Lesson 12: Linear Approximation and Differentials

Section 2.8Linear Approximation andDifferentialsV63.0121.027, Calculus IOctober 13, 2009AnnouncementsMidterm Thursday on Sections 1.1–2.4OutlineThe linear approximation of a function near a pointExamplesDifferentialsThe not-so-big ideaUsing differentials to estimate errorMidterm ReviewAdvanced ExamplesAnswerThe 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 bestapproximates f near 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 bestapproximates f near 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 bestapproximates f near 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 bestapproximates f near 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)√3 1 ()+x − π2230.87475√3212So L(x) =Thus ( )61πsin≈1800.87462.Solution (i)Solution (ii)If f( )(x) = sin x, then f(0) = 0We have f π =andand f′3)(0) = 1.f′ (π.3=So the linear approximationnear 0 isL(x) = 0 + 1 · x = x.Thus()61π61πsin≈≈ 1.06465180180Calculator check: sin(61◦) ≈ExampleExampleEstimate sin(61◦) by using a linear approximation(i) about a = 0(ii) about a = 60◦ = π/3.√3 1 ()+x − π2230.87475√3212So L(x) =Thus ( )61πsin≈1800.87462.Solution (ii) ( )We have f πand3=)f′ (π.3=Calculator 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 isL(x) = 0 + 1 · x = x.Thus()61π61πsin≈≈ 1.06465180180√3 1 ()+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)If f( )(x) = sin x, then f(0) = 0We have f π =andand f′3)(0) = 1.f′ (π.3=So the linear approximationnear 0 isL(x) = 0 + 1 · x = x.Thus()61π61πsin≈≈ 1.06465180180√3 1 ()+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)√If f( )(x) = sin x, then f(0) = 0We have f π = 3 andand f′32)(0) = 1.f′ (π.3=So the linear approximationnear 0 isL(x) = 0 + 1 · x = x.Thus()61π61πsin≈≈ 1.06465180180Document Outline

• Announcements
• The linear approximation of a function near a point
  • Examples
• Differentials
  • The not-so-big idea
  • Using differentials to estimate error
• Midterm Review
• Advanced Examples