Linear approximation and differentials ( Section 2.9)

1. Linear approximation and differentials(Section 2.9) Alex Karassev

2. Linear approximation

3. Problem of computation • How do calculators and computers know that √5 ≈ 2.236 or sin 10o≈ 0.173648 ? • 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 isy = 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 • √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 y = L(x) = 2 + ¼ (x - 4) 2.25 √5 y = √x 2 a = 4 5

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 ≈ 0.1745 • Calculator gives:sin 10o ≈ 0.1736

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 f(x) ∆y f(a) x x a 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 x a 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 • Find the differential if a = 9 and x = 8

23. Solution of 1.

24. Solution of 2.

25. Application of differentials:estimation of errors • ProblemThe 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%