M53 Lec2.3.2 Local Linear Approx and Differentials.pdf

7/30/2019 M53 Lec2.3.2 Local Linear Approx and Differentials.pdf

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For today

1 Local Linear Approximation and Differentials

Institute of Mathematics (UP Diliman) Local Linear Approx and Differentials Mathematics 53 2 / 21

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For today

1 Local Linear Approximation and Differentials


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Local Linear Approximation and Differentials

Recall:


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Recall:

f (x0)


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Recall:

f (x0) = limx→x0

f (x) − f (x0)

x − x0


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Recall:

f (x0) = limx→x0

f (x) − f (x0)

x − x0

= lim∆x→0

∆y

∆x.


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Recall:

f (x0) = limx→x0

f (x) − f (x0)

x − x0

= lim∆x→0

∆y

∆x.

If ∆x is small enough,


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Recall:

f (x0) = limx→x0

f (x) − f (x0)

x − x0

= lim∆x→0

∆y

∆x.


=

⇒

∆y

∆x ≈f (x0)


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Recall:

f (x0) = limx→x0

f (x) − f (x0)

x − x0

= lim∆x→0

∆y

∆x.


=

⇒

∆y

∆x ≈f (x0)

=⇒ ∆y ≈ f (x0)∆x


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Recall:

f (x0) = limx→x0

f (x) − f (x0)

x − x0

= lim∆x→0

∆y

∆x.


=

⇒

∆y

∆x ≈f (x0)

=⇒ ∆y ≈ f (x0)∆xy = f (x)

P

x0

f (x0)

Q

x

f (x)

S

dx=∆x

= x0 + dx

∆y

R

dy


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Definitions


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DefinitionsLet the function y = f (x) be differentiable at x.


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1 The differential dx of the independent variable x denotes an

arbitrary increment of x.


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2 The differential dy of the dependent variable y associated with

x is given by dy = f (x)dx.


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If dx

= 0,


l d ff l

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If dx

= 0, then

dy = f (x)dx


L l Li A i i d Diff i l

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If dx

= 0, then

dy = f (x)dx =⇒ dy

dx= f (x).


L l Li A i i d Diff i l

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Remark


L l Li A i ti d Diff ti l

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Remark

The symboldy

dxmay be interpreted as:


L l Li A i ti d Diff ti l

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Remark

The symboldy


the derivative of y = f (x) with respect to x


L l Li A i ti d Diff ti ls

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Remark

The symboldy


the derivative of y = f (x) with respect to x

the quotient of the differential of y by the differential of x



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Theorem



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Theorem

Let u and v be differentiable functions of x and c a constant.



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Theorem

Let u and v be differentiable functions of x and c a constant.

1 d(c) = 0

2 d(xn) = nxn−1dx

3 d(cu) = cdu

4 d(uv) = udv − vdu

5 d(uv

) = vdu−udvv2

6 d(un

) = nun−

1du



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Example

Find dy if

y = x5

− x3

+ 2x.



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Example

Find dy if

y = x5

− x3

+ 2x.

Solution.



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Example

Find dy if

y = x5

− x3

+ 2x.

Solution.

dy



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Example

Find dy if

y = x5

− x3

+ 2x.

Solution.

dy = (5x4 − 3x2 + 2)



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Example

Find dy if

y = x5

− x3

+ 2x.

Solution.

dy = (5x4 − 3x2 + 2) dx



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Example

Find dy if

y =√ x3 + 3x2.



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pp

Example

Find dy if

y =√ x3 + 3x2.

Solution.



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pp

Example

Find dy if

y =√ x3 + 3x2.

Solution.

dy



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pp

Example

Find dy if

y =√ x3 + 3x2.

Solution.

dy =1

2√ x3 + 3x2



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pp

Example

Find dy if

y =√ x3 + 3x2.

Solution.

dy =1

2√ x3 + 3x2

· (3x2 + 6x)



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Example

Find dy if

y =√ x3 + 3x2.

Solution.

dy =1

2√ x3 + 3x2

· (3x2 + 6x) dx



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Example

Find dy if

xy2 = y + x.



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Example

Find dy if

xy2 = y + x.

Solution.



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Example

Find dy if

xy2 = y + x.

Solution.

x



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Example

Find dy if

xy2 = y + x.

Solution.

x(2ydy)



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Example

Find dy if

xy2 = y + x.

