3.11 Linear Approximations and differentials

Linear Approximations

According to the graph, f' (x) = dy / dx
dy = f' (x) dx, dy = f' (a) dx
y-y₁ = f' (x) (x-x), y =y₁ + f' (x) (x-x),
y = f(a) + f' (a) (x-a),
L(x) = f(a) + f' (a)(x-a) or Δy = f(a + Δx) – f(a)

When used over short periods of time, these approximations can be very accurate.

It was very exciting in class to finnally see why it is dy / dx - the slope of the tangent line.

This tangent line approximation or linear approximation of a at b is also referred to as linearization.



Differentials

Linear approximations are often formulated via differentials.

Let y = f(x) be a differentiable function.
The differential of x, dx, is an independent variable.
The differential of y is defined as

dy = f' (x) dx

The variable dy depends on the values of x and dx

Below is a geometric and visual way of approaching differentials

Δy = f(x + Δx) – f(x)

Tangent line PR is, as we know, the derivative of f' (x). So, dy is the rise, the distance from S to R. We can see dy ≠ Δy. Δy is the amount that the curve rises when the x changes by the amount of dx.

Relative error is found by dividing the error by the total (ΔV / V). When multiplied by 100, we can find the percentage error.



Example

Find the value of y = √(25.001) through approximation.
y = √(x + 25) .001 ≈ 0 so our a-value = 0 (that makes it easy!)
f(x) = √(x + 25)
f' (x) = (1/2) (x + 25)^(-1/2)

f(a) = √(0 + 25) = 5
f' (a) = (1/2) (0 + 25)^(-1/2) = 1/10 (*For these equations go to the f(x) and f' (x) equations found above and plug in x with a and the a-value. Do not find the derivative of f' (a) by differentiating the f(a)-value - that will just be zero.)

L(x) = 5 + (1/10)(.001-0)
L(x) = 5 + .0001 = 5.0001
√(25.001) ≈ 5.0001

Another type of problem you might see will ask for a prediction (probably extrapolating) after giving some data. With the data, we can plot a graph or make a table. When the question asks for the y-value after giving an x-value, the most accurate estimate will be found by finding the derivative of the closest given x-value. Then the y-value of the new x will approximately equal the y-value of our old x plus the derivative of the new x times the difference between the new and the old: f(x) ≈ f(a) + f' (a) (x - a) (which is the same as the definition for L(x))



Additional Internet Sources
Flash
Visual Calculus - Differentials
Visual Calculus - Linear Approximations
Powerpoint
Linear Approximations and Differentials



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