Linearization

When we have a complicated function that can be impossible to make a perfactly accurate value, we use the linearization that are based on its tangent lines.
Consider the tangent to the curve y = f (x) that lies closed to the curve near the point of tangency. For a small interval to either side of the point of tangency, the y-values along the tangent line give good approximations to the y-values of the curve.
For example, we can see this phenomenon by zooming in on the graph of y = x2 + 1⁄2.
fig 9.4-1

fig 9.4-2

Curve y = x2 + 1⁄2 and its tangent y = x + 1⁄4 are very close near (1⁄2, 1⁄4)

The equation of the tangent to the curve f (x) at x = 0, where f is differentiable and passes through the point (a, f (a)) is
y = f (a) + f ' (a)(x - a).
This tangent line is the graph of the linear function
L (x) = f (a) + f ' (a)(x - a).
fig 9.2-2

For as long as L(x) is close to the graph of f, the function l (x) gives a good approximation to f (x).

Definition
If f is differentiable at x = a, then the approximation function
L (x) = f (a) + f ' (a)(x - a).
is the linearization of f at a. The approximation
f (x) ≈ L (x)
of f by L is the standard linear approximation of f at a. The point at x = a is the center of the approximation.
The difference of f (x) and L (x is the error of the linearization.

Example 17.
Find the linearization of f (x) = x2 + 2x at x = 2. Compare the approximate value and true value at x = 2.5, x = 2.05, and x = 2.005.
Solution
f (x) = x2 + 2x,
f ' (x) = 2x + 2
Since f (2) = 8 and f ' (2) = 6, the linearization is
L (x) = f (2) + f ' (2)(x - 2)
= 8 + 6(x - 2) = 6x - 4.
That is, x2 + 2x ≈ 6x - 4.
At x = 2.5, the tre value gives x2 + 2x
= (2.5)2 + 2(2.5) = 11.25 and
the linearization gives 6x - 4 = 6(2.05) - 4 = 8.3.
Therefore the approximate value differs from the true value by less than 0.25 × 10-2.
At x = 2.005, the true value gives
x2 + 2x = (2.005)2 + 2(2.005) = 8.030025 and
the linearization gives 6x - 4 = 6(2.005) - 4 = 8.03.
Therefore, the approximate value differs from the true value by less than 0.25 × 10-4.

Note:   From the above example, we see that how accurate the approximation is for some value of x near 2. It turns out that, if the variable x moves a small increment, then the true value and approximate value become close.


Example 18.
If the radius of a circle increases from r = 5 cm to 5.01 cm, find the approximate increase in the area. Estimate the area of the enlarge circle and find the error.
Solution
r=5 δr = 0.01
Let A be the area of the circle of radius r.
A(r) = πr2,
A ' (r) = 2πr.
Since A (5) = 25π and A ' (5) = 10π, the linearization is
L (r) = A (5) + A ' (5)(r - 5) = 10πr - 25π.
That is, πr2 ≈ 10πr - 25π.
At r = 5.01, L (r) = L (5.01)
= 10π (5.01) - 25π = 25.1π
The area of circle of radius 5.01 cm is approximately 25.1π cm2. Hence the approxinate increase is 0.1π cm2.
The true area is
A (5.01)= π(5.01)2 = 25.1001π cm2
Therefore
Error = True area - Approximate area
= 0.0001π cm2.

Example 19.
Given that f (x) = x1⁄2 , determine the approximate value for √ 101   by using approximation.
Solution
f (x) = x1⁄2 .
When x = 100, f (100) = 1001⁄2 = 10, we have to approximate √ 101  .
$$ f'(x) = \frac{1}{2\sqrt{x}} $$ $$ f'(100) = \frac{1}{20}. $$ The linearization is
L (x) = f (100) + f ' (100)(x - 100)
= x⁄20 + 5.
That is, √ x   ≈ x⁄20 + 5.
At x = 101, L (x) = L (101) = 10.05.
Hence,
√ 101   ≈ 10.05.

Exercise 9.4
1. Find the linearization of
(a) f (x) = √ 1 + x   at x = 3. Compare the approximate value and true value at x = 3.2, x = 3.02 and x = 3.002.
(b) f (x) = x ⁄ x + 1 at x = 1.
Compare the approximate value and true value at x = 1, x = 1.01, x = 1.001.

