M53 Lec2.3B Local Linear Approximation and Differentials.pdf

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Local Linear Approximation and Differentials

Mathematics 53

Institute of Mathematics (UP Diliman)

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For today

1 _{Local Linear Approximation}

2 _{Differentials}

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For today

2 _{Differentials}

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Local Linear Approximation

Suppose we are asked to find√38.03.

Most (if not all) of us cannot give an

accurate decimal expansion of this. We will give an estimate using the fact that a

tangent line to a curve at a pointPapproximates the curve nearP. That is,R

approximatesQwhenxis nearx0.

y=f(x)

P

x0

f(x0)

Q

x f(x)

S

| {z }

dx=∆x

=x0+dx

R

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Local Linear Approximation

Suppose we are asked to find√38.03. Most (if not all) of us cannot give an

accurate decimal expansion of this.

We will give an estimate using the fact that a

y=f(x)

P

x0

f(x0)

Q

x f(x)

S

| {z }

dx=∆x

=x0+dx

R

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Suppose we are asked to find√38.03. Most (if not all) of us cannot give an

accurate decimal expansion of this. We will give an estimate using the fact that a

y=f(x)

P

x0

f(x0)

Q

x f(x)

S

| {z }

dx=∆x

=x0+dx

R

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Local Linear Approximation

Recall:

f0(x0) = _x→xlim 0

f(x)− f(x0) x−x0

= lim

∆x→0

∆y

∆x,

where∆x= f(x)−f(x0)and∆x=x−x0.

If∆xis small enough, then

∆y

∆x ≈ f 0₍_x

0) =⇒∆y≈ f0(x0)∆x

The expression on the right, which approximates the change inydue to a change

inx, is called adifferential.

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Recall:

f0(x0) = _x→xlim 0

f(x)− f(x0) x−x0

= lim

∆x→0

∆y

∆x,

where∆x = f(x)−f(x0)and∆x=x−x0.

∆y

∆x ≈ f 0₍_x

0) =⇒∆y≈ f0(x0)∆x

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Local Linear Approximation

Recall:

f0(x0) = _x→xlim 0

f(x)− f(x0) x−x0

= lim

∆x→0

∆y

∆x,

If∆xis small enough,

then

∆y

∆x ≈ f 0₍_x

0) =⇒∆y≈ f0(x0)∆x

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Recall:

f0(x0) = _x→xlim 0

f(x)− f(x0) x−x0

= lim

∆x→0

∆y

∆x,

∆y

∆x ≈ f 0₍_x

0)

=⇒∆y≈ f0(x0)∆x

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Local Linear Approximation

Recall:

f0(x0) = _x→xlim 0

f(x)− f(x0) x−x0

= lim

∆x→0

∆y

∆x,

∆y

∆x ≈ f 0₍_x

0) =⇒∆y≈ f0(x0)∆x

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Recall:

f0(x0) = _x→xlim 0

f(x)− f(x0) x−x0

= lim

∆x→0

∆y

∆x,

∆y

∆x ≈ f 0₍_x

0) =⇒∆y≈ f0(x0)∆x

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Local Linear Approximation

Definitions

Let the functiony= f(x)be differentiable atx.

1 _The_differentialdx_{of the independent variable}x_{denotes an arbitrary}

increment ofx.

2 _The_differential_dy_{of the dependent variable}_y_{associated with}_x_{is given}

bydy= f0(x)dx.

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Definitions

increment ofx.

bydy= f0(x)dx.

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Local Linear Approximation

Definitions

increment ofx.

bydy= f0(x)dx.

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Example

Finddyif

y=x5−x3+2x.

Solution.

dy= (5x4−3x2+2)dx

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Local Linear Approximation

Example

Finddyif

y=x5−x3+2x.

Solution.

dy

= (5x4−3x2+2)dx

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Local Linear Approximation

Example

Finddyif

y=x5−x3+2x.

Solution.

dy= (5x4−3x2+2)

dx

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Local Linear Approximation

Example

Finddyif

y=x5−x3+2x.

Solution.

dy= (5x4−3x2+2)dx

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Local Linear Approximation

Example

Finddyif

y=√x3₊₃_x2_.

Solution.

dy= 1

2√x3₊₃_x2·(3x

2₊₆_x₎_dx

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Local Linear Approximation

Example

Finddyif

y=√x3₊₃_x2_.

Solution.

dy

= 1

2√x3₊₃_x2·(3x

2₊₆_x₎_dx

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Local Linear Approximation

Example

Finddyif

y=√x3₊₃_x2_.

Solution.

dy= 1

2√x3₊₃_x2

·(3x2+6x)dx

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Local Linear Approximation

Example

Finddyif

y=√x3₊₃_x2_.

