Lesson 13: Linear Approximation

Section 2.8Linear Approximation andDifferentialsV63.0121, Calculus IFebruary 26/March 2, 2009AnnouncementsMidterm I Wednesday March 4 in class.OH this week: MT 1–2pm, W 2–3pmGet half of additional ALEKS points through March 22, 11:59pmImage credit: cobalt123OutlineThe linear approximation of a function near a pointExamplesDifferentialsThe not-so-big ideaAnswerThe tangent line, of course!QuestionWhat is the equation for the line tangent to y = f(x) at (a, f(a))?AnswerL(x) = f(a) + f′(a)(x − a)The Big IdeaQuestionLet f be differentiable at a. What linear function best approximates fnear a?QuestionWhat is the equation for the line tangent to y = f(x) at (a, f(a))?AnswerL(x) = f(a) + f′(a)(x − a)The Big IdeaQuestionLet f be differentiable at a. What linear function best approximates fnear a?AnswerThe tangent line, of course!AnswerL(x) = f(a) + f′(a)(x − a)The Big IdeaQuestionLet f be differentiable at a. What linear function best approximates fnear a?AnswerThe tangent line, of course!QuestionWhat is the equation for the line tangent to y = f(x) at (a, f(a))?The Big IdeaQuestionLet f be differentiable at a. What linear function best approximates fnear a?AnswerThe tangent line, of course!QuestionWhat is the equation for the line tangent to y = f(x) at (a, f(a))?AnswerL(x) = f(a) + f′(a)(x − a)√()31+x − π2230.87475√3212So L(x) =Thus ( )61πsin≈1800.87462.Solution (i)Solution (ii)( )πIf f(x) = sin x, then f(0) = 0We have f=and3( )πand f′(0) = 1.f′= .3So the linear approximationnear 0 is L(x) = 0 + 1 · x = x.Thus()61πsin≈ 61π ≈ 1.06465180180Calculator check: sin(61◦) ≈ExampleExampleEstimate sin(61◦) by using a linear approximation(i) about a = 0(ii) about a = 60◦ = π/3.√()31+x − π2230.87475√3212So L(x) =Thus ( )61πsin≈1800.87462.Solution (ii)( )We have f π =and3( )f′ π = .3Calculator check: sin(61◦) ≈ExampleExampleEstimate sin(61◦) by using a linear approximation(i) about a = 0(ii) about a = 60◦ = π/3.Solution (i)If f(x) = sin x, then f(0) = 0and f′(0) = 1.So the linear approximationnear 0 is L(x) = 0 + 1 · x = x.Thus()61πsin≈ 61π ≈ 1.06465180180√()31+x − π2230.874750.87462.√3212So L(x) =Thus ( )61πsin≈180Calculator check: sin(61◦) ≈ExampleExampleEstimate sin(61◦) by using a linear approximation(i) about a = 0(ii) about a = 60◦ = π/3.Solution (i)Solution (ii)( )πIf f(x) = sin x, then f(0) = 0We have f=and3( )πand f′(0) = 1.f′= .3So the linear approximationnear 0 is L(x) = 0 + 1 · x = x.Thus()61πsin≈ 61π ≈ 1.06465180180√()31+x − π2230.874750.87462.12So L(x) =Thus ( )61πsin≈180Calculator check: sin(61◦) ≈ExampleExampleEstimate sin(61◦) by using a linear approximation(i) about a = 0(ii) about a = 60◦ = π/3.Solution (i)Solution (ii)( )√π3If f(x) = sin x, then f(0) = 0We have f=and32( )πand f′(0) = 1.f′= .3So the linear approximationnear 0 is L(x) = 0 + 1 · x = x.Thus()61πsin≈ 61π ≈ 1.06465180180Document Outline

• Announcements
• The linear approximation of a function near a point
  • Examples
• Differentials
  • The not-so-big idea