Function Linearization and Estimation

Linearization of a Function and Estimation

• Linearization helps approximate function values, especially when direct calculation is difficult.
• Example: Estimating the square root of 3.99 (close to square root of 4) or the square root of 37 (close to square root of 36).
• The tangent line at a point 'a' on a function f(x) serves as the linearization.
• The tangent line is a good approximation of the actual function very close to the point of interest
• As you move further away from point a, the tangent line's y-value differs significantly from f(x)'s y-value.

Finding the Linearization

• The linearization of a function is found by determining the tangent line to the function at a specific point
• Use the point-slope formula: y - y1 = m(x - x1).
• x1 is 'a', and y1 is f(a).
• The slope 'm' is the derivative of the function evaluated at 'a', i.e., f'(a).
• Linearization function: l(x) = f(a) + f'(a) * (x - a)
• l(x) represents the linearization of the function at x

Example Problem: Linearization of Square Root Function

• Function: f(x) = √x, a = 36
• Goal: Find l(x) and use it to estimate √37 (f(37)).
• Rewrite f(x) = √x as x^(1/2).
• Derivative: f'(x) = (1/2) * x^(-1/2) = 1 / (2√x)
• Evaluate f'(a): f'(36) = 1 / (2√36) = 1/12.
• Evaluate f(a): f(36) = √36 = 6.
• Substitute into linearization formula: l(x) = 6 + (1/12) * (x - 36).
• Simplify: l(x) = (1/12)x + 3.
• Estimate √37: l(37) = (37/12) + 3 ≈ 6.083.
• Actual value of √37 ≈ 6.08276, showing the approximation is close.

Second Example: Linearization of x Cubed Function

• Function: f(x) = x³, a = 2
• Goal: Find l(x) and use it to estimate (2.1)³ (f(2.1)).
• Derivative: f'(x) = 3x².
• Evaluate f'(a): f'(2) = 3 * 2² = 12.
• Evaluate f(a): f(2) = 2³ = 8.
• Linearization formula: l(x) = 8 + 12 * (x - 2).
• Simplify: l(x) = 12x - 16.
• Estimate (2.1)³: l(2.1) = 12 * 2.1 - 16 = 9.2.
• Actual value of (2.1)³ = 9.261.
• The approximation is decent even though 2.1 is not extremely close to 2.

Estimating Cube Root Using Linearization

• Problem: Estimate the cube root of 8.1 using local linear approximation.
• Determine f(x) = ∛x and a = 8 (since ∛8 is known).
• Rewrite f(x) as x^(1/3).
• Derivative: f'(x) = (1/3) * x^(-2/3) = 1 / (3 * x^(2/3)).
• Evaluate f'(a): f'(8) = 1 / (3 * 8^(2/3)) = 1/12.
• 8^(2/3) is equivalent to (8^(1/3))^2 which simplifies to 2^2 = 4
• Evaluate f(a): f(8) = ∛8 = 2.
• Linearization formula: l(x) = 2 + (1/12) * (x - 8).
• Simplify: l(x) = (1/12)x + 4/3.
• Estimate ∛8.1: l(8.1) = (1/12) * 8.1 + 4/3 ≈ 2.008333.
• Actual value of ∛8.1 ≈ 2.0082985, a close approximation.

Estimating Natural Logarithm Using Linearization

• Problem: Use linearization to approximate ln(1.05).
• Determine f(x) = ln(x) and a = 1 (since ln(1) is known).
• Derivative: f'(x) = 1/x.
• Evaluate f'(a): f'(1) = 1/1 = 1.
• Evaluate f(a): f(1) = ln(1) = 0.
• Linearization formula: l(x) = 0 + 1 * (x - 1) = x - 1.
• Estimate ln(1.05): l(1.05) = 1.05 - 1 = 0.05.
• Actual value of ln(1.05) ≈ 0.0488, which is approximately 0.05 when rounded.

Approximating Tangent Using Linearization

• Problem: Find the approximate value of tan(46°).
• Determine f(x) = tan(x) and a = 45° (since tan(45°) is known).
• tan(45°) = 1.
• Derivative: f'(x) = sec²(x).
• Evaluate f'(a): f'(45°) = sec²(45°).
• sec(45°) = 1 / cos(45°) = 1 / (√2 / 2)
• sec²(45°) = 2.
• When working with trigonometric functions in calculus, it's generally more appropriate to use radians rather than degrees

Radians

• Convert 45 degrees to π/4 radians and 46 degrees to 23π/90.
• Linearization formula: ( l(x) = 1 + 2 \left( x - \frac{\pi}{4} \right) ).
• The approximation demonstrates the utility of linearization when used with the correct units (radians for calculus).

Approximating with Linearization

• The exact answer of tan 46 is approximately 1.03553.
• Using linearization, the estimation of tan 46 is approximately 1.034907.
• When estimating trig functions with linearization, ensure angles are in radians, not degrees.

Differentials

• If y = f(x), then differentiating both sides with respect to x yields dy/dx = f'(x).
• Multiplying both sides by dx gives the differential y: dy = f'(x)dx.
• dy is approximately equal to delta y (Δy).
• dx is equal to delta x (Δx).
• Δx represents the change in the x variable, like x2 - x1.
• Δy represents the change in the y variable, like y2 - y1 or f(x + Δx) - f(x).

Graphical Representation

• Δx and Δy are associated with the function f(x).
• dx and dy are associated with the tangent line.
• dx is defined as Δx (x2 - x1).
• dy is the difference between y1 and the new y-value of the tangent line at x2.
• Δy and dy are not the same.
• For an increasing function, Δy is greater than dy.
• When points are very close, Δy is approximately equal to dy.

Calculating Differentials

• The derivative of x cubed is 3x squared times dx.
• The derivative of r to the fourth is 4r cubed times dr.
• To find the differential, differentiate each term and add its respective differential (dx, dr, etc.).
• Product Rule in Differentials: d(uv) = u dv + v du

Application to Volume of a Sphere

• The volume V of a sphere is V = (4/3)πr^3.
• The differential dv is dv = 4πr^2 dr.
• The differential can also be found by differentiating V with respect to r.

Evaluating Differentials

• Given y = x^3 + 6x, find dy: dy = (3x^2 + 6) dx.
• Given a function, x value, and dx value, substitute to find dy.

Comparing Δy and dy

• To find dy, compute dy = f'(x) * dx.
• To find Δy, calculate Δy = f(x + Δx) - f(x).
• If the change in x is small, Δy is approximately equal to dy.
• As Δx increases, dy and Δy deviate from each other.

Graphical Explanation of Δx Increase

• As Δx increases, Δy and dy diverge from each other.
• As Δx decreases, approaching the intersection point, Δy is approximately equal to dy.