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# A find dydx given that x2 y 2 4x 7y 15 b under

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Unformatted text preview: , we know f (2) = 5. The point (2.1, 5.3) is on the tangent line, so 0.3 5.3 − 5 = = 3. Slope tangent = 2.1 − 2 0.1 Thus, f (2) = 3. Since g is the inverse function of f and f (2) = 5, we know f −1 (5) = 2, so g (5) = 2. Diﬀerentiating, we have 1 1 1 = =. g (2) = f (g (5)) f (2) 3 9. (a) Find dy/dx given that x2 + y 2 − 4x + 7y = 15. (b) Under what conditions on x and/or y is the tangent line to this curve horizontal? Vertical? Solution: (a) By implicit diﬀerentiation, we have: 2x + 2y dy dy −4+7 dx dx dy (2y + 7) dx dy dx = 0 = 4 − 2x = 4 − 2x . 2y + 7 (b) The curve has a horizontal tangent line when dy/dx = 0, which occurs when 4 − 2x = 0 or x = 2. The curve has a horizontal tangent line at all points where x = 2. The curve has a vertical tangent line when dy/dx is undeﬁned, which occurs when 2y + 7 = 0 or when y = −7/2. The curve has a vertical tangent line at all points where y = −7/2. 10. In the following problems ﬁnd the local linearization of f (x) near x = 0 and use this to approximate the value of a. (a) f (x) = (1 + x)r , a = (1.2)3/5 Solution: We have f (x) = (1 + x)r , so f (x) = r(1 + x)r−1 . Thus f (0) = r so the local linearization near x = 0 is f (x) ≈ f (0) + f (0)x = 1 + rx. Thus (1 + x)r ≈ 1 + rx for small values of x. Using the linearization with r = 3/5 and x = 0.2, we have 3 1.23/5 = 1 + 0.2 = 1 + 0.12 = 1.12. 5 The actual value is 1.23/5 = 1.117. (b) f (x) = ekx , a = e0.3 Solution: We have f (x) = ekx , so f (x) = kekx . Thus f (0) = k so the local linearization near x = 0 is f (x) ≈ f (0) + f (0)x = 1 + kx. Thus ekx ≈ 1 + kx for small values of x. Using the linearization with k = 0.3 and x = 1, we have e0.3 ≈ 1 + 0.3 = 1.3 The actual value is e0.3 = 1.350. √ √ (c) f (x) = b2 + x, a = 26 Solution: We have f (x) = (b2 + x)1/2 , so f (x) = (1/2)(b2 + x)−1/2 . Thus f (0) = 1/(2b) so the local Math 115 / Exam 2 (March 20, 2012) page 4 linearization near x = 0 is 1 f (x) ≈ f (0) + f (0)x = b + x. 2b 3. [12 points] The following questions relate to the implicit function Thus 1 Math 115 / Exam 2 (March 20, 2012) page 4 2x b2y+ + ≈x = 4xy 2 . for small values of x. 4 b+ x 2b Using the linearizationdy with b = 5 and x = 1, we have 3. [12. points] The following questions relate to the implicit function . a [4 points] Compute dx √ 1 26 ≈ 5 + 1 = 5.1. 2 y + 4x = 4xy 210 . Solution: Diﬀerentiating the equation with respect to x, we have Math 115 / Exam 2 (March 20, 2012) √ dy dy 2y + 4 = 4y 2 + 8xy . . a. [4 points] Compute dx dx dx 11. 3Let y 2points] 4The following questions relate to the implicit function . [12 + 4x = xy 2 Solution: Diﬀerentiating dy equation the respect becomes the Gathering terms involving dx to one side,with equationto x, we have dy (a) Compute dx . y 2 + 4x = 4xy 2 . dy this curve 2 (1/3, 2) (b) Find the equation for the tangent line to + 4 = 4y at 2 xy dy . dy dy + 8− 4 22y dx 8 dx dy of all y dx −atxy dx = 4ytangent line to this curve is vertical. (c) a. [4 points] Compute Find the x and y -coordinates points...
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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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