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
Unformatted text preview: saliyev (is4663) Homework 4 knopf (55420) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. This assignment covers Sections 3.6 through 3.9. 001 10.0 points Find dy dx when 2 x + 1 y = 7 . 1. dy dx = 2( xy ) 1 / 2 2. dy dx = 1 2 parenleftBig x y parenrightBig 3 / 2 3. dy dx = 2 parenleftBig y x parenrightBig 3 / 2 correct 4. dy dx = 1 2 ( xy ) 1 / 2 5. dy dx = 2 parenleftBig y x parenrightBig 3 / 2 6. dy dx = 1 2 parenleftBig x y parenrightBig 3 / 2 Explanation: Differentiating implicitly with respect to x , we see that 1 2 parenleftBig 2 x x + 1 y y dy dx parenrightBig = 0 . Consequently, dy dx = 2 parenleftBig y x parenrightBig 3 / 2 . 002 10.0 points The curve with equation y 2 = x 3 + 24 x 2 is called a Tschirnhausen cubic . Its graph looks like x y Find an equation for the tangent line to this curve at the point (1 , 5). 1. y = 51 10 x + 1 10 2. y = 4 5 x 51 10 3. y = 51 10 x 1 10 correct 4. y = 45 10 x 101 10 5. y = 45 10 x 101 10 Explanation: Differentiating y 2 = x 3 + 24 x 2 implicitly we see that 2 y dy dx = 3 x 2 + 24(2 x ) . Thus dy dx = 3 x 2 + 48 x 2 y , and so at the point (1 , 5) the tangent line has slope = 3 (1) 2 + 48(1) 2 (5) = 51 10 . saliyev (is4663) Homework 4 knopf (55420) 2 Now we can use the pointslope formula to find an equation for the tangent line: y 5 = 51 10 ( x 1) . After simplification this becomes y = 51 10 x 1 10 . 003 10.0 points Find dy dx when 2 x + 3 y = 1 . 1. dy dx = 2 9 parenleftBig 1 + 2 x parenrightBig 2. dy dx = 2 9 parenleftBig 2 1 x parenrightBig correct 3. dy dx = 1 2 x 4. dy dx = 2 + 1 x 5. dy dx = 2 3 parenleftBig 1 2 x parenrightBig 6. dy dx = 2 3 parenleftBig 2 1 x parenrightBig Explanation: Differentiating implicitly with respect to x we see that 1 2 parenleftBig 2 x + 3 y dy dx parenrightBig = 0 . Thus dy dx = 2 3 parenleftBig y x parenrightBig . But y = 1 2 x 3 , so dy dx = 2 9 parenleftBig 2 x 1 x parenrightBig . Consequently, dy dx = 2 9 parenleftBig 2 1 x parenrightBig . 004 10.0 points If y = y ( x ) is defined implicitly by y 2 + xy + 3 = 0 , find the value of dy/dx at the point (4 , 1). 1. dy dx vextendsingle vextendsingle vextendsingle (4 , 1) = 1 2 2. dy dx vextendsingle vextendsingle vextendsingle (4 , 1) = 1 3. dy dx vextendsingle vextendsingle vextendsingle (4 , 1) = 1 4. dy dx vextendsingle vextendsingle vextendsingle (4 , 1) = 1 2 correct 5. dy dx vextendsingle vextendsingle vextendsingle (4 , 1) = 3 2 6. dy dx vextendsingle vextendsingle vextendsingle (4 , 1) = 3 2 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 ....
View
Full
Document
 Spring '06
 McAdam
 Calculus

Click to edit the document details