Solution
(a) f (x) = √ 1 + x  
$$f'(x) = \frac{1}{2} (1 + x)^{- \frac{1}{2}}$$ $$= \frac{1}{2 \sqrt{1 + x}}$$ $$f(3) = \sqrt{1+3} = \sqrt{4} = 2 $$ $$f'(3) = \frac{1}{2 \times 2} = \frac{1}{4}$$ $$L(x) = f(3) + f(3) (x - 3)$$ $$= 2 + \frac{1}{4}(x-3)$$ $$= 2+\frac{x}{4} - \frac{3}{4} $$ $$= \frac{5}{4} + \frac{x}{4} $$ $$= \frac{1}{4}(x+5)$$ $$f(x) \approx L(x)$$ $$\sqrt{1+x} = \frac{1}{4}(x+5) $$ $$x=3.2, \, \, \, f(3.2) = \sqrt{1+3.2}$$ $$= \sqrt{4.2} = 2.049390 = \text{True\, value}$$ $$L(3.2) = \frac{1}{4}(8.2) = 2.05 = \text{Approximate \, value}$$ $$\text{Error} = 2.049390 - 2.050000 = -0.000610$$ Therefore, the true value differs from the approximate value by less than 0.000610.
$$x = 3.02, $$ $$f(3.02) = \sqrt{1+3.02} $$ $$= \sqrt{4.02} = 2.0049390 = \text{True \ value}$$ $$L(3.02) = \frac{1}{4} (8.02)$$ $$ = 2.005 = \text{Approximate \ value}$$

Error = 2.0049390 - 2.005 = -0.000006
Therefore, the true value differs from the approximate value by less than 0.000006.

$$x = 3.002,$$ $$f(3.002) = \sqrt{1 + 3.002}$$ $$= \sqrt{4.002} = 2.0005 = \text{True \ value}$$ $$L(3.002) = \frac{1}{4} (8.002) $$ $$= 2.0005 = \text{Approximate \ value} $$ $$\text{Error} \ = 2.0005 - 2.0005 = 0$$
(b) f (x) = x ⁄ x + 1
$$f'(x) = \frac{(x+1)1 - x \sdot 1}{(x + 1)^2}$$ $$= \frac{x + 1 - x}{(x + 1)^2} $$ $$= \frac{1}{(x + 1)^2} $$
At x = 1,

$$f(1) = \frac{1}{1+1} = \frac{1}{2},$$ $$f'(1) = \frac{1}{(1+1)^2} = \frac{1}{4} $$ $$L(x) = f(1) + f'(1) (x - 1)$$ $$= \frac{1}{2} + \frac{1}{4} (x - 1) $$ $$= \frac{2}{4} + \frac{1}{4} x - \frac{1}{4}$$ $$= \frac{1}{4} + \frac{1}{4} x$$ $$= \frac{1}{4} (x + 1)$$ $$\frac{x}{x+1} \approx \frac{1}{4} (x + 1) $$ $$L(1) = \frac{1}{4} (1 + 1) = \frac{1}{2}$$ $$\text{At} \ x = 1, $$ $$\text{Error} \ = f(1) - L(1) $$ $$= \frac{1}{2} - \frac{1}{2} = 0 $$ $$\text{At} \ x = 1.01, \, \, \, f(1.01) = \frac{1.01}{2.01} $$ $$= 0.502488 $$ $$L(1.01) = \frac{1}{4} (2.01) = 0.502500 $$ $$\text{Error}\ = 0.502488 - 0.5-2500 $$ $$= -0.000012 $$ Therefore, the true value differs from the approximate value by less than 0.000012.
$$\text{at} \ x = 1.001, \ \ \ f(1.001) = \frac{1.001}{2.001} $$ $$= 0.500250 $$ $$L(1.001) = \frac{2.001}{4} = 0.500250 $$ $$\text{Error}\ = 0.500250 - 0.500250 = 0. $$
2. Show that the linearization of f (x) = (1 + x)k at x = 0 is L (x) = 1 + kx.
Solution
f (x) = (1 + x)k
f ' (x) = k (1 + x)k-1
At x = 0,
f (0) = (1 + 0)k = 1,
' (0) = (1 + 0)k-1 = k
L (x) = f (0) + f ' (0) (x - 0)
= 1 + kx.

