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Sec on 2.8
    Linear Approxima ons and
           Differen als
          V63.0121.011: Calculus I
        Professor Ma hew Leingang
               New York University


              March 2, 2011

.
Announcements

   Quiz in recita on this
   week on 1.5, 1.6, 2.1, 2.2
   Midterm March 7 in class
   on 1.1–2.5
   No quiz in recita on next
   week
Midterm FAQ
 Ques on
 What sec ons are covered on the midterm?
Midterm FAQ
 Ques on
 What sec ons are covered on the midterm?

 Answer
 The midterm will cover Sec ons 1.1–2.5 (The Chain Rule).
Midterm FAQ
 Ques on
 What sec ons are covered on the midterm?

 Answer
 The midterm will cover Sec ons 1.1–2.5 (The Chain Rule).

 Ques on
 Is Sec on 2.6 going to be on the midterm?
Midterm FAQ
 Ques on
 What sec ons are covered on the midterm?

 Answer
 The midterm will cover Sec ons 1.1–2.5 (The Chain Rule).

 Ques on
 Is Sec on 2.6 going to be on the midterm?

 Answer
 The midterm will cover Sec ons 1.1–2.5 (The Chain Rule).
Midterm FAQ, continued

 Ques on
 What format will the exam take?
Midterm FAQ, continued

 Ques on
 What format will the exam take?

 Answer
 There will be both fixed-response (e.g., mul ple choice) and
 free-response ques ons.
Midterm FAQ, continued

 Ques on
 Will explana ons be necessary?
Midterm FAQ, continued

 Ques on
 Will explana ons be necessary?

 Answer
 Yes, on free-response problems we will expect you to explain
 yourself. This is why it was required on wri en homework.
Midterm FAQ, continued
 Ques on
 Is (topic X) going to be tested?
Midterm FAQ, continued
 Ques on
 Is (topic X) going to be tested?

 Answer
      Everything covered in class or on homework is fair game for the
      exam.
      No topic that was not covered in class nor on homework will be
      on the exam.
      (This is not the same as saying all exam problems are similar to
      class examples or homework problems.)
Midterm FAQ, continued

 Ques on
 Will there be a review session?
Midterm FAQ, continued

 Ques on
 Will there be a review session?

 Answer
 The recita ons this week will review for the exam.
Midterm FAQ, continued

 Ques on
 Will calculators be allowed?
Midterm FAQ, continued

 Ques on
 Will calculators be allowed?

 Answer
 No. The exam is designed for pencil and brain.
Midterm FAQ, continued
 Ques on
 How should I study?
Midterm FAQ, continued
 Ques on
 How should I study?
 Answer
     The exam has problems; study by doing problems. If you get
     one right, think about how you got it right. If you got it wrong
     or didn’t get it at all, reread the textbook and do easier
     problems to build up your understanding.
     Break up the material into chunks. (related) Don’t put it all off
     un l the night before.
     Ask ques ons.
Midterm FAQ, continued

 Ques on
 How many ques ons are there?
Midterm FAQ, continued

 Ques on
 How many ques ons are there?

 Answer
 Does this ques on contribute to your understanding of the
 material?
Midterm FAQ, continued

 Ques on
 Will there be a curve on the exam?
Midterm FAQ, continued

 Ques on
 Will there be a curve on the exam?

 Answer
 Does this ques on contribute to your understanding of the
 material?
Midterm FAQ, continued

 Ques on
 When will you grade my get-to-know-you and photo extra credit?
Midterm FAQ, continued

 Ques on
 When will you grade my get-to-know-you and photo extra credit?

 Answer
 Does this ques on contribute to your understanding of the
 material?
Objectives
  Use tangent lines to make linear
  approxima ons to a func on.
      Given a func on and a point in
      the domain, compute the
      lineariza on of the func on at
      that point.
      Use lineariza on to approximate
      values of func ons
  Given a func on, compute the
  differen al of that func on
  Use the differen al nota on to
  es mate error in linear
  approxima ons.
Outline
 The linear approxima on of a func on near a point
    Examples
    Ques ons

 Differen als
    Using differen als to es mate error

 Advanced Examples
The Big Idea
 Ques on
 What linear func on best approximates f near a?
The Big Idea
 Ques on
 What linear func on best approximates f near a?

 Answer
 The tangent line, of course!
The Big Idea
 Ques on
 What linear func on best approximates f near a?

 Answer
 The tangent line, of course!

 Ques on
 What is the equa on for the line tangent to y = f(x) at (a, f(a))?
The Big Idea
 Ques on
 What linear func on best approximates f near a?

 Answer
 The tangent line, of course!

 Ques on
 What is the equa on for the line tangent to y = f(x) at (a, f(a))?

 Answer
                      L(x) = f(a) + f′ (a)(x − a)
tangent line = linear approximation
                        y


 The func on
                                 L(x)
   L(x) = f(a) + f′ (a)(x − a)
                                 f(x)
 is a decent approxima on to f
 near a.                         f(a)
                                                x−a


                                        .                 x
                                            a         x
tangent line = linear approximation
                        y


 The func on
                                  L(x)
   L(x) = f(a) + f′ (a)(x − a)
                                  f(x)
 is a decent approxima on to f
 near a.                          f(a)
                                                 x−a
 How decent? The closer x is to
 a, the be er the
 approxima on L(x) is to f(x)            .                 x
                                             a         x
Example
 Example
 Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on
 (i) about a = 0     (ii) about a = 60◦ = π/3.
Example
 Example
 Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on
 (i) about a = 0     (ii) about a = 60◦ = π/3.
 Solu on (i)

