Linear approximation and differentials (Section 2.9) Alex Karassev

Linear approximation

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

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

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

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

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

Linearization

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

Approximation of √5

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

Example Find approximate value of sin 10o

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

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

Differentials

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

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

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

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

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

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

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

Solution of 1.

Solution of 2.

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.

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%