final-sample-solutions

# final-sample-solutions - Math 170 Section 01 Final Exam...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

## This document was uploaded on 10/29/2011 for the course MATH 170 at Vanderbilt.

### Page1 / 6

final-sample-solutions - Math 170 Section 01 Final Exam...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online