      If f(x) = sin x, then f(0) = 0
      and f′ (0) = 1.
      So the linear approxima on
      near 0 is L(x) = 0 + 1 · x = x.
          (     )
            61π       61π
      sin          ≈      ≈ 1.06465
            180       180
Example
 Example
 Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on
 (i) about a = 0     (ii) about a = 60◦ = π/3.
 Solu on (i)
                                          Solu on (ii)
                                                         ( )
                                                We have f π =
                                                  ( )     3
                                                                   and
      If f(x) = sin x, then f(0) = 0
                                                f′ π = .
      and f′ (0) = 1.                              3


      So the linear approxima on
      near 0 is L(x) = 0 + 1 · x = x.
          (     )
            61π       61π
      sin          ≈      ≈ 1.06465
            180       180
Example
 Example
 Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on
 (i) about a = 0     (ii) about a = 60◦ = π/3.
 Solu on (i)
                                          Solu on (ii)
                                                         ( )       √
                                                                    3
                                                We have f π =
                                                  ( )     3        2
                                                                        and
      If f(x) = sin x, then f(0) = 0
                                                f′ π = .
      and f′ (0) = 1.                              3


      So the linear approxima on
      near 0 is L(x) = 0 + 1 · x = x.
          (     )
            61π       61π
      sin          ≈      ≈ 1.06465
            180       180
Example
 Example
 Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on
 (i) about a = 0     (ii) about a = 60◦ = π/3.
 Solu on (i)
                                          Solu on (ii)
                                                          ( )      √
                                                                    3
                                                We have f π =
                                                  ( )      3       2
                                                                        and
      If f(x) = sin x, then f(0) = 0
                                                f′ π = 1 .
      and f′ (0) = 1.                              3    2


      So the linear approxima on
      near 0 is L(x) = 0 + 1 · x = x.
          (     )
            61π       61π
      sin          ≈      ≈ 1.06465
            180       180
Example
 Example
 Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on
 (i) about a = 0     (ii) about a = 60◦ = π/3.
 Solu on (i)
                                          Solu on (ii)
                                                          ( )      √
                                                                    3
                                                We have f π =
                                                  ( )      3       2
                                                                        and
      If f(x) = sin x, then f(0) = 0
                                                f′ π = 1 .
      and f′ (0) = 1.                              3    2


      So the linear approxima on                So the linear approxima on is
      near 0 is L(x) = 0 + 1 · x = x.           L(x) =
          (     )
            61π       61π
      sin          ≈      ≈ 1.06465
            180       180
Example
 Example
 Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on
 (i) about a = 0     (ii) about a = 60◦ = π/3.
 Solu on (i)
                                          Solu on (ii)
                                                          ( )      √
                                                                    3
                                                We have f π =
                                                  ( )      3       2
                                                                        and
      If f(x) = sin x, then f(0) = 0
                                                f′ π = 1 .
      and f′ (0) = 1.                              3    2


      So the linear approxima on                So the linear approxima on is
                                                        √
                                                          3 1(        π)
      near 0 is L(x) = 0 + 1 · x = x.           L(x) =      +     x−
          (     )                                        2     2      3
            61π       61π
      sin          ≈      ≈ 1.06465
            180       180
Example
 Example
 Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on
 (i) about a = 0     (ii) about a = 60◦ = π/3.
 Solu on (i)
                                          Solu on (ii)
                                                          ( )      √
                                                                    3
                                                We have f π =
                                                  ( )      3       2
                                                                        and
      If f(x) = sin x, then f(0) = 0
                                                f′ π = 1 .
      and f′ (0) = 1.                              3    2


      So the linear approxima on                So the linear approxima on is
                                                        √
                                                          3 1(        π)
      near 0 is L(x) = 0 + 1 · x = x.           L(x) =      +     x−
          (     )                                        2     2      3
            61π       61π                           (     )
      sin          ≈      ≈ 1.06465                   61π
            180       180                       sin         ≈
                                                      180
Example
 Example
 Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on
 (i) about a = 0     (ii) about a = 60◦ = π/3.
 Solu on (i)
                                          Solu on (ii)
                                                          ( )      √
                                                                    3
                                                We have f π =
                                                  ( )      3       2
                                                                        and
      If f(x) = sin x, then f(0) = 0
                                                f′ π = 1 .
      and f′ (0) = 1.                              3    2


      So the linear approxima on                So the linear approxima on is
                                                        √
                                                          3 1(        π)
      near 0 is L(x) = 0 + 1 · x = x.           L(x) =      +     x−
          (     )                                        2     2      3
            61π       61π                           (     )
      sin          ≈      ≈ 1.06465                   61π
            180       180                       sin         ≈ 0.87475
                                                      180
Example
 Example
 Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on
 (i) about a = 0     (ii) about a = 60◦ = π/3.
 Solu on (i)
                                          Solu on (ii)
                                                          ( )      √
                                                                    3
                                                We have f π =
                                                  ( )      3       2
                                                                        and
      If f(x) = sin x, then f(0) = 0
                                                f′ π = 1 .
      and f′ (0) = 1.                              3    2


      So the linear approxima on                So the linear approxima on is
                                                        √
                                                          3 1(        π)
      near 0 is L(x) = 0 + 1 · x = x.           L(x) =      +     x−
          (     )                                        2     2      3
            61π       61π                           (     )
      sin          ≈      ≈ 1.06465                   61π
            180       180                       sin         ≈ 0.87475
                                                      180
Example
 Example
 Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on
 (i) about a = 0     (ii) about a = 60◦ = π/3.
 Solu on (i)
                                          Solu on (ii)
                                                          ( )      √
                                                                    3
                                                We have f π =
                                                  ( )      3       2
                                                                        and
      If f(x) = sin x, then f(0) = 0
                                                f′ π = 1 .
      and f′ (0) = 1.                              3    2


