Use a linear approximation (or differentials) to estimate the given number.

Question

√100.8

Answer 1

The linear approximation uses the tangent line to curve at a point. The tangent line at x = a can approximate every point close to x = a.

From Algebra, a point-slope form of a line:

y - y₁ = m(x - x₁)

y = y₁ + m(x - x₁)

Let (x₁, y₁) = (a, f(a)) and m = f'(a) [slope at x = a]

Choose a = 100 (perfect square)

We have,

y(x) = f(a) + f'(a)(x - a)

y = f(x) = √x

f'(x) = (1/2)x^-1/2

y(100.8) = f(100) + (1/2)(100)^-1/2[100.8 - 100]

= √100 + (1/20)(0.8)

= 10 + 0.4

= 10.04

10.04 is an approximation for √100.8.

√100.8 = 10.03992032 (Calculator answer)

Answer 2

Let f be the function defined on [0, ∞) by f(x) = √x.

The derivative of f is given on (0, ∞) by:

f'(x) = 1 / (2√x).

Let us consider the linearization of the function f around a = 100.

f(100) = √100 = 10

f'(100) = 1 / (2√100) = 1 / (2 * 10) = 1/20

The linearization of the function f around a = 100 is given by:

L(x) - f(a) = f'(a)(x - a)

⇒ L(x) - f(100) = f'(100)(x - 100)

⇒ L(x) - 10 = (1/20)(x - 100)

⇒ L(x) - 10 = (1/20)x - 5

⇒ L(x) - 10 + 10 = (1/20)x - 5 + 10

⇒ L(x) = (1/20)x + 5

Using this linear approximation, an estimate for √100.8 is given by:

√100.8 = f(100.8) ≈ L(100.8) = (1/20) * 100.8 + 5 = 10.04