1. Linear approximation and differentials (Section 3.9)
Alex Karassev

2. Linear approximation

3. Problem of computation
How do calculators and computers know that √5 ≈ or sin 10o ≈ ?
They use various methods of approximation, one of which is Taylor polynomial approximation
A simplest case of Taylor polynomial approximation is linear approximation or linearization

4. Linear approximation
Equation of tangent line to y=f(x) at a is y = f(a) + f′(a) (x - a) y
y = f(x)
f(a) x a

5. Linear approximation
If x is near a, we have: f(x) ≈ f(a) + f′(a) (x - a) y
f(a) + f′(a) (x - a)
y = f(x)
f(x)
f(a) x a x

6. Linear approximation
Function L(x) = f(a) + f′(a) (x - a) is called linear approximation (or linearization) of f(x) at a y
y = L(x)
L(x)
y = f(x)
f(x)
f(a) x a x

7. Example Find linearization of f(x) = √x at a
Use it to find approximate value of √5

8. Linearization

9. Approximation of √5 Find a such that Use linearization at a
√a is easy to compute
a is close to 5
Use linearization at a
Take a = 4 and compute linear approximation

10. Approximation of √5

11. Approximation of √5 y = L(x) = 2 + ¼ (x - 4) y y = √x a = 4 5 √5 2.25

12. Example
Find approximate value of sin 10o

13. Example We measure x in radians So, 10o = 10 (π/180) = π/18 radians
Consider f(x) = sin x
Find a such that
sin(a) is easy to compute
a is near π/18
Take a = 0 and compute linear approximation

14. Solution f(x) ≈ f(a) + f′(a) (x - a) = f(0) + f′(0) (x - 0)
f(x) = sin x, f′(x) = (sin x) ′ = cos x
Therefore we obtain: sin x ≈ sin(0) + cos(0) (x - 0) = 0 +1(x – 0) = x
Thus sin x ≈ x (when x is near 0)
For x = π/18 we obtain: sin 10o = sin (π/18) ≈ π/18 ≈
Calculator gives: sin 10o ≈

15. Differentials

16. Differentials Compare f(x) and f(a) Change in y: ∆y = f(x) – f(a)
f(x) ≈ f(a) + f′(a) (x - a)
Therefore ∆y = f(x) – f(a) ≈ (f(a) + f′(a) (x - a)) – f(a) = f′(a) (x - a)
Let x – a = ∆x = dx
Then ∆y ≈ f′(a) (x - a) = f′(a) dx

17. Differentials Definition
dy = f′(a) dx is called the differential of function x at a
Thus, ∆y ≈ dy
Note: dx = x – a

18. Differentials dy = f′(a) dx y y = L(x) L(x) y = f(x) dy ∆y x a x
dx = x – a

19. Differential as a linear function
dy = f′(a) dx
For fixed a, dy is a linear function of dx y dy
y = L(x)
y = f(x) dy
dy = f′(a) dx dx dx x a x
dx = x – a

20. Differential at arbitrary point
We can let a vary
Then, differential of function f at any number x is dy = f′(x) dx
For each x, we obtain a linear function with slope f′(x)
df(x) = f′(x) dx

21. Differentials and linear approximation
dy = f′(a) dx
∆y = f(x) – f(a)
Therefore f(x) = f(a) + ∆y
∆y ≈ dy
Thus f(x) ≈ f(a) + dy

22. Example Let f(x) = √x Find the differential if a = 4 and x = 5

23. Solution of 1.

24. Solution of 2.

25. Application of differentials: estimation of errors
Problem The edge of a cube was found to be 30 cm with a possible error in measurement of 0.1 cm. Estimate the maximum possible error in computing the volume of the cube.

26. Solution
Suppose that the exact length of the edge is x and the "ideal" value is a = 30 cm.
Then the volume of the cube is V(x) = x3
Possible error is the absolute value of the difference between the "ideal" volume and "real" volume: ∆V = V(x) – V(a)
∆V ≈ dV = V'(a) dx
dx = ± 0.1
V'(x) = (x3)' = 3x2
∆V ≈ dV = V'(a) dx =3a2 dx = ± 3(30)2 (0.1) = ± 270 cm3
So error = |∆V| ≈ 270 cm3
Relative error = |∆V| / V ≈ 270 / 303 = 0.01 = 1%