# Linear Approximation and Differentials in Calculus

*Ray is a licensed engineer in the Philippines. He loves to write about mathematics and civil engineering.*

Linear Approximation, or sometimes referred to as the Linearization or Tangent Line Approximation, is a calculus method that uses the tangent line to approximate another point on a curve. Linear Approximation is an excellent method to estimate f (x) values as long as it is near x = a.

The given figure below shows a curve that lies very close to its tangent line near the point of tangency. In fact, by zooming in toward an end on a differentiable function graph, the graph looks more and more like its tangent line. This method is the basis for a way of finding approximate values of a function.

The idea is that it might be easy to calculate a value of f (a) of a function but challenging to compute nearby values of the function *f*. So we settle for the easily calculated values of the linear function L, whose graph is the tangent line of the function f at (a, f (a)).

**Linear Approximation Formula**

Use the tangent line at (a, f (a)) as an approximation to the curve y=f(x) when x is near a. The tangent line equation is shown below.

y = f (a) + f’ (a) (x-a)

The Approximation given by the equation below is called the linear Approximation or tangent line approximation of f at a. In calculus, a linear approximation is an approximation of a general equation using a linear function.

f (x) = f (a) + f’ (a) (x-a)

The linear function whose graph is this tangent line, shown below, is called the Linearization of the function f at a.

L (x) = f (a) + f’ (a) (x - a) [equation of tangent line]

## How to Perform Linear Approximation

There are three easy steps on how to perform Linear Approximation.

- Plug in the given value of the given variable
*a*for*x,*then solve for the value of y to find the ordered pair. - Take the derivative of the given function
*f*to find the slope of the tangent line*f'*. - Next, there are two options you can go through in finding the Linearization of the given function.

**Option 1**: Plug in the ordered pair from the first step and solve for the slope *m* or dy/dx*.* The Linearization is found by substituting the ordered pair and slope obtained from the previous actions into a point-slope equation.

y – y1 = m (x – x1)

**Option 2**: Use the given formula of the equation of the tangent line in finding the Linearization.

L (x) = f (a) + f’ (a) (x - a)

## Example 1: Finding the Linear Approximation of a Natural Logarithmic Function

Find the linearization of the natural logarithm ln(x) at x = 1.

**Solution**

Consider the natural logarithm function ln(x). Calculate the derivative of the function and its value at point a.

f(a) = ln (x)

f(1) = ln (1)

f(1) = 0

f’(x) = f’(a) = 1/x

f’(1) = 1/1

f’(1) = 1

In getting the linearization of the given function, use the linear approximation function L(x).

L(x) = f(a) + f’(a) (x - a)

L(x) = f(1) + f’(1) (x - 1)

L(x) = 0 + (1)(x – 1)

L(x) = x - 1

**Final Answer**

Therefore, the linearization of the given natural logarithm is L(x) = x – 1.

## Example 2: Finding the Linear Approximation of an Equation to the Fourth Power

Find the linear approximation of the function f(x) = x^{4} – 6t^{3} + 3t – 7 at x = -3.

**Solution**

First, substitute the value of x = -3 to the given equation.

f(x) = x^{4} – 6t^{3} + 3t – 7

f(-3) = (-3)^{4} – 6(-3)^{3} + 3(-3) – 7

f(-3) = 227

Get the derivative of the given equation and substitute the value of x = -3.

f(x) = x^{4} – 6t^{3} + 3t – 7

f'(x) = 4x^{3} – 18t^{2} + 3

f'(-3) = 4(-3)^{3} – 18(-3)^{2} + 3

f'(-3) = -267

Substitute the values of f(-3) and f'(-3) to the linearization formula of a function.

L(x) = f (a) + f’(a) (x - a)

L(x) = 227 + (-267) (x – (-3))

L(x) = 227 + (-267) (x + 3)

L(x) = -267x - 574

**Final Answer**

The linear approximation of the function f(x) = x^{4} – 6t^{3} + 3t – 7 at x = -3 is L(x) = -267x – 574.

## Example 3: Finding the Linearization of a Square Root Function

Find the linearization of the function f(x) = √x + 3 at a = 1 and use it to approximate the numbers √3.98 and √4.1. Are these approximations overestimates or underestimates?

**Solution**

Find the derivative of the function f(x) = (x + 3)1/2.

f(x) = (x + 3)1/2

f'(x) = (1/2) (x + 3)-1/2

f'(x) = 1 / (2√(x + 3))

Substitute a = 1 to the equation and the derivative of the function

f(1) = (1 + 3)1/2

f(1) = 2

f'(1) = 1 / (2√(1 + 3))

f'(1) = ¼

Put the obtained values of f (1) and f' (1) to the equation of the tangent line. Therefore, the linearization is given below.

L(x) = f (1) + f’(1) (x - 1)

L(x) = 2 + (1/4) (x – 1)

L(x) = (7/4) + (x/4)

After solving the linearization of the given square root function, approximate the value of the given square root numbers when x is near 1. As seen from the graph, the approximations are overestimated since the tangent line lies above the curve.