Solution.

dy= 1

2√x3₊₃_x2·(3x 2₊₆_x₎

dx

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Local Linear Approximation

Example

Finddyif

y=√x3₊₃_x2_.

Solution.

dy= 1

2√x3₊₃_x2·(3x

2₊₆_x₎_dx

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Local Linear Approximation

Letdx=∆x =x−x0,∆y= f(x)−f(x0).

Ifdx≈0

=⇒∆y ≈ f0(x0)dx=dy

=⇒ f(x) ≈ f(x0) +f0(x0)(x−x0)

Equation of the tangent line to the graph ofy= f(x)at the point(x0,f(x0)):

y−f(x0) = f0(x0)(x−x0)

∴y= f(x0) + f0(x0)(x−x0)

Thus, whendx≈0, orx≈x0, f(x)is approximated by the tangent line to f atx0.

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Ifdx≈0

=⇒∆y ≈ f0(x0)dx

=dy

=⇒ f(x) ≈ f(x0) +f0(x0)(x−x0)

y−f(x0) = f0(x0)(x−x0)

∴y= f(x0) + f0(x0)(x−x0)

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Local Linear Approximation

Ifdx≈0

=⇒∆y ≈ f0(x0)dx=dy

=⇒ f(x) ≈ f(x0) + f0(x0)(x−x0)

y−f(x0) = f0(x0)(x−x0)

∴y= f(x0) + f0(x0)(x−x0)

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Ifdx≈0

=⇒∆y ≈ f0(x0)dx=dy

=⇒ f(x) ≈ f(x0) + f0(x0)(x−x0)

y−f(x0) = f0(x0)(x−x0)

∴y= f(x0) + f0(x0)(x−x0)

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Local Linear Approximation

Ifdx≈0

=⇒∆y ≈ f0(x0)dx=dy

=⇒ f(x) ≈ f(x0) + f0(x0)(x−x0)

y−f(x0) = f0(x0)(x−x0)

∴y= f(x0) + f0(x0)(x−x0)

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Ifdx≈0

=⇒∆y ≈ f0(x0)dx=dy

=⇒ f(x) ≈ f(x0) + f0(x0)(x−x0)

y−f(x0) = f0(x0)(x−x0)

∴

y= f(x0) + f0(x0)(x−x0)

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Local Linear Approximation

Ifdx≈0

=⇒∆y ≈ f0(x0)dx=dy

=⇒ f(x) ≈ f(x0) + f0(x0)(x−x0)

y−f(x0) = f0(x0)(x−x0)

∴y= f(x0) + f0(x0)(x−x0)

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Ifdx≈0

=⇒∆y ≈ f0(x0)dx=dy

=⇒ f(x) ≈ f(x0) + f0(x0)(x−x0)

y−f(x0) = f0(x0)(x−x0)

∴y= f(x0) + f0(x0)(x−x0)

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Local Linear Approximation

x0 x f(x)

f(x0) +f0(x0)(x−x0)

y= f(x)

`

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Remarks

The tangent lineL(x) = f(x0) + f0(x0)(x−x0)is also called thelocal

linear approximation of f(x)atx0.

The local linear approximation is the “best” linear approximation of f nearx0.

If dx=∆x =x−x0, then x=x0+dx.

Sincef(x)≈ f(x0) + f0(x0)(x−x0), we have

f(x0+dx)≈ f(x0) + f0(x0)dx

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Remarks

f(x0+dx)≈ f(x0) + f0(x0)dx

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Local Linear Approximation

Remarks

f(x0+dx)≈ f(x0) + f0(x0)dx

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Remarks

If dx=∆x =x−x0,

then x=x0+dx.

f(x0+dx)≈ f(x0) + f0(x0)dx

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Local Linear Approximation

Remarks

f(x0+dx)≈ f(x0) + f0(x0)dx

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Remarks

f(x0+dx)≈ f(x0) + f0(x0)dx

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Local Linear Approximation

Remarks

f(x0+dx)≈ f(x0) + f0(x0)dx

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Remarks

dx=∆x≈0

=⇒∆y≈dy

dy is easier to compute than ∆y

∴ dy is used to approximate ∆y when dx≈0

Ifdx6=0, then

dy= f0(x)dx =⇒ dy

dx = f 0₍_x₎_.

The symbol dy

dx may be interpreted as:

the derivative ofy= f(x)with respect tox

the quotient of the differential ofyby the differential ofx

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Local Linear Approximation

Remarks

dx=∆x≈0=⇒

∆y≈dy

Ifdx6=0, then

dy= f0(x)dx =⇒ dy

dx = f 0₍_x₎_.

The symbol dy

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Remarks

dx=∆x≈0=⇒∆y≈dy

Ifdx6=0, then

dy= f0(x)dx =⇒ dy

dx = f 0₍_x₎_.