Solution.

x(2ydy) + y2



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Example

Find dy if

xy2 = y + x.

Solution.

x(2ydy) + y2dx



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Example

Find dy if

xy2 = y + x.

Solution.

x(2ydy) + y2dx = dy



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Example

Find dy if

xy2 = y + x.

Solution.

x(2ydy) + y2dx = dy + dx



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Example

Find dy if

xy2 = y + x.

Solution.

x(2ydy) + y2dx = dy + dx =⇒ dy =



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Example

Find dy if

xy2 = y + x.

Solution.

x(2ydy) + y2dx = dy + dx =⇒ dy =1 − y2

2xy

−1dx



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∆x ≈ 0



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∆x ≈ 0=

⇒



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∆x ≈ 0=

⇒∆y

≈f (x0)∆x



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∆x ≈ 0=

⇒∆y

≈f (x0)∆x

=⇒ f (x) ≈ f (x0) + f (x0)(x − x0)



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∆x ≈ 0=

⇒∆y

≈f (x0)∆x

=⇒ f (x) ≈ f (x0) + f (x0)(x − x0)

Equation of the

tangent line to the

graph of y = f (x) atthe point (x0, f (x0)):



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∆x ≈ 0=

⇒∆y

≈f (x0)∆x

=⇒ f (x) ≈ f (x0) + f (x0)(x − x0)

Equation of the

tangent line to the


y − f (x0) = f (x0)(x − x0)



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∆x ≈ 0=

⇒∆y

≈f (x0)∆x

=⇒ f (x) ≈ f (x0) + f (x0)(x − x0)

Equation of the

tangent line to the


y − f (x0) = f (x0)(x − x0)

∴



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∆x ≈ 0=

⇒∆y

≈f (x0)∆x

=⇒ f (x) ≈ f (x0) + f (x0)(x − x0)

Equation of the

tangent line to the


y − f (x0) = f (x0)(x − x0)

∴ y = f (x0) + f (x0)(x−x0)



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∆x ≈ 0=

⇒∆y

≈f (x0)∆x

=⇒ f (x) ≈ f (x0) + f (x0)(x − x0)

Equation of the

tangent line to the


y − f (x0) = f (x0)(x − x0)

∴ y = f (x0) + f (x0)(x−x0)

x0 x

f (x)

f (x) + f

(x0)(x−

x0)

y = f (x)



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Remarks



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Remarks

L(x) = f (x0) + f (x0)(x − x0) is the local linear approximation of

f (x) at x0.



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Remarks


f (x) at x0.

(The tangent line to the graph of f at x0 approximates the graph of f when

x is near x0.)



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Remarks


f (x) at x0.


x is near x0.)

It can be shown that the local linear approximation is the “best” linear

approximation of f near x0.



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Remarks


f (x) at x0.


x is near x0.)



If dx = ∆x = x − x0,



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Remarks


f (x) at x0.


x is near x0.)



If dx = ∆x = x − x0, then x = x0 + dx.



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Remarks


f (x) at x0.


x is near x0.)




Since f (x) ≈ f (x0) + f

(x0)(x − x0), we have



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Remarks


f (x) at x0.


x is near x0.)




Since f (x) ≈ f (x0) + f

(x0)(x − x0), we have

f (x0 + dx) ≈ f (x0) + f (x0)dx



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Remarks



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Remarks

dx = ∆x ≈ 0



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Remarks

dx = ∆x ≈ 0=⇒



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Remarks

dx = ∆x ≈ 0=⇒ dy = f (x)dx



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Remarks

dx = ∆x ≈ 0=⇒ dy = f (x)dx = f (x)∆x



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Remarks

dx = ∆x ≈ 0=⇒ dy = f (x)dx = f (x)∆x ≈ ∆y



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Remarks

dx = ∆x ≈ 0=⇒ dy = f (x)dx = f (x)∆x ≈ ∆y

∴ dy ≈ ∆y



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Remarks

dx = ∆x ≈ 0=⇒ dy = f (x)dx = f (x)∆x ≈ ∆y

∴ dy ≈ ∆y

dy is easier to compute than ∆y



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Remarks

dx = ∆x ≈ 0=⇒ dy = f (x)dx = f (x)∆x ≈ ∆y

∴ dy ≈ ∆y


∴ dy is used to approximate ∆y when dx

≈0



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Remarks

dx = ∆x ≈ 0=⇒ dy = f (x)dx = f (x)∆x ≈ ∆y

∴ dy ≈ ∆y


∴ dy is used to approximate ∆y when dx

≈0



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Example

Find the local linear approximation of f (x) = 3√ x at x0 = 8.



