This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: jiwani (amj566) Homework07 Fouli (58395) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find y when xy + 5 x + 2 x 2 = 4 . 1. y = y + 5 + 2 x x 2. y = 5 + 2 x y x 3. y = ( y + 5 + 4 x ) 4. y = y + 5 + 2 x x 5. y = y + 5 + 4 x x 6. y = y + 5 + 4 x x correct Explanation: Differentiating implicitly with respect to x we see that d dx ( xy + 5 x + 2 x 2 ) = d dx (4) . Thus ( xy + y ) + 5 + 4 x = 0 , and so xy = y 5 4 x . Consequently, y = y + 5 + 4 x x . 002 10.0 points Find dy dx when 4 x + 1 y = 7 . 1. dy dx = 4 parenleftBig y x parenrightBig 3 / 2 2. dy dx = 4( xy ) 1 / 2 3. dy dx = 1 4 ( xy ) 1 / 2 4. dy dx = 1 4 parenleftBig x y parenrightBig 3 / 2 5. dy dx = 1 4 parenleftBig x y parenrightBig 3 / 2 6. dy dx = 4 parenleftBig y x parenrightBig 3 / 2 correct Explanation: Differentiating implicitly with respect to x , we see that 1 2 parenleftBig 4 x x + 1 y y dy dx parenrightBig = 0 . Consequently, dy dx = 4 parenleftBig y x parenrightBig 3 / 2 . 003 10.0 points Find dy dx when tan(2 x + y ) = 2 x . 1. dy dx = 4 x 2 2 + y 2 2. dy dx = 8 x 2 1 + 4 x 2 3. dy dx = 4 y 2 2 + x 2 4. dy dx = 8 x 2 1 + 4 x 2 correct 5. dy dx = 8 y 2 2 + x 2 jiwani (amj566) Homework07 Fouli (58395) 2 6. dy dx = 8 x 2 1 + 4 y 2 Explanation: Differentiating tan(2 x + y ) = 2 x implicitly with respect to x we see that sec 2 (2 x + y ) parenleftbigg 2 + dy dx parenrightbigg = 2 . Thus sec 2 (2 x + y ) dy dx = 2(1 sec 2 (2 x + y )) , in which case dy dx = 2(1 sec 2 (2 x + y )) sec 2 (2 x + y ) . But sec 2 = 1 + tan 2 , so sec 2 (2 x + y ) = 1 + tan 2 (2 x + y ) = 1 + 4 x 2 , while 1 sec 2 (2 x + y ) = tan 2 (2 x + y ) = 4 x 2 . Consequently, dy dx = 8 x 2 1 + 4 x 2 . 004 10.0 points Find the equation of the tangent line to the graph of y 2 xy 20 = 0 , at the point P = (1 , 5). 1. y = 5 x 2. 9 y + 5 x = 40 3. 8 y + 5 x = 35 4. 9 y = 5 x + 40 correct 5. 8 y = 5 x + 35 Explanation: Differentiating implicitly with respect to x we see that 2 y dy dx y x dy dx = 0 , so dy dx = y 2 y x . At P = (1 , 5), therefore, dy dx vextendsingle vextendsingle vextendsingle P = 5 9 . Thus by the point slope formula, the equation of the tangent line at P is given by y 5 = 5 9 ( x 1) . Consequently, 9 y = 5 x + 40 . 005 10.0 points Find the slope of the tangent line to the graph of 2 x 3 + y 3 + xy = 0 at the point P (1 , 1). 1. slope = 3 2 2. slope = 2 3 3. slope = 4 5 4. slope = 4 5 5. slope = 5 4 6. slope = 5 4 correct jiwani (amj566) Homework07 Fouli (58395) 3 Explanation: Differentiating implicitly with respect to x we see that 6 x 2 + 3 y 2 dy dx + y + x dy dx = 0 ....
View
Full
Document
This note was uploaded on 07/10/2011 for the course KIN 321M taught by Professor Jensen during the Spring '11 term at University of Texas at Austin.
 Spring '11
 Jensen

Click to edit the document details