The symbol dy

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Local Linear Approximation

Remarks

Ifdx6=0, then

dy= f0(x)dx =⇒ dy

dx = f 0₍_x₎_.

The symbol dy

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Remarks

Ifdx6=0, then

dy= f0(x)dx =⇒ dy

dx = f 0₍_x₎_.

The symbol dy

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Local Linear Approximation

Remarks

Ifdx6=0,

then

dy= f0(x)dx =⇒ dy

dx = f 0₍_x₎_.

The symbol dy

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Remarks

Ifdx6=0, then

dy= f0(x)dx

=⇒ dy

dx = f 0₍_x₎_.

The symbol dy

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Local Linear Approximation

Remarks

Ifdx6=0, then

dy= f0(x)dx =⇒ dy

dx = f 0₍_x₎_.

The symbol dy

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Remarks

Ifdx6=0, then

dy= f0(x)dx =⇒ dy

dx = f 0₍_x₎_.

The symbol dy

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Local Linear Approximation

Remarks

Ifdx6=0, then

dy= f0(x)dx =⇒ dy

dx = f 0₍_x₎_.

The symbol dy

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Remarks

Ifdx6=0, then

dy= f0(x)dx =⇒ dy

dx = f 0₍_x₎_.

The symbol dy

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Local Linear Approximation and Differentials

Example

Find the local linear approximation of f(x) =√3 _x_at_x

0=8and use this to

estimate√38.03.

Solution.

We have f0(x) = 1

3√3 x2.

L(x) = f(x0) + f0(x0)(x−x0)

∴Atx0=8:

L(x) = f(8) + f0(8)(x−8) =2+ 1

12(x−8).

Thus,√3 8.03= f(8.03)≈L(8.03) =2+ 1

12(8.03−8) =2.0025.

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Local Linear Approximation and Differentials

Example

0=8and use this to

estimate√38.03.

Solution.

We have f0(x) = 1

3√3 x2.

L(x) = f(x0) + f0(x0)(x−x0)

∴Atx0=8:

L(x) = f(8) + f0(8)(x−8) =2+ 1

12(x−8).

Thus,√3 8.03= f(8.03)≈L(8.03) =2+ 1

12(8.03−8) =2.0025.

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Local Linear Approximation and Differentials

Example

0=8and use this to

estimate√38.03.

Solution.

We have f0(x) = 1

3√3 x2.

L(x) = f(x0) + f0(x0)(x−x0)

∴Atx0=8:

L(x) = f(8) + f0(8)(x−8) =2+ 1

12(x−8).

Thus,√3 8.03= f(8.03)≈L(8.03) =2+ 1

12(8.03−8) =2.0025.

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Local Linear Approximation and Differentials

Example

0=8and use this to

estimate√38.03.

Solution.

We have f0(x) = 1

3√3 x2.

L(x) = f(x0) + f0(x0)(x−x0)

∴Atx0=8:

L(x) = f(8) + f0(8)(x−8) =2+ 1

12(x−8).

Thus,√3 8.03= f(8.03)≈L(8.03) =2+ 1

12(8.03−8) =2.0025.

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Local Linear Approximation and Differentials

Example

0=8and use this to

estimate√38.03.

Solution.

We have f0(x) = 1

3√3 x2.

L(x) = f(x0) + f0(x0)(x−x0)

∴Atx0=8:

L(x)

= f(8) + f0(8)(x−8) =2+ 1

12(x−8).

Thus,√3 8.03= f(8.03)≈L(8.03) =2+ 1

12(8.03−8) =2.0025.

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Local Linear Approximation and Differentials

Example

0=8and use this to

estimate√38.03.

Solution.

We have f0(x) = 1

3√3 x2.

L(x) = f(x0) + f0(x0)(x−x0)

∴Atx0=8:

L(x) = f(8)

+ f0(8)(x−8) =2+ 1

12(x−8).

Thus,√3 8.03= f(8.03)≈L(8.03) =2+ 1

12(8.03−8) =2.0025.

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Local Linear Approximation and Differentials

Example

0=8and use this to

estimate√38.03.

Solution.

We have f0(x) = 1

3√3 x2.

L(x) = f(x0) + f0(x0)(x−x0)

∴Atx0=8:

L(x) = f(8) + f0(8)

(x−8) =2+ 1

12(x−8).

Thus,√3 8.03= f(8.03)≈L(8.03) =2+ 1

12(8.03−8) =2.0025.

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Local Linear Approximation and Differentials

Example

0=8and use this to

estimate√38.03.

Solution.

We have f0(x) = 1

3√3 x2.