      So the linear approxima on                So the linear approxima on is
                                                        √
                                                          3 1(        π)
      near 0 is L(x) = 0 + 1 · x = x.           L(x) =      +     x−
          (     )                                        2     2      3
            61π       61π                           (     )
      sin          ≈      ≈ 1.06465                   61π
            180       180                       sin         ≈ 0.87475
                                                      180
Illustration
    y



                     y = sin x




     .                       x
               61◦
Illustration
    y
                     y = L1 (x) = x


                      y = sin x




     .                         x
         0     61◦
Illustration
    y
                                    y = L1 (x) = x



             big difference!          y = sin x




     .                                        x
         0                    61◦
Illustration
    y
                           y = L1 (x) = x

                                          √            (     )
                           y = L2 (x) =   2
                                           3
                                               +   1
                                                   2    x− π
                                                           3
                            y = sin x




     .                               x
         0     π/3   61◦
Illustration
    y
                             y = L1 (x) = x

                                             √            (     )
                              y = L2 (x) =   2
                                              3
                                                  +   1
                                                      2    x− π
                                                              3
                                y = sin x
                     very li le difference!




     .                                 x
         0     π/3   61◦
Another Example
 Example
           √
 Es mate    10 using the fact that 10 = 9 + 1.
Another Example
 Example
           √
 Es mate    10 using the fact that 10 = 9 + 1.

 Solu on
                                                       √
 The key step is to use a linear approxima on to f(x) = x near a = 9 to es mate
         √
 f(10) = 10.
Another Example
 Example
           √
 Es mate    10 using the fact that 10 = 9 + 1.

 Solu on
                                                       √
 The key step is to use a linear approxima on to f(x) = x near a = 9 to es mate
         √
 f(10) = 10.

                    f(10) ≈ L(10) = f(9) + f′ (9)(10 − 9)
                                          1          19
                                  =3+          (1) =    ≈ 3.167
                                         2·3         6
Another Example
 Example
              √
 Es mate       10 using the fact that 10 = 9 + 1.

 Solu on
                                                       √
 The key step is to use a linear approxima on to f(x) = x near a = 9 to es mate
         √
 f(10) = 10.

                            f(10) ≈ L(10) = f(9) + f′ (9)(10 − 9)
                                                  1          19
                                          =3+          (1) =    ≈ 3.167
                                                 2·3         6
          (        )2
              19
 Check:                 =
              6
Another Example
 Example
              √
 Es mate       10 using the fact that 10 = 9 + 1.

 Solu on
                                                       √
 The key step is to use a linear approxima on to f(x) = x near a = 9 to es mate
         √
 f(10) = 10.

                            f(10) ≈ L(10) = f(9) + f′ (9)(10 − 9)
                                                  1          19
                                          =3+          (1) =    ≈ 3.167
                                                 2·3         6
          (        )2
              19            361
 Check:                 =       .
              6             36
Dividing without dividing?
 Example
 A student has an irra onal fear of long division and needs to
 es mate 577 ÷ 408. He writes
             577            1             1  1
                 = 1 + 169     = 1 + 169 × ×    .
             408           408            4 102
                             1
 Help the student es mate       .
                            102
Dividing without dividing?
 Solu on
           1
 Let f(x) = . We know f(100) and we want to es mate f(102).
           x
                                         1   1
      f(102) ≈ f(100) + f′ (100)(2) =      −     (2) = 0.0098
                                        100 1002
                               577
                          =⇒       ≈ 1.41405
                               408
                     577
 Calculator check:       ≈ 1.41422.
                     408
Questions
 Example
 Suppose we are traveling in a car and at noon our speed is 50 mi/hr.
 How far will we have traveled by 2:00pm? by 3:00pm? By midnight?
Questions
 Example
 Suppose we are traveling in a car and at noon our speed is 50 mi/hr.
 How far will we have traveled by 2:00pm? by 3:00pm? By midnight?

 Answer
     100 mi
     150 mi
     600 mi (?) (Is it reasonable to assume 12 hours at the same
     speed?)
Questions
 Example
 Suppose our factory makes MP3 players and the marginal cost is
 currently $50/lot. How much will it cost to make 2 more lots? 3
 more lots? 12 more lots?
Questions
 Example
 Suppose our factory makes MP3 players and the marginal cost is
 currently $50/lot. How much will it cost to make 2 more lots? 3
 more lots? 12 more lots?

 Answer
     $100
     $150
     $600 (?)
Questions
 Example
 Suppose a line goes through the point (x0 , y0 ) and has slope m. If
 the point is moved horizontally by dx, while staying on the line, what
 is the corresponding ver cal movement?
Questions
 Example
 Suppose a line goes through the point (x0 , y0 ) and has slope m. If
 the point is moved horizontally by dx, while staying on the line, what
 is the corresponding ver cal movement?

 Answer
 The slope of the line is
                                     rise
                               m=
                                     run
 We are given a “run” of dx, so the corresponding “rise” is m dx.
Outline
 The linear approxima on of a func on near a point
    Examples
    Ques ons

 Differen als
    Using differen als to es mate error

 Advanced Examples
Differentials are derivatives
 The fact that the the tangent line is an
 approxima on means that
                                                 y
          f(x + ∆x) − f(x) ≈ f′ (x) ∆x
                 ∆y               dy


 Rename ∆x = dx, so we can write this as
                                                                    dy
              ∆y ≈ dy = f′ (x)dx.                              ∆y


                                                     dx = ∆x
 Note this looks a lot like the Leibniz-Newton
 iden ty
                    dy                           .                       x
                        = f′ (x)                     x x + ∆x
                    dx
Using differentials to estimate error
 Es ma ng error with
 differen als                    y
 If y = f(x), x0 and ∆x is
 known, and an es mate of ∆y
 is desired:
       Approximate: ∆y ≈ dy                   ∆y
                                                   dy


