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

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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...
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This note was uploaded on 11/19/2012 for the course ECON CALC 1 taught by Professor Fisherselin during the Fall '11 term at NYU.

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