L(x) = f(x0) + f0(x0)(x−x0)

∴Atx0=8:

L(x) = f(8) + f0(8)(x−8)

=2+ 1

12(x−8).

Thus,√3 8.03= f(8.03)≈L(8.03) =2+ 1

12(8.03−8) =2.0025.

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Local Linear Approximation and Differentials

Example

0=8and use this to

estimate√38.03.

Solution.

We have f0(x) = 1

3√3 x2.

L(x) = f(x0) + f0(x0)(x−x0)

∴Atx0=8:

L(x) = f(8) + f0(8)(x−8) =2

+ 1

12(x−8).

Thus,√3 8.03= f(8.03)≈L(8.03) =2+ 1

12(8.03−8) =2.0025.

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Local Linear Approximation and Differentials

Example

0=8and use this to

estimate√38.03.

Solution.

We have f0(x) = 1

3√3 x2.

L(x) = f(x0) + f0(x0)(x−x0)

∴Atx0=8:

L(x) = f(8) + f0(8)(x−8) =2+ 1

12

(x−8).

Thus,√3 8.03= f(8.03)≈L(8.03) =2+ 1

12(8.03−8) =2.0025.

(62)

Local Linear Approximation and Differentials

Example

0=8and use this to

estimate√38.03.

Solution.

We have f0(x) = 1

3√3 x2.

L(x) = f(x0) + f0(x0)(x−x0)

∴Atx0=8:

L(x) = f(8) + f0(8)(x−8) =2+ 1

12(x−8).

Thus,√3 8.03= f(8.03)≈L(8.03) =2+ 1

12(8.03−8) =2.0025.

(63)

Local Linear Approximation and Differentials

Example

0=8and use this to

estimate√38.03.

Solution.

We have f0(x) = 1

3√3 x2.

L(x) = f(x0) + f0(x0)(x−x0)

∴Atx0=8:

L(x) = f(8) + f0(8)(x−8) =2+ 1

12(x−8).

Thus,√3 8.03

= f(8.03)≈L(8.03) =2+ 1

12(8.03−8) =2.0025.

(64)

Local Linear Approximation and Differentials

Example

0=8and use this to

estimate√38.03.

Solution.

We have f0(x) = 1

3√3 x2.

L(x) = f(x0) + f0(x0)(x−x0)

∴Atx0=8:

L(x) = f(8) + f0(8)(x−8) =2+ 1

12(x−8).

Thus,√3 8.03= f(8.03)

≈L(8.03) =2+ 1

12(8.03−8) =2.0025.

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Local Linear Approximation and Differentials

Example

0=8and use this to

estimate√38.03.

Solution.

We have f0(x) = 1

3√3 x2.

L(x) = f(x0) + f0(x0)(x−x0)

∴Atx0=8:

L(x) = f(8) + f0(8)(x−8) =2+ 1

12(x−8).

Thus,√3 8.03= f(8.03)≈L(8.03) =2+ 1

12(8.03−8)

=2.0025.

(66)

Local Linear Approximation and Differentials

Example

0=8and use this to

estimate√38.03.

Solution.

We have f0(x) = 1

3√3 x2.

L(x) = f(x0) + f0(x0)(x−x0)

∴Atx0=8:

L(x) = f(8) + f0(8)(x−8) =2+ 1

12(x−8).

Thus,√3 8.03= f(8.03)≈L(8.03) =2+ 1

12(8.03−8) =2.0025.

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Local Linear Approximation and Differentials

Example

Approximate√3 27.027.

Solution.

Let f(x) =√3 _x_{. Then} _f0₍_x_{) =} 1

3√3 x2 .

f(x0+dx)≈ f(x0) + f0(x0)dx

Thus,

3 √

27.027 = f(27+0.027)

≈ f(27) +f0(27)·(0.027)

= 3+ 1

3·9 ·(0.027)

= 3.01.

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Local Linear Approximation and Differentials

Example

Solution.

Let f(x) =√3 _x_{. Then} _f0₍_x_{) =} 1

3√3 x2 .

f(x0+dx)≈ f(x0) + f0(x0)dx

Thus,

3 √

27.027 = f(27+0.027)

≈ f(27) +f0(27)·(0.027)

= 3+ 1

3·9 ·(0.027)

= 3.01.

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Local Linear Approximation and Differentials

Example

Solution.

Let f(x) =√3 _x_.

Then f0(x) = 1

3√3 x2 .

f(x0+dx)≈ f(x0) + f0(x0)dx

Thus,

3 √

27.027 = f(27+0.027)

≈ f(27) +f0(27)·(0.027)

= 3+ 1

3·9 ·(0.027)

= 3.01.

(70)

Local Linear Approximation and Differentials

Example

Solution.