       Differen ate:
       dy = f′ (x) dx
                                    dx = ∆x



       Evaluate at x = x0 and   .                       x
       dx = ∆x.                     x x + ∆x
Using differentials to estimate error
  Example
  A regular sheet of plywood
  measures 8 ft × 4 ft. Suppose
  a defec ve plywood-cu ng
  machine will cut a rectangle
  whose width is exactly half its
  length, but the length is prone
  to errors. If the length is off by
  1 in, how bad can the area of
  the sheet be off by?
Solution
 Solu on
             1
 Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and
             2
 ∆ℓ = 1 in.
Solution
 Solu on
             1
 Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and
             2
 ∆ℓ = 1 in.         ( )
                      97      9409         9409
  (I) A(ℓ + ∆ℓ) = A        =       So ∆A =      − 32 ≈ 0.6701.
                      12       288         288
Solution
 Solu on
                1
 Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and
                2
 ∆ℓ = 1 in.           ( )
                        97      9409            9409
   (I) A(ℓ + ∆ℓ) = A          =       So ∆A =        − 32 ≈ 0.6701.
                        12       288             288
       dA
  (II)     = ℓ, so dA = ℓ dℓ, which should be a good es mate for ∆ℓ.
       dℓ
Solution
 Solu on
                1
 Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and
                2
 ∆ℓ = 1 in.           ( )
                        97       9409           9409
   (I) A(ℓ + ∆ℓ) = A          =       So ∆A =         − 32 ≈ 0.6701.
                        12        288            288
       dA
  (II)     = ℓ, so dA = ℓ dℓ, which should be a good es mate for ∆ℓ.
       dℓ
       When ℓ = 8 and dℓ = 12 , we have dA = 12 = 2 ≈ 0.667.
                               1                8
                                                     3
Solution
 Solu on
                1
 Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and
                2
 ∆ℓ = 1 in.           ( )
                        97       9409           9409
   (I) A(ℓ + ∆ℓ) = A          =       So ∆A =         − 32 ≈ 0.6701.
                        12        288            288
       dA
  (II)     = ℓ, so dA = ℓ dℓ, which should be a good es mate for ∆ℓ.
       dℓ
       When ℓ = 8 and dℓ = 12 , we have dA = 12 = 2 ≈ 0.667.
                               1                8
                                                     3
 So we get es mates close to the hundredth of a square foot.
Why should we care?
 Why use linear approxima ons dy when the actual difference ∆y is
 known?
     Linear approxima on is quick and reliable. Finding ∆y exactly
     depends on the func on.
     With more complicated func ons, linear approxima on much
     simpler. See the “Advanced Examples” later.
     In real life, some mes only f(a) and f′ (a) are known, and not
     the general f(x).
Outline
 The linear approxima on of a func on near a point
    Examples
    Ques ons

 Differen als
    Using differen als to es mate error

 Advanced Examples
Gravitation
 Example

    Drop a 1 kg ball off the roof of the Silver Center (50m high). We usually say
    that a falling object feels a force F = −mg from gravity.
Gravitation
 Example

    Drop a 1 kg ball off the roof of the Silver Center (50m high). We usually say
    that a falling object feels a force F = −mg from gravity.
                                       GMm
    In fact, the force felt is F(r) = − 2 , where M is the mass of the earth and
                                        r
    r is the distance from the center of the earth to the object. G is a constant.
Gravitation
 Example

    Drop a 1 kg ball off the roof of the Silver Center (50m high). We usually say
    that a falling object feels a force F = −mg from gravity.
                                       GMm
    In fact, the force felt is F(r) = − 2 , where M is the mass of the earth and
                                        r
    r is the distance from the center of the earth to the object. G is a constant.
                                             GMm                        GM
    At r = re the force really is F(re ) =     2
                                                 , and g is defined to be 2
                                              re                         re
Gravitation
 Example

    Drop a 1 kg ball off the roof of the Silver Center (50m high). We usually say
    that a falling object feels a force F = −mg from gravity.
                                       GMm
    In fact, the force felt is F(r) = − 2 , where M is the mass of the earth and
                                        r
    r is the distance from the center of the earth to the object. G is a constant.
                                             GMm                        GM
    At r = re the force really is F(re ) =     2
                                                 , and g is defined to be 2
                                              re                         re
    What is the maximum error in replacing the actual force felt at the top of
    the building F(re + ∆r) by the force felt at ground level F(re )? The rela ve
    error? The percentage error?
Gravitation Solution
 Solu on
 We wonder if ∆F = F(re + ∆r) − F(re ) is small.
     Using a linear approxima on,
                                dF           GMm
                   ∆F ≈ dF =          dr = 2 3 dr
                                dr            re
                                ( re )
                                 GMm dr            ∆r
                              =              = 2mg
                                   r2
                                    e     re       re

                            ∆F      ∆r
     The rela ve error is      ≈ −2
                             F      re
Solution continued

  re = 6378.1 km. If ∆r = 50 m,
   ∆F      ∆r        50
      ≈ −2    = −2         = −1.56 × 10−5 = −0.00156%
    F      re      6378100
Systematic linear approximation
    √        √
    2 is irra onal, but   9/4   is ra onal and 9/4 is close to 2.
Systematic linear approximation
    √        √
    2 is irra onal, but 9/4 is ra onal and 9/4 is close to 2. So
            √     √              √         1             17
              2=     9/4 − 1/4 ≈   9/4 +        (−1/4) =
                                         2(3/2)          12
Systematic linear approximation
    √        √
     2 is irra onal, but 9/4 is ra onal and 9/4 is close to 2. So
             √     √              √         1             17
               2=     9/4 − 1/4 ≈   9/4 +        (−1/4) =
                                          2(3/2)          12

   This is a be er approxima on since (17/12)2 = 289/144
Systematic linear approximation
    √        √
    2 is irra onal, but 9/4 is ra onal and 9/4 is close to 2. So
            √     √              √         1             17
              2=     9/4 − 1/4 ≈   9/4 +        (−1/4) =
                                         2(3/2)          12

