Homework 4-solutions

# Homework 4-solutions - saliyev(is4663 – Homework 4 –...

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Unformatted text preview: saliyev (is4663) – Homework 4 – knopf – (55420) 1 This print-out should have 20 questions. Multiple-choice 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 point-slope 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 ....
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Homework 4-solutions - saliyev(is4663 – Homework 4 –...

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