3. Find the approximate change in the volume of a sphere when its radius decrease from 5 cm to 4.97 cm . Estimate the area of the contract circle and find the error.
Solution
Let v be the volume of the sphere with radius r cm.
$$V(r) = \frac{4}{3} \ \pi r^3 $$ $$V'(r) = 4 \pi r^2 $$ $$\text{At}\ r = 5, $$ $$V(5) = \frac{4}{3} \times \pi \times 5^3$$ $$= \frac{4}{3} \times \pi \times 125 $$ $$= \frac{500 \ \pi }{3} \ \ cm^3 $$ $$V'(5) = 4 \pi \times 5^2 $$ $$= 100 \ \pi \ \ cm^3 $$ $$L(r) = V(5) + V'(5) (r - 5) $$ $$= \frac{500 \ \pi}{3} + 100 \ \pi (r - 5) $$ $$= \frac{500 \ \pi}{3} + 100 \ \pi r - 500 \ \pi $$ $$= 100 \ \pi r - \frac{1000 \ \pi}{3} $$ $$V(r) \approx L(r) $$ $$\frac{4}{3} \ \pi \ r^3 \approx 100 \ \pi r - \frac{1000 \ \pi }{3} $$ $$\text{At} \ r = 4.97, $$ $$V(4.97) = \frac{4}{3} \times \pi \times (4.97)^3 $$ $$= 514.230433 \ \ cm^3 $$ $$L(4.97) = 100 \pi \times 4.97 - \frac{1000 \ \pi}{3} $$ $$= 514.173998 \ \ cm^3 $$
Error = V (4.97) - L (4.97)
= 514.230433 - 514.173998
= 0.056435 cm3.