   This is a be er approxima on since (17/12)2 = 289/144
   Do it again!
    √      √                  √             1
     2 = 289/144 − 1/144 ≈ 289/144 + 17 (−1/144) = 577/408
                                         2( /12)
         (     )2
           577       332, 929             1
   Now            =           which is          away from 2.
           408       166, 464          166, 464
Illustration of the previous
example




            .
Illustration of the previous
example




            .
Illustration of the previous
example




            .
                  2
Illustration of the previous
example




            .
                  2
Illustration of the previous
example




            .
                  2
Illustration of the previous
example


                (2, 17 )
                    12




            .
                           2
Illustration of the previous
example


                (2, 17 )
                    12




            .
                           2
Illustration of the previous
example



             (2, 17/12)
                          (9, 3)
                           4 2
Illustration of the previous
example



             (2, 17/12)
                          ( 289     ()9 , 3 )
                                  17 4 2
                           144 , 12
Illustration of the previous
example



             (2, 17/12)          9 3
              ( 577 ) ( 289 17()4 , 2 )
               2, 408   144 , 12
Summary
  Linear approxima on: If f is differen able at a, the best linear
  approxima on to f near a is given by

                    Lf,a (x) = f(a) + f′ (a)(x − a)

  Differen als: If f is differen able at x, a good approxima on to
  ∆y = f(x + ∆x) − f(x) is
                                dy        dy
                   ∆y ≈ dy =       · dx =    · ∆x
                                dx        dx
  Don’t buy plywood from me.

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Lesson 12: Linear Approximations and Differentials (slides)