√3.98 = (7/4) + (x/4)

√3.98 = (7/4) + (0.98/4)

√3.98 = 1.995

√4.1 = (7/4) + (x/4)

√4.1 = (7/4) + (1.1/4)

√4.1 = 2.025

The table below shows and compare the estimates from the linear approximation with real values. According to the graph, the tangent line approximation gives a reliable estimate when x is close to 1, but the approximate linear accuracy decreases when x moves farther away from 1.

Square Root Equations | x | L(x) Estimates | Actual Value |
---|---|---|---|

√3.9 | 0.9 | 1.975 | 1.97484176… |

√3.98 | 0.98 | 1.995 | 1.99499373… |

√4 | 1 | 2 | 2.00000000… |

√4.05 | 1.05 | 2.0125 | 2.012462117… |

√4.1 | 1.1 | 2.025 | 2.02484567… |

√5 | 2 | 2.25 | 2.23606797… |

√6 | 3 | 2.5 | 2.44948974… |

**Final Answer**

The corresponding linear approximation of the function (x + 3)1/2 is equal to (7/4) + (x/4). Also, the overestimated approximate values of √3.98 and √4.1 are 1.995 and 2.025, respectively.

## Example 4: Finding the Linear Approximation of √x

Find the linear approximate of the square root function f(x) = √x at x = 16. Then, use the approximation to estimate √16.1.

**Solution**

Solve for the value of the given function for f (16) and f' (16).

f(x) = √x

f(16) = √16

f(16) = 4

f'(x) = √x

f'(16) = 1/(2√x)

f'(16) = 1/(2√16)

f'(16) = 1/(2√x)

f'(16) = 1/8

Substitute the obtained values to the linearization formula.

L(x) = f (16) + f’ (16) (x - 16)

L(x) = 4 + (1/8) (x – 16)

Using the equation obtained, use the linear approximation to estimate √16.1.

L(16.1) = 4 + (1/8) (16.1 – 16)

L(16.1) = 4.0125

**Final Answer**

The linear approximate of the function f(x) = √x at x = 16 is L (x) = 4 + (1/8) (x – 16). Using this, the estimated value of √16.1 is 4.0125.

## Example 5: Predicting a Temperature From a Linear Approximation

Suppose that after you stuff a chicken, its temperature is ten degrees Celsius, you put it in a 165°C oven. After an hour, the meat thermometer indicates that the chicken's temperature is 34°C, and after two hours, it displays 54°C. Predict the temperature of the chicken after three hours.

**Solution**

The function T (t) represents the temperature of the chicken after *t* hours. Given the temperature of the chicken in different conditions, then T (0) = 10, T (1) = 34, and T (2) = 54. To make a linear approximation with a = 2, then estimate the derivative of T at 2.

T'(2) = limx→2 [T (t) – T (2)] / (t – 2)

Use the difference quotient with t = 1 in estimating the value of T' (2).

T'(2) = [T (1) – T (2)] / (1 – 2)

T'(2) = [34 – 54] / (-1)

T'(2) = 20

The temperature change between t = 1 and t = 2 is 20°C/h. With the obtained estimate, the linear approximation for the temperature after three hours is given below.

T(3) = T(2) + T’(2) (3 - 2)

T(3) = 54 + 20 (1)

T(3) = 74

**Final Answer**

The predicted temperature of the chicken after three hours is 74°C.

## Example 6: Linear Approximation of a Sine Function

Find the linear approximation for sin (θ) at θ = 0.

**Solution**

Compute the tangent line equation to the given equation sin (θ) at θ = 0.

f(θ) = sin(θ)

f(0) = 0

f'(θ) = cos(θ)

f'(0) = 1

Compute the linear approximation by substituting the formula for the tangent line equation.

L(x) = f(a) + f’(a) (x-a)

L(θ) = f(0) + f'(0) (θ – 0)

L(θ) = 0 + 1(θ)

L(θ) = θ

**Final Answer**

Therefore, sin (θ) = θ considering θ stays small.

## Example 7: Linear Approximation of Roots and Powers

Find the linear approximation of f(x) = (1 + x)^{n} at x = 0. Use the resulted linear approximation to estimate (1.05)^{5}.

**Solution**

Given the function f(x) = (1 + x)^{n}, solve for f(x) and f’(x) at x = 0.

f(x) = (1 + x)^{n}

f(0) = (1 + 0)^{n}

f(0) = 1

f’(x) = n(1 + x)^{n-1}

f’(0) = n(1 + 0)^{n-1}

f’(0) = n

In getting the approximate value of f(x) = (1 + x)^{n}, apply the approximate formula or the tangent line equation.

L(x) = f(a) + f’(a) (x - a)

L(x) = f(0) + f’(0) (x – 0)

L(x) = 1 + nx

Next, evaluate (1.05)^{5} by evaluating L(1.05) when n = 5.

L(0.05) = 1 + (0.05)(5)

L(0.05) = 1 + 0.25

L(0.05) = 1.25

**Final Answer**

Therefore, the estimated value of (1.05)^{5} is 1.25.