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Example


Solution.



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Example


Solution.

We have f (x) =1

33√ x2

.



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Example


Solution.

We have f (x) =1

33√ x2

.

L(x) = f (x0) + f (x0)(x − x0)



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Example


Solution.

We have f (x) =1

33√ x2

.

L(x) = f (x0) + f (x0)(x − x0)

∴ At x0 = 8:



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Example


Solution.

We have f (x) =1

33√ x2

.

L(x) = f (x0) + f (x0)(x − x0)

∴ At x0 = 8:

L(x)



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Example


Solution.

We have f (x) =1

33√ x2

.

L(x) = f (x0) + f (x0)(x − x0)

∴ At x0 = 8:

L(x) = f (8)



l

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Example


Solution.

We have f (x) =1

33√ x2

.

L(x) = f (x0) + f (x0)(x − x0)

∴ At x0 = 8:

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



E l

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Example


Solution.

We have f (x) =1

33√ x2

.

L(x) = f (x0) + f (x0)(x − x0)

∴ At x0 = 8:

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

−8)



E l

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Example

Find the local linear approximation of f (x) =3√ x at x0 = 8.

Solution.

We have f (x) =1

33√ x2

.

L(x) = f (x0) + f (x0)(x − x0)

∴ At x0 = 8:

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

−8)

= 2



E l

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Example


Solution.

We have f (x) =1

33√ x2

.

L(x) = f (x0) + f (x0)(x − x0)

∴ At x0 = 8:

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

−8)

= 2 + 112



E l

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Example


Solution.

We have f (x) =1

33√ x2

.

L(x) = f (x0) + f (x0)(x − x0)

∴ At x0 = 8:

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

−8)

= 2 + 112

(x − 8).



Example

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Approximate 3√

27.027.



Example

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Approximate 3√

27.027.

Solution.



Example√

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Approximate 3√

27.027.

Solution.

Let f (x) = 3√ x.



Example3√

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Approximate 3√

27.027.

Solution.

Let f (x) = 3√ x. Then f (x) =

1

33√ x2

.



Example3√

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Approximate 3√

27.027.

Solution.


1

33√ x2

.

f (x0 + dx)

≈f (x0) + f (x0)dx



Example

A i 3√

7 7

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Approximate 3√

27.027.

Solution.


1

33√ x2

.

f (x0 + dx)

≈f (x0) + f (x0)dx

Thus,

3√

27.027



Example

A i 3√

27 027

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Approximate 3√

27.027.

Solution.


1

33√ x2

.

f (x0 + dx)

≈f (x0) + f (x0)dx

Thus,

3√

27.027 = f (27 + 0.027)



Example

A i t 3√

27 027

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Approximate 3√

27.027.

Solution.


1

33√ x2

.

f (x0 + dx)

≈f (x0) + f (x0)dx

Thus,

3√

27.027 = f (27 + 0.027)

≈f (27) + f (27)

·(0.027)



Example

A i t 3√

27 027

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Approximate 3√

27.027.

Solution.


1

33√ x2

.

f (x0 + dx)

≈f (x0) + f (x0)dx

Thus,

3√

27.027 = f (27 + 0.027)

≈f (27) + f (27)

·(0.027)

= 3



Example

Approximate 3√

27 027

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Approximate√

27.027.

Solution.


1

33√ x2

.

f (x0 + dx)

≈f (x0) + f (x0)dx

Thus,

3√

27.027 = f (27 + 0.027)

≈f (27) + f (27)

·(0.027)

= 3 + 13 · 9



Example

Approximate 3√

27 027

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Approximate√

27.027.

Solution.


1

33√ x2

.

f (x0 + dx)

≈f (x0) + f (x0)dx

Thus,

3√

27.027 = f (27 + 0.027)

≈f (27) + f (27)

·(0.027)

= 3 + 13 · 9

· (0.027)



Example

Approximate 3√

27 027

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Approximate√

27.027.

Solution.


1

33√ x2

.

f (x0 + dx)

≈f (x0) + f (x0)dx

Thus,

3√

27.027 = f (27 + 0.027)

≈f (27) + f (27)

·(0.027)

= 3 + 13 · 9

· (0.027)

= 3.01.