  • 1. Sec on 2.8 Linear Approxima ons and Differen als V63.0121.011: Calculus I Professor Ma hew Leingang New York University March 2, 2011 .
  • 2. Announcements Quiz in recita on this week on 1.5, 1.6, 2.1, 2.2 Midterm March 7 in class on 1.1–2.5 No quiz in recita on next week
  • 3. Midterm FAQ Ques on What sec ons are covered on the midterm?
  • 4. Midterm FAQ Ques on What sec ons are covered on the midterm? Answer The midterm will cover Sec ons 1.1–2.5 (The Chain Rule).
  • 5. Midterm FAQ Ques on What sec ons are covered on the midterm? Answer The midterm will cover Sec ons 1.1–2.5 (The Chain Rule). Ques on Is Sec on 2.6 going to be on the midterm?
  • 6. Midterm FAQ Ques on What sec ons are covered on the midterm? Answer The midterm will cover Sec ons 1.1–2.5 (The Chain Rule). Ques on Is Sec on 2.6 going to be on the midterm? Answer The midterm will cover Sec ons 1.1–2.5 (The Chain Rule).
  • 7. Midterm FAQ, continued Ques on What format will the exam take?
  • 8. Midterm FAQ, continued Ques on What format will the exam take? Answer There will be both fixed-response (e.g., mul ple choice) and free-response ques ons.
  • 9. Midterm FAQ, continued Ques on Will explana ons be necessary?
  • 10. Midterm FAQ, continued Ques on Will explana ons be necessary? Answer Yes, on free-response problems we will expect you to explain yourself. This is why it was required on wri en homework.
  • 11. Midterm FAQ, continued Ques on Is (topic X) going to be tested?
  • 12. Midterm FAQ, continued Ques on Is (topic X) going to be tested? Answer Everything covered in class or on homework is fair game for the exam. No topic that was not covered in class nor on homework will be on the exam. (This is not the same as saying all exam problems are similar to class examples or homework problems.)
  • 13. Midterm FAQ, continued Ques on Will there be a review session?
  • 14. Midterm FAQ, continued Ques on Will there be a review session? Answer The recita ons this week will review for the exam.
  • 15. Midterm FAQ, continued Ques on Will calculators be allowed?
  • 16. Midterm FAQ, continued Ques on Will calculators be allowed? Answer No. The exam is designed for pencil and brain.
  • 17. Midterm FAQ, continued Ques on How should I study?
  • 18. Midterm FAQ, continued Ques on How should I study? Answer The exam has problems; study by doing problems. If you get one right, think about how you got it right. If you got it wrong or didn’t get it at all, reread the textbook and do easier problems to build up your understanding. Break up the material into chunks. (related) Don’t put it all off un l the night before. Ask ques ons.
  • 19. Midterm FAQ, continued Ques on How many ques ons are there?
  • 20. Midterm FAQ, continued Ques on How many ques ons are there? Answer Does this ques on contribute to your understanding of the material?
  • 21. Midterm FAQ, continued Ques on Will there be a curve on the exam?
  • 22. Midterm FAQ, continued Ques on Will there be a curve on the exam? Answer Does this ques on contribute to your understanding of the material?
  • 23. Midterm FAQ, continued Ques on When will you grade my get-to-know-you and photo extra credit?
  • 24. Midterm FAQ, continued Ques on When will you grade my get-to-know-you and photo extra credit? Answer Does this ques on contribute to your understanding of the material?
  • 25. Objectives Use tangent lines to make linear approxima ons to a func on. Given a func on and a point in the domain, compute the lineariza on of the func on at that point. Use lineariza on to approximate values of func ons Given a func on, compute the differen al of that func on Use the differen al nota on to es mate error in linear approxima ons.
  • 26. Outline The linear approxima on of a func on near a point Examples Ques ons Differen als Using differen als to es mate error Advanced Examples
  • 27. The Big Idea Ques on What linear func on best approximates f near a?
  • 28. The Big Idea Ques on What linear func on best approximates f near a? Answer The tangent line, of course!
  • 29. The Big Idea Ques on What linear func on best approximates f near a? Answer The tangent line, of course! Ques on What is the equa on for the line tangent to y = f(x) at (a, f(a))?
  • 30. The Big Idea Ques on What linear func on best approximates f near a? Answer The tangent line, of course! Ques on What is the equa on for the line tangent to y = f(x) at (a, f(a))? Answer L(x) = f(a) + f′ (a)(x − a)
  • 31. tangent line = linear approximation y The func on L(x) L(x) = f(a) + f′ (a)(x − a) f(x) is a decent approxima on to f near a. f(a) x−a . x a x
  • 32. tangent line = linear approximation y The func on L(x) L(x) = f(a) + f′ (a)(x − a) f(x) is a decent approxima on to f near a. f(a) x−a How decent? The closer x is to a, the be er the approxima on L(x) is to f(x) . x a x
  • 33. Example Example Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on (i) about a = 0 (ii) about a = 60◦ = π/3.
  • 34. Example Example Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on (i) about a = 0 (ii) about a = 60◦ = π/3. Solu on (i) If f(x) = sin x, then f(0) = 0 and f′ (0) = 1. So the linear approxima on near 0 is L(x) = 0 + 1 · x = x. ( ) 61π 61π sin ≈ ≈ 1.06465 180 180
  • 35. Example Example Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on (i) about a = 0 (ii) about a = 60◦ = π/3. Solu on (i) Solu on (ii) ( ) We have f π = ( ) 3 and If f(x) = sin x, then f(0) = 0 f′ π = . and f′ (0) = 1. 3 So the linear approxima on near 0 is L(x) = 0 + 1 · x = x. ( ) 61π 61π sin ≈ ≈ 1.06465 180 180
  • 36. Example Example Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on (i) about a = 0 (ii) about a = 60◦ = π/3. Solu on (i) Solu on (ii) ( ) √ 3 We have f π = ( ) 3 2 and If f(x) = sin x, then f(0) = 0 f′ π = . and f′ (0) = 1. 3 So the linear approxima on near 0 is L(x) = 0 + 1 · x = x. ( ) 61π 61π sin ≈ ≈ 1.06465 180 180
