B are same. Therefore the moving distance of A and B during first two minutes are same and the cars are side by side.

3 MATH 7 Final Spring 6 - Moon () (8 pts) Find the volume of a cap of a sphere with radius r and height h. Let x be the variable describing the height, and let the x-coordinate of the center of the sphere is. Then the cross section is a circular disk. By Pythagorean theorem, if its radius is t, then t + x = r t = r x. So the area A(x) of the cross section is Therefore the volume is r A(x) = π( r x ) = π(r x ). r Volume = A(x) dx = π(r x ) dx r h r h = πr x π ] r x = πr r π r h r (πr (r h) π ) (r h) = πr h + π ( (r h) r ) = πr h π r h + π rh π h = πrh π h Describing the area of the cross section π(r x ): 4 pts. Describing the volume as an integral Getting the answer πrh π h : 8 pts. r r h π(r x ) dx: 6 pts.

4 MATH 7 Final Spring 6 - Moon (4) Below is the graph of a function h(x) = e (x). (a) ( pts) Find the domain and range of h. domain = (, ), range = (, ) Each item worths pt. (b) ( pts) Sketch the graph of h on the above plane. (c) ( pts) Find h (e). x = h (e) h(x) = e e x = e x = x = So h (e) =. Knowing the definition of the inverse function: pt. Getting the answer h (e) = : pts. 4

5 MATH 7 Final Spring 6 - Moon (d) ( pts) By using the inverse function theorem, find (h ) (e). h (x) = e (x) x h () = e = e (h ) (e) = h () = e Stating inverse function theorem: pt. Applying inverse function theorem and getting the answer e : pts. Getting the answer without using inverse function theorem: pt. (e) ( pts) Find h. y = h(x) y = e x ln y = x x = (ln y) h (y) = (ln y) (5) (5 pts) Test the series for convergence or divergence. n + n n n+ n n Because (n + )n = n n n= n= n + n = n n = n <, by it comparison test, n n= n + n <. Setting up an appropriate it n Getting a nonzero it: 4 pts. Making a correct conclusion: 5 pts. n+ n n : pts. + n n = + = 5

6 MATH 7 Final Spring 6 - Moon (6) If we let T (t) be the temperature of an object at time t and T s be the (constant) temperature of the surroundings, then we can formulate Newton s Law of Cooling as a differential equation dt dt = k(t T s) for some constant k. A roast turkey is taken from an oven when its temperature has reached 85 F and is placed on a table in a room where the temperature is F. (a) (5 pts) If the temperature of the turkey is 5 F after half an hour, what is the temperature after 45 minutes? d dt (T T s) = dt dt = k(t T s) T T s = Ce kt T (t) = T s + Ce kt T s = T (t) = + Ce kt 85 = T () = + C C = T (t) = + e kt 5 = T (.5) = + e.5k e.5k = e.5k =.5k = ln k = T (t) = + e t T (.) = + e. 6.9 The temperature is approximately 6.9 F. By solving the differential equation and obtaining the general solution T (t) = T s + Ce kt : pts. By using given conditions and getting the solution T (t) = +e 4 pts. Getting the answer 6.9 F : 5 pts. t : (b) (5 pts) When will the turkey have cooled to F? = T (t) = + e t e t = 5 e t = ln t = ln t = So it will take approximately.94 hours..94 Setting up the equation = + e t : pts. Solving the equation and getting the answer.94: 4 pts. Writing the answer using an appropriate unit: 5 pts. If the solution obtained in (a) was wrong, one can get at most pts. 6

7 MATH 7 Final Spring 6 - Moon (7) Consider the it x sin x x x cos x. (a) (5 pts) By using L Hospital s rule, evaluate the it. x sin x x x cos x = = cos x = cos x + x sin x cos x cos x + cos x x sin x = sin x sin x + sin x + x cos x Knowing the statement of L Hospital s rule and applying it in an appropriate way: pts. Applying it several times correctly and getting the correct answer : 5 pts. (b) (5 pts) By using Taylor series, evaluate the it. Below is a list of Taylor series of several functions. function Taylor series x sin x x x cos x = e x sin x cos x ( x x n n! = + x + x! + x! + n= ( ) n x n+ (n + )! = x x! + x5 5! n= ( ) n xn (n)! = x! + x4 4! n= ) x x + x5! 5! ) = 4! x x ( x! + x4 x x5! x x5! x x + x! x5 5! + x x + x! x5 4! + + 5! = + = x +! 5! x + =! = 4!! 4!! By using Taylor series, describing the given it as a ratio of two power series: pts. Dividing the numerator and the denominator by x and obtaining x +! 5! x + : 4 pts.! 4! Getting the answer : 5 pts. 7