Let f(x) =√3 _x_{. Then} _f0₍_x_{) =} 1

3√3 x2 .

f(x0+dx)≈ f(x0) + f0(x0)dx

Thus,

3 √

27.027 = f(27+0.027)

≈ f(27) +f0(27)·(0.027)

= 3+ 1

3·9 ·(0.027)

= 3.01.

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Local Linear Approximation and Differentials

Example

Solution.

Let f(x) =√3 _x_{. Then} _f0₍_x_{) =} 1

3√3 x2 .

f(x0+dx)≈ f(x0) + f0(x0)dx

Thus,

3 √

27.027 = f(27+0.027)

≈ f(27) +f0(27)·(0.027)

= 3+ 1

3·9 ·(0.027)

= 3.01.

(72)

Local Linear Approximation and Differentials

Example

Solution.

Let f(x) =√3 _x_{. Then} _f0₍_x_{) =} 1

3√3 x2 .

f(x0+dx)≈ f(x0) + f0(x0)dx

Thus,

3 √

27.027

= f(27+0.027)

≈ f(27) +f0(27)·(0.027)

= 3+ 1

3·9 ·(0.027)

= 3.01.

(73)

Local Linear Approximation and Differentials

Example

Solution.

Let f(x) =√3 _x_{. Then} _f0₍_x_{) =} 1

3√3 x2 .

f(x0+dx)≈ f(x0) + f0(x0)dx

Thus,

3 √

27.027 = f(27+0.027)

≈ f(27) +f0(27)·(0.027)

= 3+ 1

3·9 ·(0.027)

= 3.01.

(74)

Local Linear Approximation and Differentials

Example

Solution.

Let f(x) =√3 _x_{. Then} _f0₍_x_{) =} 1

3√3 x2 .

f(x0+dx)≈ f(x0) + f0(x0)dx

Thus,

3 √

27.027 = f(27+0.027)

≈ f(27) +f0(27)·(0.027)

= 3+ 1

3·9 ·(0.027)

= 3.01.

(75)

Local Linear Approximation and Differentials

Example

Solution.

Let f(x) =√3 _x_{. Then} _f0₍_x_{) =} 1

3√3 x2 .

f(x0+dx)≈ f(x0) + f0(x0)dx

Thus,

3 √

27.027 = f(27+0.027)

≈ f(27) +f0(27)·(0.027)

= 3

+ 1

3·9 ·(0.027)

= 3.01.

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Local Linear Approximation and Differentials

Example

Solution.

Let f(x) =√3 _x_{. Then} _f0₍_x_{) =} 1

3√3 x2 .

f(x0+dx)≈ f(x0) + f0(x0)dx

Thus,

3 √

27.027 = f(27+0.027)

≈ f(27) +f0(27)·(0.027)

= 3+ 1

3·9

·(0.027)

= 3.01.

(77)

Local Linear Approximation and Differentials

Example

Solution.

Let f(x) =√3 _x_{. Then} _f0₍_x_{) =} 1

3√3 x2 .

f(x0+dx)≈ f(x0) + f0(x0)dx

Thus,

3 √

27.027 = f(27+0.027)

≈ f(27) +f0(27)·(0.027)

= 3+ 1

3·9 ·(0.027)

= 3.01.

(78)

Local Linear Approximation and Differentials

Example

Solution.

Let f(x) =√3 _x_{. Then} _f0₍_x_{) =} 1

3√3 x2 .

f(x0+dx)≈ f(x0) + f0(x0)dx

Thus,

3 √

27.027 = f(27+0.027)

≈ f(27) +f0(27)·(0.027)

= 3+ 1

3·9 ·(0.027)

= 3.01.

(79)

Local Linear Approximation and Differentials

Example

Approximate√15.96.

Solution.

Let f(x) =√x. Then f0(x) = 1

2√x.

Thus,

√

15.96 = f(16−0.04)

≈ f(16) + f0(16)·(−0.04)

= 4+ 1

2·4·(−0.04)

= 4−0.005

= 3.995.

(80)

Local Linear Approximation and Differentials

Example

Solution.

2√x.

Thus,

√

15.96 = f(16−0.04)

≈ f(16) + f0(16)·(−0.04)

= 4+ 1

2·4·(−0.04)

= 4−0.005

= 3.995.

(81)

Local Linear Approximation and Differentials

Example

Solution.

Let f(x) =√x.

Then f0(x) = 1

2√x.

Thus,

√

15.96 = f(16−0.04)

≈ f(16) + f0(16)·(−0.04)

= 4+ 1

2·4·(−0.04)

= 4−0.005

= 3.995.

(82)

Local Linear Approximation and Differentials

Example

Solution.

2√x.

