Differentials and Linear Approximations

Linear Approximation

Linear approximation can be a useful method to calculate values that are “tricky” to calculate without a calculator.

For example, if we want to know what the value of y is on the line, y = √(x) for an x-value of x = 1.1 we will struggle to calculate √(1.1) without a calculator.

We can calculate y = √(1) and say that this value will give us a y-value that is approximate to y = √(1.1), as 1 and 1.1 are close.

Definition - Linear Approximation

If a function f is differentiable at a, and the point x is close to a, then the tangent line at point a is close to the line of the function f.

In this lesson, we will be calculating the equation for tangents lines (not just the equation for the slope of a tangent line at a point, but the actual equation for the tangent line) Lets recall, how we calculate this:

To calculate the equation for a straight line, we use,

y - y0 = m(x - x0)

(x0, y0) being a point that we know on that line.

So, say we have a line of function f(x) and we want to know the equation to the tangent line (T(x)) that touches the line f(x) at a point where x is some number a, x = a, then y = f(a).

So, for our tangent line T(x) we know a point on that tangent line (a, f(a)) [which is also the point on the original line f(x)]

For calculating the equation of our tangent line T(x) we know a point that is on that line (a, f(a)). Now we need to calculate the slope, m, of T(x). We do this by calculating the derivative of f(x) and putting in the point x = a which would be f’(a). f’(a) will give us the slope of the tangent line T(x). Once we have that, we can input our point (a, f(a)) and slope (m = f’(a))

y - y0 = m(x - x0)

With,

x0 = a

y0 = f(a)

m = f’(a)

So, we can rewrite y - y0 = m(x - x0) as,

y - f(a) = f’(a)(x - a)

Or,

y - f(a) + f(a) = f’(a)(x - a) + f(a)

y = f(a) + f’(a)(x - a)

Definition - Equation for Tangent Line

With this giving us our equation for T(x) as y as a function of x, we can write,

T(x) = f(a) + f’(a)(x - a)

Below is an illustration to help visualize exactly how we get to this result,Graph of function f(x) curved showing a tangent annotated with T(x) = f(a) + f’(a)(x - a)

Beginner Linear Approximation

Normally we use linear approximation in order to find values that are hard to calculate without a calculator. For the first example we will use a function that is “easy” to calculate without a calculator to help familiarize ourselves with the method.

Use Linear Approximation with the function f(x) = 3x2 to approximate the values of 3(2.1)2 and 3(1.9)2.

Use the tangent line T(x) at the point a = (2, 12) to linearly approximate 3(2.1)2 and 3(1.9)2.

T(x) = f(a) + f'(a)(x - a)

So, with Linear Approximation (LA), we can roughly calculate values of numbers if they are "close" to our a-value for which we have a function.

We have our a-value of point (2, 4) in our function f(x) = 3x2. As 2.1 and 1.9 are "close" to 2, we can use LA to calculate f(2.1) = 3(2.1)2 and f(1.9) = 3(1.9)2.

Our first step in LA is to find the Tangent Line, T(x) at our original a-value, x = 2. T(x) = f(2) + f'(2)(x - 2)

To find T(x) at point (2, 12) we must first find the slope, f'(2)

f(x) = 3x2

f'(x) = (2)3x = 6x

f'(2) = 6(2) = 12

f(2) = 3(2)2 = 12   ,    f'(2) = 12

T(x) = f(a) + f'(a)(x - a)

= f(2) + f'(2)(x - 2)

= 12 + 12(x - 2)

= 12 + 12x - 24 = 12x - 12

So T(x) = 12 x - 12

Now, with LA, we can input T(2.1) and T(1.9) to approximate the values of 3(2.1)2 and 3(1.9)2

T(2.1) = 12(2.1) - 12 = 24.12 - 12 = 12.12

T(1.9) = 12(1.9) - 12 = 23.88 - 12 = 11.88

So, 3(2.1)2 roughly equals T(2.1) = 12.12

And, 3(1.9)2 roughly equals T(1.9) = 11.88 by use of Linear Approximation.

More Linear Approximation Examples

Definition - Linear Approximation Method

To review the Linear Approximation Method, if we have that f is a function that is differentiable at a, with a being the point that is in the function f(x) and is the point for a Tangent Line T(x),

Then, for the equation for the Tangent Line T(x),

1) Calculate our equation for the tangent line at the given point, a.

2) Check that the point we are approximating, say x0, is close to the value of a

3) Input the value x0 into T(x0)

Why Linear Approximation?

Often, it can be difficult when we don't have a calculator, calculating some values in a function that involves an advanced operation (√(x), (x), cos(x)), especially when they are not whole numbers, [i.e. x = 1.11, x = π, etc.]

When we calculate the equation of a Tangent, we are calculating the equation of a straight line. Straight lines inherently are easier to work with as they only involve an x-value that is to the power of 1 [i.e. 2x + 1, x - 1 etc.]

If we know that an x-value is close to the x-point that the tangent line is based off, then we know that for values close to that x-value, if we input them into the easier to calculate T(x) rather than the harder to calculate f(x), those values will be linearly approximate to the actual value of the function f(x). Except they are easier to calculate without a calculator!Graph of function f(x) curved showing a tangent as described in the three paragraphs below

Note how in the visual, on the graph, the values of f(x1) and T(x1) are very close to each other and the values of f(x0) and T(x0) are close to each other - or linearly approximate.

Looking on the x-axis, we see that x0 and x1 are relatively close to a, the point for the tangent line of the function. The values of x0, x1 and those between will be given linearly approximate values for T(x) and f(x).

Note, the value b, it is far away from the value of a, and so the difference on the y-axis of T(b) and f(b) is great and therefore is not linearly approximate.

Linear Approximation

 

part 1 of the solution to using linear approximation to find the value of e^0.9
part 2 of the solution to using linear approximation to find the value of e^0.9

Linear Approximation

part 1 of the solution to using linear approximation to calculate the  value of (2.025)^6 being the money accrued on $2 over 6 years with an annual growth of 2.5%
part 2 of the solution to using linear approximation to calculate the  value of (2.025)^6 being the money accrued on $2 over 6 years with an annual growth of 2.5%