Instit te of Mathematics (UP Diliman) Local Linea A o and Diffe entials Mathematics 53 15 / 21


Example

A i√

1

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Approximate√

15.96.

I tit t f M th ti (UP Dili ) L l Li A d Diff ti l M th ti 53 16 / 21


Example

A i√

15 96

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Approximate√

15.96.

Solution.



Example

A i t√

15 96

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Approximate√

15.96.

Solution.

Let f (x) =√ x.



Example

A i t√

15 96

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Approximate√

15.96.

Solution.

Let f (x) =√ x. Then f (x) =

1

2√ x

.



Example

Approximate√

15 96

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Approximate√

15.96.

Solution.


1

2√ x

.

Thus,

√ 15.96



Example

Approximate√

15 96

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Approximate√

15.96.

Solution.


1

2√ x

.

Thus,

√ 15.96 = f (16 − 0.04)



Example

Approximate√

15 96

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Approximate√

15.96.

Solution.


1

2√ x

.

Thus,

√ 15.96 = f (16 − 0.04)

≈ f (16) + f (16) · (−0.04)



Example

Approximate√

15 96

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Approximate√

15.96.

Solution.


1

2√ x

.

Thus,

√ 15.96 = f (16 − 0.04)

≈ f (16) + f (16) · (−0.04)

= 4



Example

Approximate√

15 96

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Approximate√

15.96.

Solution.


1

2√ x

.

Thus,

√ 15.96 = f (16 − 0.04)

≈ f (16) + f (16) · (−0.04)

= 4 +1

2

·4



Example

Approximate√

15.96

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Approximate√

15.96.

Solution.


1

2√ x

.

Thus,

√ 15.96 = f (16 − 0.04)

≈ f (16) + f (16) · (−0.04)

= 4 +1

2

·4

· (−0.04)



Example

Approximate√

15.96.

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Approximate√

15.96.

Solution.


1

2√ x

.

Thus,

√ 15.96 = f (16 − 0.04)

≈ f (16) + f (16) · (−0.04)

= 4 +1

2

·4

· (−0.04)

= 4 − 0.005



Example

Approximate√

15.96.

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pp√

5 96

Solution.


1

2√ x

.

Thus,

√ 15.96 = f (16 − 0.04)

≈ f (16) + f (16) · (−0.04)

= 4 +1

2

·4

· (−0.04)

= 4 − 0.005= 3.995.



Example1

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A ball 5 in in diameter is to be covered by a rubber material which is

1

16 inthick. Use differentials to estimate the volume of the rubber material that

will be used.



Example1

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A ball 5 in in diameter is to be covered by a rubber material which is

1

16 inthick. Use differentials to estimate the volume of the rubber material that

will be used.

Solution.



Example1

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A ball 5 in in diameter is to be covered by a rubber material which is 16 inthick. Use differentials to estimate the volume of the rubber material that

will be used.

Solution.

Volume of the ball with radius r:



Example1

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will be used.

Solution.

Volume of the ball with radius r: V (r) =

4

3πr

3



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will be used.

Solution.

Volume of the ball with radius r: V (r

) =

4

3πr3

=⇒dV



Example

A ball 5 in in diameter is to be covered by a rubber material which is 1 in

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A ball 5 in in diameter is to be covered by a rubber material which is16

in

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

will be used.

Solution.


) =

4

3πr3

=⇒dV

= 4πr2 dr



Example


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in


will be used.

Solution.


) =

4

3πr3

=⇒dV

= 4πr2 dr

Volume of rubber material



Example


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in


will be used.

Solution.

Volume of the ball with radius r: V (r) =4

3πr3 =

⇒dV = 4πr2 dr

Volume of rubber material = V ( 52 + 1

16 ) − V ( 52 )



Example


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y16thick. Use differentials to estimate the volume of the rubber material that

will be used.

Solution.


3πr3 =

⇒dV = 4πr2 dr


16 ) − V ( 52 )

= ∆V



Example


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will be used.

Solution.


3πr3 =

⇒dV = 4πr2 dr


16 ) − V ( 52 )

= ∆V dr = 116



Example


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will be used.

Solution.


3πr3 =

⇒dV = 4πr2 dr


16 ) − V ( 52 )

= ∆V dr = 116

≈ dV



Example


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16thick. Use differentials to estimate the volume of the rubber material that

will be used.