Thus,

√

15.96 = f(16−0.04)

≈ f(16) + f0(16)·(−0.04)

= 4+ 1

2·4·(−0.04)

= 4−0.005

= 3.995.

(83)

Local Linear Approximation and Differentials

Example

Solution.

2√x.

Thus,

√

15.96

= f(16−0.04)

≈ f(16) + f0(16)·(−0.04)

= 4+ 1

2·4·(−0.04)

= 4−0.005

= 3.995.

(84)

Local Linear Approximation and Differentials

Example

Solution.

2√x.

Thus,

√

15.96 = f(16−0.04)

≈ f(16) + f0(16)·(−0.04)

= 4+ 1

2·4·(−0.04)

= 4−0.005

= 3.995.

(85)

Local Linear Approximation and Differentials

Example

Solution.

2√x.

Thus,

√

15.96 = f(16−0.04)

≈ f(16) + f0(16)·(−0.04)

= 4+ 1

2·4·(−0.04)

= 4−0.005

= 3.995.

(86)

Local Linear Approximation and Differentials

Example

Solution.

2√x.

Thus,

√

15.96 = f(16−0.04)

≈ f(16) + f0(16)·(−0.04)

= 4

+ 1

2·4·(−0.04)

= 4−0.005

= 3.995.

(87)

Local Linear Approximation and Differentials

Example

Solution.

2√x.

Thus,

√

15.96 = f(16−0.04)

≈ f(16) + f0(16)·(−0.04)

= 4+ 1

2·4

·(−0.04)

= 4−0.005

= 3.995.

(88)

Local Linear Approximation and Differentials

Example

Solution.

2√x.

Thus,

√

15.96 = f(16−0.04)

≈ f(16) + f0(16)·(−0.04)

= 4+ 1

2·4·(−0.04)

= 4−0.005

= 3.995.

(89)

Local Linear Approximation and Differentials

Example

Solution.

2√x.

Thus,

√

15.96 = f(16−0.04)

≈ f(16) + f0(16)·(−0.04)

= 4+ 1

2·4·(−0.04)

= 4−0.005

= 3.995.

(90)

Local Linear Approximation and Differentials

Example

Solution.

2√x.

Thus,

√

15.96 = f(16−0.04)

≈ f(16) + f0(16)·(−0.04)

= 4+ 1

2·4·(−0.04)

= 4−0.005

= 3.995.

(91)

For today

2 _{Differentials}

(92)

Differentials

Example

A ball 10 inches in diameter is to be covered by a rubber material which is₁₆1 in

thick. Use differentials to estimate the volume of the rubber material that will be

used.

Solution.

Volume of a sphere with radiusr: V(r) = 4

3πr

3 ₌_⇒ _dV₌₄_π_r2_dr

Volume of rubber material = V(5+₁₆1)−V(5)

= ∆V r=5, dr= ₁₆1

≈ dV

= 4πr2dr = 4π(5)2·(₁₆1) = 25π

4 in3

(93)

Differentials

Example

used.

Solution.

Volume of a sphere with radiusr:

V(r) = 4

3πr

3 ₌_⇒ _dV₌₄_π_r2_dr

= ∆V r=5, dr= ₁₆1

≈ dV

= 4πr2dr = 4π(5)2·(₁₆1) = 25π

4 in3

(94)

Differentials

Example

used.

Solution.

3πr

3

=⇒ dV=4πr2dr

= ∆V r=5, dr= ₁₆1

≈ dV

= 4πr2dr = 4π(5)2·(₁₆1) = 25π

4 in3

(95)

Differentials

Example

used.

Solution.

3πr

3 ₌_⇒ _dV

=4πr2dr

= ∆V r=5, dr= ₁₆1

≈ dV

= 4πr2dr = 4π(5)2·(₁₆1) = 25π

4 in3

(96)

Differentials

Example

used.

Solution.

3πr

3 ₌_⇒ _dV₌₄_π_r2_dr

= ∆V r=5, dr= ₁₆1

≈ dV

= 4πr2dr = 4π(5)2·(₁₆1) = 25π

4 in3

(97)

Differentials

Example

used.

Solution.

3πr

3 ₌_⇒ _dV₌₄_π_r2_dr

Volume of rubber material

= V(5+₁₆1)−V(5)

= ∆V r=5, dr= ₁₆1

≈ dV

= 4πr2dr = 4π(5)2·(₁₆1) = 25π

4 in3

(98)

Differentials

Example

used.

Solution.

3πr

3 ₌_⇒ _dV₌₄_π_r2_dr

= ∆V r=5, dr= ₁₆1

≈ dV

= 4πr2dr = 4π(5)2·(₁₆1) = 25π

4 in3

(99)

Differentials

Example

used.

Solution.