  • 37. Example Example Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on (i) about a = 0 (ii) about a = 60◦ = π/3. Solu on (i) Solu on (ii) ( ) √ 3 We have f π = ( ) 3 2 and If f(x) = sin x, then f(0) = 0 f′ π = 1 . and f′ (0) = 1. 3 2 So the linear approxima on near 0 is L(x) = 0 + 1 · x = x. ( ) 61π 61π sin ≈ ≈ 1.06465 180 180
  • 38. Example Example Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on (i) about a = 0 (ii) about a = 60◦ = π/3. Solu on (i) Solu on (ii) ( ) √ 3 We have f π = ( ) 3 2 and If f(x) = sin x, then f(0) = 0 f′ π = 1 . and f′ (0) = 1. 3 2 So the linear approxima on So the linear approxima on is near 0 is L(x) = 0 + 1 · x = x. L(x) = ( ) 61π 61π sin ≈ ≈ 1.06465 180 180
  • 39. Example Example Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on (i) about a = 0 (ii) about a = 60◦ = π/3. Solu on (i) Solu on (ii) ( ) √ 3 We have f π = ( ) 3 2 and If f(x) = sin x, then f(0) = 0 f′ π = 1 . and f′ (0) = 1. 3 2 So the linear approxima on So the linear approxima on is √ 3 1( π) near 0 is L(x) = 0 + 1 · x = x. L(x) = + x− ( ) 2 2 3 61π 61π sin ≈ ≈ 1.06465 180 180
  • 40. Example Example Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on (i) about a = 0 (ii) about a = 60◦ = π/3. Solu on (i) Solu on (ii) ( ) √ 3 We have f π = ( ) 3 2 and If f(x) = sin x, then f(0) = 0 f′ π = 1 . and f′ (0) = 1. 3 2 So the linear approxima on So the linear approxima on is √ 3 1( π) near 0 is L(x) = 0 + 1 · x = x. L(x) = + x− ( ) 2 2 3 61π 61π ( ) sin ≈ ≈ 1.06465 61π 180 180 sin ≈ 180
  • 41. Example Example Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on (i) about a = 0 (ii) about a = 60◦ = π/3. Solu on (i) Solu on (ii) ( ) √ 3 We have f π = ( ) 3 2 and If f(x) = sin x, then f(0) = 0 f′ π = 1 . and f′ (0) = 1. 3 2 So the linear approxima on So the linear approxima on is √ 3 1( π) near 0 is L(x) = 0 + 1 · x = x. L(x) = + x− ( ) 2 2 3 61π 61π ( ) sin ≈ ≈ 1.06465 61π 180 180 sin ≈ 0.87475 180
  • 42. Example Example Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on (i) about a = 0 (ii) about a = 60◦ = π/3. Solu on (i) Solu on (ii) ( ) √ 3 We have f π = ( ) 3 2 and If f(x) = sin x, then f(0) = 0 f′ π = 1 . and f′ (0) = 1. 3 2 So the linear approxima on So the linear approxima on is √ 3 1( π) near 0 is L(x) = 0 + 1 · x = x. L(x) = + x− ( ) 2 2 3 61π 61π ( ) sin ≈ ≈ 1.06465 61π 180 180 sin ≈ 0.87475 180
  • 43. Example Example Es mate sin(61◦ ) = sin(61π/180) by using a linear approxima on (i) about a = 0 (ii) about a = 60◦ = π/3. Solu on (i) Solu on (ii) ( ) √ 3 We have f π = ( ) 3 2 and If f(x) = sin x, then f(0) = 0 f′ π = 1 . and f′ (0) = 1. 3 2 So the linear approxima on So the linear approxima on is √ 3 1( π) near 0 is L(x) = 0 + 1 · x = x. L(x) = + x− ( ) 2 2 3 61π 61π ( ) sin ≈ ≈ 1.06465 61π 180 180 sin ≈ 0.87475 180
  • 44. Illustration y y = sin x . x 61◦
  • 45. Illustration y y = L1 (x) = x y = sin x . x 0 61◦
  • 46. Illustration y y = L1 (x) = x big difference! y = sin x . x 0 61◦
  • 47. Illustration y y = L1 (x) = x √ ( ) y = L2 (x) = 2 3 + 1 2 x− π 3 y = sin x . x 0 π/3 61◦
  • 48. Illustration y y = L1 (x) = x √ ( ) y = L2 (x) = 2 3 + 1 2 x− π 3 y = sin x very li le difference! . x 0 π/3 61◦
  • 49. Another Example Example √ Es mate 10 using the fact that 10 = 9 + 1.
  • 50. Another Example Example √ Es mate 10 using the fact that 10 = 9 + 1. Solu on √ The key step is to use a linear approxima on to f(x) = x near a = 9 to es mate √ f(10) = 10.
  • 51. Another Example Example √ Es mate 10 using the fact that 10 = 9 + 1. Solu on √ The key step is to use a linear approxima on to f(x) = x near a = 9 to es mate √ f(10) = 10. f(10) ≈ L(10) = f(9) + f′ (9)(10 − 9) 1 19 =3+ (1) = ≈ 3.167 2·3 6
  • 52. Another Example Example √ Es mate 10 using the fact that 10 = 9 + 1. Solu on √ The key step is to use a linear approxima on to f(x) = x near a = 9 to es mate √ f(10) = 10. f(10) ≈ L(10) = f(9) + f′ (9)(10 − 9) 1 19 =3+ (1) = ≈ 3.167 2·3 6 ( )2 19 Check: = 6
  • 53. Another Example Example √ Es mate 10 using the fact that 10 = 9 + 1. Solu on √ The key step is to use a linear approxima on to f(x) = x near a = 9 to es mate √ f(10) = 10. f(10) ≈ L(10) = f(9) + f′ (9)(10 − 9) 1 19 =3+ (1) = ≈ 3.167 2·3 6 ( )2 19 361 Check: = . 6 36
  • 54. Dividing without dividing? Example A student has an irra onal fear of long division and needs to es mate 577 ÷ 408. He writes 577 1 1 1 = 1 + 169 = 1 + 169 × × . 408 408 4 102 1 Help the student es mate . 102
  • 55. Dividing without dividing? Solu on 1 Let f(x) = . We know f(100) and we want to es mate f(102). x 1 1 f(102) ≈ f(100) + f′ (100)(2) = − (2) = 0.0098 100 1002 577 =⇒ ≈ 1.41405 408 577 Calculator check: ≈ 1.41422. 408
  • 56. Questions Example Suppose we are traveling in a car and at noon our speed is 50 mi/hr. How far will we have traveled by 2:00pm? by 3:00pm? By midnight?
  • 57. Questions Example Suppose we are traveling in a car and at noon our speed is 50 mi/hr. How far will we have traveled by 2:00pm? by 3:00pm? By midnight? Answer 100 mi 150 mi 600 mi (?) (Is it reasonable to assume 12 hours at the same speed?)
  • 58. Questions Example Suppose our factory makes MP3 players and the marginal cost is currently $50/lot. How much will it cost to make 2 more lots? 3 more lots? 12 more lots?
  • 59. Questions Example Suppose our factory makes MP3 players and the marginal cost is currently $50/lot. How much will it cost to make 2 more lots? 3 more lots? 12 more lots? Answer $100 $150 $600 (?)
  • 60. Questions Example Suppose a line goes through the point (x0 , y0 ) and has slope m. If the point is moved horizontally by dx, while staying on the line, what is the corresponding ver cal movement?
  • 61. Questions Example Suppose a line goes through the point (x0 , y0 ) and has slope m. If the point is moved horizontally by dx, while staying on the line, what is the corresponding ver cal movement? Answer The slope of the line is rise m= run We are given a “run” of dx, so the corresponding “rise” is m dx.
  • 62. Outline The linear approxima on of a func on near a point Examples Ques ons Differen als Using differen als to es mate error Advanced Examples
