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**Unformatted text preview: **Math 170, Section 01, Final Exam, Practice Solutions Wednesday, December 9th, 2009 Name (PRINT clearly): You will receive full credit only if you show all work and explain all steps. Disclaimer : This is a practice final exam for Math 170. No claim is being made that it will resemble the actual final exam. In particular, you should not assume that every “type” of problem is listed below. 1. Compute the length of the curve y = ln(sec( x )) from x = 0 to x = π/ 4. Solution :We have y = sec( x ) tan( x ) sec( x ) = tan( x ). Hence, p 1 + ( y ) 2 = p 1 + tan 2 ( x ) = sec( x ). The arc-length is given by Z π/ 4 sec( x ) dx = ln | sec( x ) + tan( x ) | π/ 4 = ln( √ 2 + 1)- ln(1) = ln( √ 2 + 1) . 2. Find the ares of the surface obtained by rotating the graph of x = 1 3 ( y 2 + 2) 3 / 2 about the x-axis for 1 ≤ y ≤ 2. Solution :We have dx dy = y ( y 2 + 2) 1 / 2 . The surface area is 2 π Z 2 1 y s 1 + dx dy 2 dy = 2 π Z 2 1 y p 1 + y 2 ( y 2 + 2) dy = 2 π Z 2 1 y p 1 + y 4 + 2 y 2 dy = 2 π Z 2 1 y (1 + y 2 ) dy = 1 2 π (1 + y 2 ) 2 2 1 = 21 π 2 . 3. The masses m i are located at the points P i , for i = 1 , 2 , 3. Find the moments M x and M y and locate the centre of mass ( x, y ) of the system. m 1 = 6 ,m 2 = 5 ,m 3 = 9 ,P 1 = (1 , 5) ,P 2 = (3 ,- 1) ,P 3 = (- 2 ,- 1) . Solution :We have M x = m 1 y 1 + m 2 y 2 + m 3 y 3 = 6(5) + 5(- 1) + 9(- 1) x = m 1 y 1 + m 2 y 2 + m 3 y 3 m 1 + m 2 + m 3 = (6)(5) + (5)(- 1) + (9)(- 1) 6 + 5 + 9 = 4 5 . Similarly, y = (6)(1) + (5)(3) + (9)(- 2) 20 = 3 20 . 4. Find the centroid of the region bounded by y = x 2 , x = y 2 , x = 0 and x = 1. Solution :The area A = Z 1 √ x- x 2 dx = 2 3 x 3 / 2- 1 3 x 3 1 = 2 3- 1 3 = 1 3 . x = 1 A Z 1 x ( √ x- x 2 ) dx = 3 2 5 x 5 / 2- x 4 4 1 = 3( 2 5- 1 4 ) = 9 20 . y = 1 2 A Z 1 ( √ x ) 2- ( x 2 ) 2 dx = 3 1 2 x 2- x 5 5 1 = 3 1 2- 1 5 = 9 20 . 5. Suppose that the waiting times in a restaurant line are exponentially distributed with a mean waiting time of 4 minutes. 1 (a) Write down the formula for the density function. Solution :The density function is of the form f ( x ) = ce- cx where c is the reciprocal of the mean. So in this case f ( x ) = 1 4 e- x/ 4 . (b) What is the probability that you will have to wait in line for 4 minutes or longer. Solution :We have P ( X > 4) = Z ∞ 4 ce- cx dx = lim t →∞- e- cx t 4 = lim t →∞ e- 4 c- e- ct = e- 4 c = e- 1 = 1 e ....

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