3πr

3 ₌_⇒ _dV₌₄_π_r2_dr

= ∆V

r=5, dr= ₁₆1

≈ dV

= 4πr2dr = 4π(5)2·(₁₆1) = 25π

4 in3

(100)

Differentials

Example

used.

Solution.

3πr

3 ₌_⇒ _dV₌₄_π_r2_dr

= ∆V r=5, dr= ₁₆1

≈ dV

= 4πr2dr = 4π(5)2·(₁₆1) = 25π

4 in3

(101)

Differentials

Example

used.

Solution.

3πr

3 ₌_⇒ _dV₌₄_π_r2_dr

= ∆V r=5, dr= ₁₆1

≈ dV

= 4πr2dr = 4π(5)2·(₁₆1) = 25π

4 in3

(102)

Differentials

Example

used.

Solution.

3πr

3 ₌_⇒ _dV₌₄_π_r2_dr

= ∆V r=5, dr= ₁₆1

≈ dV

= 4πr2dr

= 4π(5)2·(₁₆1) = 25π

4 in3

(103)

Differentials

Example

used.

Solution.

3πr

3 ₌_⇒ _dV₌₄_π_r2_dr

= ∆V r=5, dr= ₁₆1

≈ dV

= 4πr2dr = 4π(5)2·(₁₆1)

= 25π 4 in3

(104)

Differentials

Example

used.

Solution.

3πr

3 ₌_⇒ _dV₌₄_π_r2_dr

= ∆V r=5, dr= ₁₆1

≈ dV

= 4πr2dr = 4π(5)2·(₁₆1) = 25π

4 in3

(105)

Example

A metal rod 15 cm long and 8 cm in diameter is to be insulated, except for the

ends, with a material 0.001 cm thick. Use differentials to estimate the volume of

the insulation.

Solution.

Volume of the rod: V=πr2h=15πr2 =⇒ dV =30πr dr

Volume of the insulation = V(4+0.001)−V(4)

= ∆V r=4, dr=0.001

≈ dV

= 30π(4)·(0.001)

= 0.12πcm3.

(106)

Example

the insulation.

Solution.

Volume of the rod:

V=πr2h=15πr2 =⇒ dV =30πr dr

= ∆V r=4, dr=0.001

≈ dV

= 30π(4)·(0.001)

= 0.12πcm3.

(107)

Example

the insulation.

Solution.

Volume of the rod: V=πr2h

=15πr2 =⇒ dV =30πr dr

= ∆V r=4, dr=0.001

≈ dV

= 30π(4)·(0.001)

= 0.12πcm3.

(108)

Example

the insulation.

Solution.

Volume of the rod: V=πr2h=15πr2

=⇒ dV =30πr dr

= ∆V r=4, dr=0.001

≈ dV

= 30π(4)·(0.001)

= 0.12πcm3.

(109)

Example

the insulation.

Solution.

Volume of the rod: V=πr2h=15πr2 =⇒ dV

=30πr dr

= ∆V r=4, dr=0.001

≈ dV

= 30π(4)·(0.001)

= 0.12πcm3.

(110)

Example

the insulation.

Solution.

= ∆V r=4, dr=0.001

≈ dV

= 30π(4)·(0.001)

= 0.12πcm3.

(111)

Example

the insulation.

Solution.

Volume of the insulation

= V(4+0.001)−V(4)

= ∆V r=4, dr=0.001

≈ dV

= 30π(4)·(0.001)

= 0.12πcm3.

(112)

Example

the insulation.

Solution.

= ∆V r=4, dr=0.001

≈ dV

= 30π(4)·(0.001)

= 0.12πcm3.

(113)

Example

the insulation.

Solution.

= ∆V

r=4, dr=0.001

≈ dV

= 30π(4)·(0.001)

= 0.12πcm3.

(114)

Example

the insulation.

Solution.

= ∆V r=4, dr=0.001

≈ dV

= 30π(4)·(0.001)

= 0.12πcm3.

(115)

Example

the insulation.

Solution.

= ∆V r=4, dr=0.001

≈ dV

= 30π(4)·(0.001)

= 0.12πcm3.

(116)

Example

the insulation.

Solution.

= ∆V r=4, dr=0.001

≈ dV

= 30π(4)·(0.001)

= 0.12πcm3.

(117)

Example

the insulation.

Solution.

= ∆V r=4, dr=0.001

≈ dV

= 30π(4)·(0.001)

= 0.12πcm3.

(118)

Example

Suppose that the side of a square is measured with a ruler to be8inches with a

measurement error of at most±1

64 of an inch. Estimate the error in the computed

area of the square.

Solution.