  • 63. Differentials are derivatives The fact that the the tangent line is an approxima on means that y f(x + ∆x) − f(x) ≈ f′ (x) ∆x ∆y dy Rename ∆x = dx, so we can write this as dy ∆y ≈ dy = f′ (x)dx. ∆y dx = ∆x Note this looks a lot like the Leibniz-Newton iden ty dy . x = f′ (x) x x + ∆x dx
  • 64. Using differentials to estimate error Es ma ng error with differen als y If y = f(x), x0 and ∆x is known, and an es mate of ∆y is desired: Approximate: ∆y ≈ dy ∆y dy Differen ate: dy = f′ (x) dx dx = ∆x Evaluate at x = x0 and . x dx = ∆x. x x + ∆x
  • 65. Using differentials to estimate error Example A regular sheet of plywood measures 8 ft × 4 ft. Suppose a defec ve plywood-cu ng machine will cut a rectangle whose width is exactly half its length, but the length is prone to errors. If the length is off by 1 in, how bad can the area of the sheet be off by?
  • 66. Solution Solu on 1 Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and 2 ∆ℓ = 1 in.
  • 67. Solution Solu on 1 Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and 2 ∆ℓ = 1 in. ( ) 97 9409 9409 (I) A(ℓ + ∆ℓ) = A = So ∆A = − 32 ≈ 0.6701. 12 288 288
  • 68. Solution Solu on 1 Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and 2 ∆ℓ = 1 in. ( ) 97 9409 9409 (I) A(ℓ + ∆ℓ) = A = So ∆A = − 32 ≈ 0.6701. 12 288 288 dA (II) = ℓ, so dA = ℓ dℓ, which should be a good es mate for ∆ℓ. dℓ
  • 69. Solution Solu on 1 Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and 2 ∆ℓ = 1 in. ( ) 97 9409 9409 (I) A(ℓ + ∆ℓ) = A = So ∆A = − 32 ≈ 0.6701. 12 288 288 dA (II) = ℓ, so dA = ℓ dℓ, which should be a good es mate for ∆ℓ. dℓ When ℓ = 8 and dℓ = 12 , we have dA = 12 = 2 ≈ 0.667. 1 8 3
  • 70. Solution Solu on 1 Write A(ℓ) = ℓ2 . We want to know ∆A when ℓ = 8 ft and 2 ∆ℓ = 1 in. ( ) 97 9409 9409 (I) A(ℓ + ∆ℓ) = A = So ∆A = − 32 ≈ 0.6701. 12 288 288 dA (II) = ℓ, so dA = ℓ dℓ, which should be a good es mate for ∆ℓ. dℓ When ℓ = 8 and dℓ = 12 , we have dA = 12 = 2 ≈ 0.667. 1 8 3 So we get es mates close to the hundredth of a square foot.
  • 71. Why should we care? Why use linear approxima ons dy when the actual difference ∆y is known? Linear approxima on is quick and reliable. Finding ∆y exactly depends on the func on. With more complicated func ons, linear approxima on much simpler. See the “Advanced Examples” later. In real life, some mes only f(a) and f′ (a) are known, and not the general f(x).
  • 72. Outline The linear approxima on of a func on near a point Examples Ques ons Differen als Using differen als to es mate error Advanced Examples
  • 73. Gravitation Example Drop a 1 kg ball off the roof of the Silver Center (50m high). We usually say that a falling object feels a force F = −mg from gravity.
  • 74. Gravitation Example Drop a 1 kg ball off the roof of the Silver Center (50m high). We usually say that a falling object feels a force F = −mg from gravity. GMm In fact, the force felt is F(r) = − 2 , where M is the mass of the earth and r r is the distance from the center of the earth to the object. G is a constant.
  • 75. Gravitation Example Drop a 1 kg ball off the roof of the Silver Center (50m high). We usually say that a falling object feels a force F = −mg from gravity. GMm In fact, the force felt is F(r) = − 2 , where M is the mass of the earth and r r is the distance from the center of the earth to the object. G is a constant. GMm GM At r = re the force really is F(re ) = 2 , and g is defined to be 2 re re
  • 76. Gravitation Example Drop a 1 kg ball off the roof of the Silver Center (50m high). We usually say that a falling object feels a force F = −mg from gravity. GMm In fact, the force felt is F(r) = − 2 , where M is the mass of the earth and r r is the distance from the center of the earth to the object. G is a constant. GMm GM At r = re the force really is F(re ) = 2 , and g is defined to be 2 re re What is the maximum error in replacing the actual force felt at the top of the building F(re + ∆r) by the force felt at ground level F(re )? The rela ve error? The percentage error?
  • 77. Gravitation Solution Solu on We wonder if ∆F = F(re + ∆r) − F(re ) is small. Using a linear approxima on, dF GMm ∆F ≈ dF = dr = 2 3 dr dr re ( re ) GMm dr ∆r = = 2mg r2 e re re ∆F ∆r The rela ve error is ≈ −2 F re
  • 78. Solution continued re = 6378.1 km. If ∆r = 50 m, ∆F ∆r 50 ≈ −2 = −2 = −1.56 × 10−5 = −0.00156% F re 6378100
  • 79. Systematic linear approximation √ √ 2 is irra onal, but 9/4 is ra onal and 9/4 is close to 2.
  • 80. Systematic linear approximation √ √ 2 is irra onal, but 9/4 is ra onal and 9/4 is close to 2. So √ √ √ 1 17 2= 9/4 − 1/4 ≈ 9/4 + (−1/4) = 2(3/2) 12
  • 81. Systematic linear approximation √ √ 2 is irra onal, but 9/4 is ra onal and 9/4 is close to 2. So √ √ √ 1 17 2= 9/4 − 1/4 ≈ 9/4 + (−1/4) = 2(3/2) 12 This is a be er approxima on since (17/12)2 = 289/144
  • 82. Systematic linear approximation √ √ 2 is irra onal, but 9/4 is ra onal and 9/4 is close to 2. So √ √ √ 1 17 2= 9/4 − 1/4 ≈ 9/4 + (−1/4) = 2(3/2) 12 This is a be er approxima on since (17/12)2 = 289/144 Do it again! √ √ √ 1 2 = 289/144 − 1/144 ≈ 289/144 + 17 (−1/144) = 577/408 2( /12) ( )2 577 332, 929 1 Now = which is away from 2. 408 166, 464 166, 464
  • 83. Illustration of the previous example .
  • 84. Illustration of the previous example .
  • 85. Illustration of the previous example . 2
  • 86. Illustration of the previous example . 2
  • 87. Illustration of the previous example . 2
  • 88. Illustration of the previous example (2, 17 ) 12 . 2
  • 89. Illustration of the previous example (2, 17 ) 12 . 2
  • 90. Illustration of the previous example (2, 17/12) (9, 3) 4 2
  • 91. Illustration of the previous example (2, 17/12) ( 289 ()9 , 3 ) 17 4 2 144 , 12
  • 92. Illustration of the previous example (2, 17/12) 9 3 ( 577 ) ( 289 17()4 , 2 ) 2, 408 144 , 12
  • 93. Summary Linear approxima on: If f is differen able at a, the best linear approxima on to f near a is given by Lf,a (x) = f(a) + f′ (a)(x − a) Differen als: If f is differen able at x, a good approxima on to ∆y = f(x + ∆x) − f(x) is dy dy ∆y ≈ dy = · dx = · ∆x dx dx Don’t buy plywood from me.