Solution.


3πr3 =

⇒dV = 4πr2 dr


16 ) − V ( 52 )

= ∆V dr = 116

≈ dV

= V ( 5

2)·

( 1

16)



Example


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will be used.

Solution.


3πr3 =

⇒dV = 4πr2 dr


16 ) − V ( 52 )

= ∆V dr = 116

≈ dV

= V ( 5

2)·

( 1

16)

= 4π( 52 )2 · ( 1

16 )



Example


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will be used.

Solution.


3πr3 =

⇒dV = 4πr2 dr


16 ) − V ( 52 )

= ∆V dr = 116

≈ dV

= V ( 5

2)·

( 1

16)

= 4π( 52 )2 · ( 1

16 )= 25π

16 in3



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

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volume of the insulation.



Example



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Solution.



Example



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Solution.

Volume of the rod:



Example



f

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Solution.

Volume of the rod: V = πr2h



Example



l f h i l i

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Solution.

Volume of the rod: V = πr2h = 15πr2



Example



l f h i l i

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Solution.

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



Example



l f th i l ti

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Solution.

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



Example



l f th i l ti

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Solution.


Volume of the insulation



Example



volume of the insulation

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Solution.


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



Example




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Solution.


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



Example




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Solution.


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



Example




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Solution.



≈ dV



Example




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Solution.



≈ dV

= V (4) · (0.001)



Example




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Solution.



≈ dV

= V (4) · (0.001)

= 30π (4) · (0.001)



Example




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Solution.



≈ dV

= V (4) · (0.001)

= 30π (4) · (0.001)

= 0.12π cm3.



Example

Suppose that the side of a square is measured with a ruler to be 8 inches

with a measurement error of at most

±1

64

of an inch. Estimate the error in

the computed area of the square.

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p q



Example



±1

64 of an inch. Estimate the error in


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p q

Solution.



Example



±1



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p q

Solution.

Area of square with side x:



Example



±1



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p q

Solution.

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



Example



±1



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Solution.

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


Local Linear Approximation and DifferentialsExample



±1



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Solution.

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


Local Linear Approximation and DifferentialsExample



±1



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Solution.

Area of square with side x: A(x) = x2 =⇒ dA = 2x dxMeasurement error of at most

±1

64



Example



±1



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Solution.


±1

64 =

⇒ |dx

|= 1

64



Example



±1



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Solution.


±1

64 =

⇒ |dx

|= 1

64

Error in the computed area



Example



±1



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Solution.


±1

64 =

⇒ |dx

|= 1

64

Error in the computed area = ∆A



Example



±1



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Solution.


±1

64 =

⇒ |dx

|= 1

64


|∆A|



Example



±1



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Solution.


±1

64 =

⇒ |dx

|= 1

64


|∆A| ≈ |dA|



Example



±1



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Solution.


±1

64 =

⇒ |dx

|= 1

64


|∆A| ≈ |dA|= 2x |dx|



Example



±1



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Solution.


±1

64 =

⇒ |dx

|= 1

64


|∆A| ≈ |dA|= 2x |dx| x = 8



Example



±1



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Solution.


±1

64 =

⇒ |dx

|= 1

64


|∆A| ≈ |dA|= 2x |dx| x = 8= 2(8)



Example



±1



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Solution.


±1

64 =

⇒ |dx

|= 1

64


|∆A| ≈ |dA|= 2x |dx| x = 8= 2(8)( 1

64 )



Example



±1



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Solution.


±1

64 =

⇒ |dx

|= 1

64


|∆A| ≈ |dA|= 2x |dx| x = 8= 2(8)( 1

64 )

= 14



Example



±1



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Solution.


±1

64 =

⇒ |dx

|= 1

64


|∆A| ≈ |dA|= 2x |dx| x = 8= 2(8)( 1

64 )

= 14

∴ The propagated error in the computed area is at most ±14 of a square

inch.Institute of Mathematics (UP Diliman) Local Linear Approx and Differentials Mathematics 53 19 / 21

Exercise

1 Find dydx if

i ( ) 2 2 3

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sin(xy) = xy2 − 2x3.

2 Determine D4x [ cos(4x) ].

3 Approximate 3√

8.03.


* * * The End * * *

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Next Meeting:

Rates of Change

Rectilinear Motion


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Documents

M53 Lec2.3.2 Local Linear Approx and Differentials.pdf