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Calculus - Santowski 9/30/2016Calculus - Santowski1
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1. Define the terms explicit and implicit equations 2. Implicitly differentiate implicitly defined equations 3. Determine the equation of tangents and normals of implicitly defined equations 4. Apply implicit differentiation to real world problems 9/30/2016Calculus - Santowski2
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1. Isolate y from the equation x 2 + y 2 = 25 2. Isolate y from 2x 5 + x 4 y + y 5 = 36 3. Differentiate y = (sin2x + x 3 ) 4 4. Differentiate 36 = 2x 5 + x 4 y + y 5 on the TI-89 using the impDif( command. (You figure out the syntax. HINT: variables???) 9/30/2016Calculus - Santowski3
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Up to this point in the course, we have always defined functions by expressing one variable explicitly in terms of another i.e. y = f(x) = x 2 - 1/x + sin(x) In other courses, we have also seen functions expressed implicitly i.e. in terms of both variables i.e. x 2 + y 2 = 25 In simple implicit functions, we can always isolate the y term to rewrite the equation in explicit terms ◦ i.e. y = + (25 – x 2 ) In other cases, rewriting an implicit relation is not so easy i.e. 2x 5 + x 4 y + y 5 = 36 9/30/2016Calculus - Santowski4
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We need to agree on one convention when we see a y term in an implicitly (or explicitly defined equation), we will understand that we are really saying y(x) i.e. that y(x) is a differentiable function in x Therefore, if we see y 5, then we will interpret this expression as (y(x)) 5 it is therefore differentiable in x. 9/30/2016Calculus - Santowski5
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9/30/20166Calculus - Santowski
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Differentiate the following: 9/30/2016Calculus - Santowski7
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We apply the chain rule in that we can recognize [y(x)] 5 as a composed function with the "inner" function being y(x) and the "outer" function being x 5 So then according to the chain rule, 9/30/2016Calculus - Santowski8
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Example: Find dy/dx for 2x 5 + x 4 y + y 5 = 36 There are two strategies that must be used First, the basic rule of equations is that we can do anything to an equation, provided that we do the same thing to both sides of the equation. So it will relate to taking derivatives we will start by taking the derivative of both sides. Secondly, then, work with the various implicitly defined fcns ….. 9/30/2016Calculus - Santowski9
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9/30/2016Calculus - Santowski10
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ex 2. Find dy/dx if x + y = x 2 y 3 + 5 ex 3. Find the slope of the tangent line drawn to x 2 + 2xy + 3y 2 = 27 at x = 0. ex 4. Determine the equation of the tangent line to the ellipse 4x 2 + y 2 - 8x + 6y = 12 at x = 3. ex 5. Find d 2 y/dx 2 of x 3 + y 3 = 6xy 9/30/2016Calculus - Santowski11
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9/30/2016Calculus - Santowski12
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Differentiate the following: 9/30/2016Calculus - Santowski13
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Visual Calculus - Implicit Differentiation from UTK Visual Calculus - Implicit Differentiation from UTK Calculus I (Math 2413) - Derivatives - Implicit Differentiation from Paul Dawkins Calculus I (Math 2413) - Derivatives - Implicit Differentiation from Paul Dawkins Examples and Explanations of Implicit Differentiation from UC Davis Examples and Explanations of Implicit Differentiation from UC Davis Section 3.5 - Implicit Differentiation from OU Section 3.5 - Implicit Differentiation from OU 9/30/2016Calculus - Santowski14
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Textbook, pg359-361: (1) Algebra: Q4,8,10-16 (2) tangent eqns: Q17-27 odds (3) word probs: Q37,38,42,43 Use TI-89 to verify solutions 9/30/2016Calculus - Santowski15
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1. Find the equation of the lines that are tangent to the ellipse x 2 + 4y 2 = 16 AND also pass through the point (4,6) 2. Prove that the curves defined by x 2 – y 2 = k and xy = p intersect orthogonally for all values of of the constants k and p. Illustrate with a sketch 9/30/201616Calculus - Santowski
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3. Find the equation of the tangent line at an arbitrary point (a,b) on the curve x(2/3) + y(2/3) = 1. show that the distance between the x- and y-intercepts of the tangent line is independent of the point of tangency 9/30/201617Calculus - Santowski
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