8 MATH 7 Final Spring 6 - Moon (8) (a) (5 pts) Evaluate the integral t x(ln x) dx u=ln x,du= = x(ln x) dx. dx = x(ln x) t x dx ln t ln t u du = x(ln x) dx ln t ln = (ln t) + (ln ) t dx = t x(ln x) t u du = u ] ln t (ln t) + (ln ) = (ln ) Stating the definition of an improper integral: pts. Evaluating the ordinary integral and getting the value (ln t) + (ln ) : 4 pts. Computing the it and getting the answer (ln ) : 5 pts. ln (b) ( pts) Test the convergence or divergence of the series n(ln n). Because the sequence n= n(ln n) is positive, decreasing sequence, we can apply the integral test. The integral integral test, the series n= dx is convergent by (a). By x(ln x) is convergent, too. n(ln n) If the reasoning is incorrect, one cannot get the full credit. 8

9 MATH 7 Final Spring 6 - Moon (9) Let f(x) = n= n n x n. (a) (5 pts) Find the radius of convergence of f. a n = n n xn a n+ x n a n = n+ n+ n+ n x n n n = x n+ n n n x n n + n+ = x n n n n + = n x x = n + By ratio test, f(x) is absolutely convergent if x <, or equivalently, x <. Also f(x) is divergent if x >, or equivalently, x >. Therefore the radius of convergence is. x n+ n+ n+ Stating the it n : pts. x n n n Finding the it x : 4 pts. Getting the radius of convergence : 5 pts. (b) (5 pts) Test the convergence or divergence of f at two endpoints of the interval of convergence. When x =, n= n n n = n= n = n= n This is divergent because p-test (/ ). When x =, n n ( )n = ( ) n. n The sequence n= n= n is a decreasing positive sequence and Therefore by alternating series test, ( ) n <. n n=. n n =. Applying an appropriate test, determining the convergence at one of two endpoints: pts. Getting the answer for both endpoints: 5 pts. If there is no correct reason, one cannot get the credit. 9

10 MATH 7 Final Spring 6 - Moon () Let g(x) = x. (a) (4 pts) Find a power series representation of g(x). (Hint: use geometric series) x = + (x) + (x) + (x) + (x) 4 + = + x + x + x + 4 x 4 + = n x n n= (b) ( pts) Take the derivative of the representation in (a). ( ) d n x n = n nx n = + x + x + 4 4x + dx n= n= (c) ( pts) By using (b), evaluate the infinite sum n= + = n= ( ) d n x n = d dx dx x n= n nx n = ( x) n= n n n = ( ) = ( ) = 8 whose power series representa- ( x) Finding a rational function tion is n nx n : pts. n= By evaluating, getting the answer 8: pts. n n n.

11 MATH 7 Final Spring 6 - Moon () Let f(x) = x /. (a) (5 pts) Find the degree Taylor polynomial of f at a =. f() = / = f (x) = x / f () = f (x) = ( ) x 4/ f () = 9 f (x) = ( ) ( 4 ) x 7/ f () = 8 7 T (x) =f() + f ()(x ) + f ()! = + (x ) 9 (x ) + 4 (x ) 8 (x ) + f () (x )! Stating the formula for Taylor polynomial f()+f ()(x )+ f () (x! ) + f () (x ) : pts.! By taking derivatives, getting the answer + (x ) 9 (x ) (x ) : + pts. Mistake for finding derivatives: - pt for each. (b) (5 pts) By using (a), evaluate an approximation of (.5) / to the sixth decimal places. (.5) / =f(.5) T (.5) = + (.5 ) 9 (.5 ) + 4 (.5 ) 8.6 Finding the formula for the approximation: + (.5 ) 9 (.5 ) (.5 ) : 4 pts. Getting the answer.6: 5 pts. If the Taylor polynomial obtained in (a) is incorrect, one can get at most pts.

MATH 1207 R02 MIDTERM EXAM 2 SOLUTION

MATH 1207 R02 MIDTERM EXAM 2 SOLUTION MATH 7 R MIDTERM EXAM SOLUTION FALL 6 - MOON Name: Write your answer neatly and show steps. Except calculators, any electronic devices including laptops and cell phones are not allowed. () (5 pts) Find