4. If y = 4 √ x   + 3x2, find the approximate change in y when x changes from 9 to 8.98.
Solution
$$y(x) = 4\sqrt{x} + 3x^2 $$ $$y'(x) = 4 \sdot \frac{1}{2} x^{-\frac{1}{2}} $$ $$= \frac{2}{\sqrt{x}} + 6$$ $$\text{At}\ x = 9, $$ $$y(9) = 4\times 3 + 3(9^2) $$ $$=12 + 243 = 255 $$ $$y'(9) = \frac{2}{3} + 6(9) $$ $$= \frac{2}{3} + \frac{2 \times 54}{3} $$ $$= \frac{164}{3} $$ $$L(x) = y(9) + y'(9) (x-9) $$ $$= 255 + \frac{164}{3} (x-9) $$ $$= 255 + \frac{164}{3} x - 164(3) $$ $$= \frac{164\ x}{3} - 237 $$ $$y(x) \approx L(x) $$ $$\text{At}\ x = 8.98, $$ $$y(8.98) = 4\sqrt(8.98) + 3(8.98)^2$$ $$= 253.907859 $$ $$= L(8.98) = \frac{164 \times 8.98}{3} - 273 $$ $$= \text{Error}\ = y(8.98)- L(8.98) = 0.001192. $$
5. If y = 3√ 72 + x2   find the approximate change in y when
(i) x increases from 3 to 3.01,
(ii) x decreases from 3 to 2.98.
Find their errors.
Solution
$$\text{Let}\ y(x) = 3\sqrt{72 + x^2} $$ $$y'(x) = 3\frac{1}{2} (72 + x^2)^{-\frac{1}{2}} \sdot 2x $$ $$= \frac{3x}{\sqrt{72 + x^2}} $$ $$\text{At}\ x = 3, $$ $$y(3) = 3\sqrt{72 + 3^2}$$ $$= 3\sqrt{72 + 9} = 3 \times 9 = 27$$ $$y'(3) = \frac{3(3)}{\sqrt{72 + 3^2}}$$ $$= \frac{9}{\sqrt{81}} = \frac{9}{9} = 1$$ $$L(x) = y(3) + y'(3) (x - 3) $$ $$= 27 + x - 3 $$ $$= 24 + x$$ $$y(x) \approx L(x) $$ $$3\sqrt{72 + x^2 \approx 24 + x}$$ $$\text{(i)\ \ \ At \ x = 3.01,} $$ $$y(3.01) = 3\sqrt{72 + (3.01)^2} = 27.010015 $$ $$L(3.01) = 24 + 3.01 = 27.01 $$ $$\text{Error}\ = y(3.01) - L(3.01) $$ $$= 27.010015 - 27.01 $$ $$= 0.000015 $$ $$\text{(ii)}\ \ \ x = 2.98, $$ $$y(2.98) = 3 \sqrt{72 + (2.98)^2}$$ $$= 26.980059 $$ $$L(2.98) = 24 + 2.98 = 26.98 $$ $$\text{Error}\ = 26.980059 - 26.98 = 0.000059. $$
6. Use linearization to apprximate the following values.
$$ \text{(i)}\ \sqrt{80} $$ $$ \text{ii} \ \sqrt[3]{65} $$ $$ \text(iii) \ \frac{1}{\sqrt{1.22}} $$
Solution
$$\text{(i) \ \ Let}\ \ f(x) = \sqrt{x} $$ $$f'(x) = \frac{1}{2} \ x^{-\frac{1}{2}} $$ $$= \frac{1}{2\sqrt{x}} $$ $$x = 81, $$ $$f(81) = \sqrt{81} = 9 $$ $$f'(81) = \frac{1}{2\sqrt{81}} $$ $$= \frac{1}{2 \times 9} = \frac{1}{18} $$ $$L(x) = f(81) + f'(81) (x - 81) $$ $$ = 9 + \frac{1}{18} (x - 8) $$ $$= 9 + \frac{1}{18} x - \frac{1}{18} \sdot 81 $$ $$= 9 + \frac{1}{18} x - \frac{9}{2} $$ $$= \frac{1}{18} x + \frac{9}{2}$$ $$f(x) \approx L(x) $$ $$L(x) = \frac{1}{18} x + \frac{9}{2}$$ $$L(80) = \frac{1}{18} (80) + \frac{9}{2} $$ $$= 8.9444$$
$$\text{(ii) \ \ \ Let} \ \ f(x) = \sqrt[3]{x} = x^{\frac{1}{3}} $$ $$f'(x) = \frac{1}{3} x^{-\frac{2}{3}} $$ $$= \frac{1}{3 \sqrt[3]{x^2}} $$ $$x = 64,$$ $$f(64) = \sqrt[3]{64} $$ $$= \sqrt[3]{2^6} $$ $$= 2^{6 \times \frac{1}{3}}$$ $$= 4$$ $$f'(64) = \frac{1}{3 \sqrt[3]{64^2}} $$ $$= \frac{1}{3 \sqrt[3]{2^{12}}}$$ $$= \frac{1}{3 \sdot 2^{12 \times \frac{1}{3}} }$$ $$= \frac{1}{48} $$ $$L(x) = f(64) + f'(64) (x - 64) $$ $$= 4 \times \frac{1}{48} (x - 64) $$ $$= 4 + \frac{x}{48} - \frac{4}{3} $$ $$= \frac{8}{3} + \frac{x}{48} $$ $$x = 65, $$ $$L(65) = \frac{8}{3} + \frac{65}{48} $$ $$= 2.6666666 + 1.3541666 $$ $$= 4.0208332$$ $$\sqrt[3]{65} \approx 4.0208332 $$
$$\text{(iii) \ \ \ Let} \ \ \ f(x) = \frac{1}{\sqrt{x}} = \frac{1}{x^{\frac{1}{2}} } = x^{- \frac{1}{2}} $$ $$f'(x) = - \frac{1}{2} x^{- \frac{3}{2}} $$ $$= - \frac{1}{2 \sqrt{x^3}}$$ $$x = 1.21, $$ $$f(1.21) = \frac{1}{\sqrt{1.21}} = \frac{10}{11} $$ $$f'(1.21) = - \frac{1}{2 \sqrt{1.21^3}}$$ $$= - \frac{1}{2 \times 1.331} \sdot \frac{1000}{1000} $$ $$= - \frac{500}{1331} $$ $$L(x) = f(1.21) + f'(1.21) (x - 1.21)$$ $$= \frac{10}{11} - \frac{500}{1331} (x - 1.21) $$ $$L(1.22) = \frac{10}{11} - \frac{500}{1331} (1.22 - 1.21) $$ $$ = \frac{10}{11} - \frac{500}{1331} (0.01) $$ $$= \frac{10}{11} - \frac{5}{1331} $$ $$= \frac{1210 - 5}{1331} $$ $$= \frac{1205}{1331} = 0.905334 $$ $$\frac{1}{\sqrt{1.22}} \approx 0.905334 $$