Area of square with sidex: A(x) =x2 ₌_⇒ _dA₌₂_{x dx}

Measurement error of at most±1

64 =⇒ |dx|= 641

Error in the computed area= ∆A

We are computing error fromA(8), so we takex=8.

|∆A| ≈ |dA|=|2x| · |dx|=2(8)(₆₄1) = 1₄

∴The propagated error in the computed area is at most±1

4of a square inch.

(119)

Example

area of the square.

Solution.

Area of square with sidex:

A(x) =x2 ₌_⇒ _dA₌₂_{x dx}

64 =⇒ |dx|= 641

|∆A| ≈ |dA|=|2x| · |dx|=2(8)(₆₄1) = 1₄

4of a square inch.

(120)

Example

area of the square.

Solution.

Area of square with sidex: A(x) =x2

=⇒ dA=2x dx

64 =⇒ |dx|= 641

|∆A| ≈ |dA|=|2x| · |dx|=2(8)(₆₄1) = 1₄

4of a square inch.

(121)

Example

area of the square.

Solution.

Area of square with sidex: A(x) =x2 =⇒ dA

=2x dx

64 =⇒ |dx|= 641

|∆A| ≈ |dA|=|2x| · |dx|=2(8)(₆₄1) = 1₄

4of a square inch.

(122)

Example

area of the square.

Solution.

Area of square with sidex: A(x) =x2 =⇒ dA=2x dx

64 =⇒ |dx|= 641

|∆A| ≈ |dA|=|2x| · |dx|=2(8)(₆₄1) = 1₄

4of a square inch.

(123)

Example

area of the square.

Solution.

64

=⇒ |dx|= ₆₄1

|∆A| ≈ |dA|=|2x| · |dx|=2(8)(₆₄1) = 1₄

4of a square inch.

(124)

Example

area of the square.

Solution.

64 =⇒ |dx|= 641

|∆A| ≈ |dA|=|2x| · |dx|=2(8)(₆₄1) = 1₄

4of a square inch.

(125)

Example

area of the square.

Solution.

64 =⇒ |dx|= 641

Error in the computed area

= ∆A

|∆A| ≈ |dA|=|2x| · |dx|=2(8)(₆₄1) = 1₄

4of a square inch.

(126)

Example

area of the square.

Solution.

64 =⇒ |dx|= 641

|∆A| ≈ |dA|=|2x| · |dx|=2(8)(₆₄1) = 1₄

4of a square inch.

(127)

Example

area of the square.

Solution.

64 =⇒ |dx|= 641

|∆A| ≈ |dA|=|2x| · |dx|=2(8)(₆₄1) = 1₄

4of a square inch.

(128)

Example

area of the square.

Solution.

64 =⇒ |dx|= 641

|∆A|

≈ |dA|=|2x| · |dx|=2(8)(₆₄1) = 1₄

4of a square inch.

(129)

Example

area of the square.

Solution.

64 =⇒ |dx|= 641

|∆A| ≈ |dA|

=|2x| · |dx|=2(8)(₆₄1) = 1₄

4of a square inch.

(130)

Example

area of the square.

Solution.

64 =⇒ |dx|= 641

|∆A| ≈ |dA|=|2x| · |dx|

=2(8)(₆₄1) = 1₄

4of a square inch.

(131)

Example

area of the square.

Solution.

64 =⇒ |dx|= 641

|∆A| ≈ |dA|=|2x| · |dx|=2(8)

(₆₄1) = 1₄

4of a square inch.

(132)

Example

area of the square.

Solution.

64 =⇒ |dx|= 641

|∆A| ≈ |dA|=|2x| · |dx|=2(8)(₆₄1)

= 1₄

4of a square inch.

(133)

Example

area of the square.

Solution.

64 =⇒ |dx|= 641

|∆A| ≈ |dA|=|2x| · |dx|=2(8)(₆₄1) = 1₄

4of a square inch.

(134)

Example

area of the square.

Solution.

64 =⇒ |dx|= 641

|∆A| ≈ |dA|=|2x| · |dx|=2(8)(₆₄1) = 1₄

4of a square inch.

(135)

Exercises

1 _Finddy_ify=tan(x2−√x)_.

2 _Approximate 4

√

16.08.

3 _Approximate(0.99)3+ (0.99)2− 1

(0.99)2.

4 _{Find the local linear approximation of}_f₍_x_{) =}_sin_x_at_x₌₀_{and use this to}

approximatesin(0.001).

5 _{Estimate the change in the surface area of a cube if its side is decreased}

from4in. to3.97in.

6 _{A hemispherical dome of radius 6 m. is to be coated with a layer of paint is}

0.001 m. thick. Estimate the amount of paint needed using differentials.

7 _{A cubic box is to be made to enclose}1000_cm.3_{of space. If the metal to be}

used in making the box is0.02cm. thick, approximate the amount of metal

needed using differentials.