{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

CALCIReviewProblemsExam2sols

# Dx 2y 8xy dy dy this curve at the point 1 2 b 4 points

This preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: which the . dx dy Gathering the solution which gives terms involving dx to one side, the equation becomes Solution: Diﬀerentiating the equation4with 4 y 2 − respect to x, we have dy = . dy dy 2y dx − 82y − 8xy 4y 2 − 4dy =2 dy xy dx + 4 dx4y + 8xy . 2y = dx dx which gives the solution dy Gathering terms involving dx to one side,ythe equation becomes 4 2−4 dy = . dx 2y − 8xy dy dy this curve at the point ( 1 , 2). b. [4 points] Find the equation for the 2y tangent 8xy to = 4y 2 − 4 − line 3 dx dx Solution: The slope is which gives the solution dy 4 · 22 − 4 2 = dy = −9, 1 dx ( 1 ,2) 2 · 2 − 8 4y · − 4 . = ·3 2 3 dx 2y to this 1 b. [4 points] Find the equation for the tangent line− 8xy curve at the point ( 3 , 2). ! so by the point-slope formula, the equation is Solution: The slope is y = −9x + 5. 4 dy 4 · 22 − = = −9, dx ( 1 ,2) 2 · 2 − 8 · 1 · 2 3 The actual value is page 4 26dy 5.099. = 3 1 b. so by the point-slope equation for the tangent line to this curve at the point ( 3 , 2). [4 points] Find the formula, the equation is Solution: The slope is y = −9x + 2 . 5 dy c. [4 points] Find the x- and y -coordinates = all points 4 which 9, tangent line to this of 4 · 2 − at = − the dx ( 1 ,2) 2·2−8· 1 ·2 curve is vertical. ! 3 3 Solution:the The slope isformula, the equation is so by point-slope undeﬁned as these points, so we must have 2y − 8xy = 0. Factoring out a 2y we get 2y (1 − = −=x0+ 5. y 4x) 9 c. [4 points] Find the x- and y -coordinates of all points at which the tangent line to this which gives the solutions y = 0 or x = 1 . Plugging into the equation for the implicit 4 curve is vertical. function, y = 0 gives the point (0, 0). However, when we plug in x = 1 , we get the 4 Solution: equation y 2 +The y 2 , which undeﬁned as these points, so , 0) is the only point− 8which 0. 1 = slope is has no solutions. Therefore, (0 we must have 2y at xy = Factoring line 2 vertical. the tangentout a is y we get ! 2y (1 − 4x) = 0 c. [4 points] Find the x- and y -coordinates of all points at which the tangent line to this 1 which is vertical. curve gives the solutions y = 0 or x = 4 . Plugging into the equation for the implicit function, y = 0 gives the point (0, 0). However, when we plug in x = 1 , we get the 4 Solution: The slope is undeﬁned as these points, so we must have 2y − 8xy = 0. equation y 2 + 1 = y 2 , which has no solutions. Therefore, (0, 0) is the only point at which Factoring out a 2y we get the tangent line is vertical. 2y (1 − 4x) = 0 which gives the solutions y = 0 or x = 1 . Plugging into the equation for the implicit 4 function, y = 0 gives the point (0, 0). However, when we plug in x = 1 , we get the 4 equation y 2 + 1 = y 2 , which has no solutions. Therefore, (0, 0) is the only point at which actual 5. [12 points] value of f (x)? (Show your work.) a. [3 points] Find locallocal linearization Lunderestimates of the actual value when f (0) > Solution: The the linearization...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern