Linear Approximations and Differentials Examples

Section 3.10: Linear Approximations and Differentials
Examples
Page 255 : 2. Find the linearization L(x) of the function at a.
f(x) = sinx, a = π/6
OK, the linearization of a function at a point is just a fancy way of asking for the tangent line
approximation to the function at that point.
We know y – y1 = m(x – x1) or in the notation of this problem f(x) – f(a) = f’(a)(x – a) or
f(x) ≈ f’(a) (x – a) + f(a) which in the notation of linearization becomes:
L(x) = f’(a) (x – a) + f(a).
So,
a=

6
= x1
()=2 =y
3
f’( ) = 2 = m
f(x) = sinx
f
1
1

6
f’(x) = cosx
L(x) =

6
3
2
x -  +
 6
1
2

L(x) = 
or you may prefer
3
2

x +
6
-
#12: Find the differential of each function:
a)
y=
s
1 + 2s
b)
y = e-u cosu
 3
12
x=
1
3
16.
y = cosx,
dx = -0.02
a)
Find the differential dy, and
b)
evaluate dy for the given values of x and dx.
26.
Use a linear approximation (or differentials) to estimate
1
.
4.002
I am going to use: ynew ≈ f’(xold) (xnew – xold) + yold (where I am mixing my y and f(x)
notations).
We are trying to estimate the value of the reciprocal of a number, so our y or f(x) will be
1
f(x) = . The number we are trying to estimate is the reciprocal of 4.002, so we will let our
x
xold be a number reasonably close to 4.002 whose reciprocal we know, so xold = 4 and thus
1
1
yold = f(xold) = . xnew is 4.002 and ynew =
, so we are trying to estimate ynew.
4
4.002
f(x) = y =
f’(x) =
1
,
x
xold = 4,
yold = f(xold) = f(4) =
dy
=
dx
at xold
1
= 0.25
4
dy
= f’(xold) = f’(4) =
dx
ynew ≈ f’(xold) (xnew – xold) + yold
ynew ≈ f’(4) (4.002 – 4) +
1
4
ynew ≈ (-0.0625) (0.002) + (0.25)
ynew ≈ 0.249875 and so
-1
1
(0.002) +
16
4
→
ynew ≈
→
ynew ≈ (-0.000125) + 0.25 →
1
≈ 0.249875
4.002
→
34.
of 0.2
a)
b)
The radius of a circular disk is given as 24 cm. with a maximum error in measurement
cm.
Use differentials to estimate the maximum error in the calculated area of the disk.
What is the relative error? What is the percentage error?
36.
Use differentials to estimate the amount of paint needed to apply a coat of paint 0.05
cm thick to a hemispherical dome with a diameter of 50 feet.