## Example 8: Linear Approximation of a Cos(X)

Find the approximate value for the function f(x) = cos (x) at x = π/2 using the linear approximation formula. Use the result to approximate the value of cos(89π/180).

**Solution**

Compute the tangent line equation to the given equation cos(x) at x = π/2.

f(x) = cos(x)

f(π/2) = cos(π/2)

f(π/2) = 0

f'(x) = -sin(x)

f'(π/2) = -sin(π/2)

f'(π/2) = -1

Compute the linear approximation by substituting the formula for the tangent line equation.

L(x) = f(a) + f’(a) (x-a)

L(x) = f(π/2) + f'(π/2) (x – π/2)

L(x) = 0 – 1(x - π/2)

L(x) = -x + π/2

Substitute x = π/180 to the linearization equation to solve the value of cosine (π).

L(x) = -x + π/2

L(89π/180) = -(89π/180) + π/2

L(89π/180) = π/180 = 0.0175

**Final Answer**

The approximate value of the function f(x) = cos (89π/180) is π/180 or approximately 0.0175.

## Example 9: Approximate Value of a Square Root Function

Find the approximate value for the square root function f(x) = √50 using the linear approximation formula.

**Solution**

The function can be read as f(x) = √x at x = 50. Find the value of the function and its derivative at a = 49.

f(a) = √x

f(a) = √49

f’(a) = (√a)’ = 1 / (2√a)

f’(49) = 1 / (2√a)

f’(49) = 1 / (2√49)

f’(49) = 1 / 14

Using the tangent line equation, compute the approximate value at x = 50.

L(x) = f(a) + f’(a) (x - a)

L(x) = √49 + (1/14) (50 – 49)

L(50) = 7 + (1/14)

L(50) = 99/14

L(50) = 7.0714

**Final Answer**

The approximate value of the function f(x) = √50 is 7.0174.

## Example 10: Approximate Value of the Sum of Cubic and Cosine Function

Find the linear approximation of the function f(x) = x^{3} + 4cos(x) at a = 0.

**Solution**

Solve for the value of the given cosine function for f(0) and f'(0). Solve for the value of the function and its derivative at a = 0.

f(x) = x^{3} + 4cos(x)

f(0) = (0)^{3} + 4cos(0)

f(0) = 4

f(x) = x^{3} + 4cos(x)

f’(x) = 3x^{2} – 4sin(x)

f’(0) = 3(0)^{2} – 4sin(0)

f’(0) = 0

Substitute the value of f(0) and f’(0) to the tangent line equation.

L(x) = f(a) + f’(a) (x - a)

L(x) = f(0) + f’(0) (x - 0)

L(x) = 4 + 0 (x – 0)

L(x) = 4

**Final Answer**

The linear approximation of the given function is the horizontal line L(x) = 4.

## Example 11: Linear Approximation of a Decimal Value to the Fourth Power

Use a linear approximation (or differentials) to estimate the given number (1.999)^{4}.

**Solution**

Consider the given function (1.999)^{4} as x^{4}. Therefore, find the f(a) and f’(a) at a=2 since it is the nearest possible integer.

f(a) = a^{4}

f(2) = 2^{4}

f(2) = 16

f’(a) = 4a^{3}

f’(2) = 4(2)^{3}

f’(2) = 32

Substitute the value of f(2) and f’(2) to the tangent line equation.

L(x) = f(a) + f’(a) (x - a)

L(1.999) = 16 + 32 (1.999 – 2)

L(1.999) = 16 – 0.032

L(1.999) = 15.968

**Final Answer**

The approximate value of (1.999)^{4} is line 15.968.

## Example 12: Linear Approximation of a Natural Logarithmic Equation

Use linear approximation to estimate the value of e^{0.2}.

**Solution**

Let f(x) = e^{x} and set a = 0 in solving for f(a) and f’(a).

f(a) = e^{a}

f(0) = e^{0}

f(0) = 1

f’(a) = e^{a}

f(0) = e^{0}

f(0) = 1

In getting the approximate value of e^{0.2}, apply the approximate formula or the tangent line equation.

L(x) = f(a) + f’(a) (x - a)

L(0.2) = f(0) + f’(0) (0.2 - 0)

L(0.2) = 1 + 1 (0.2 – 0)

L(0.2) = 1.2

**Final Answer**

Therefore, the approximate value of e^{0.2} is 1.20.

## Example 13: Approximate Value of Inverse Trigonometric Function

Use a linear approximation (or differentials) to estimate the inverse trigonometric equation y = arcos (0.52).

**Solution**

Suppose that f(x) = arcos (x) and a = 0.5. Replace the increment of the given function ∆y by its differential and compute the inverse trigonometric function's approximate value.

f(a) = arcos (a)

f(0.5) = arcos (0.5)

f(a) = arcos (a)

f’(a) = -1/√(1-x2)

f’(0.5) = -1/√(1-0.52)

f’(0.5) = -1/√0.75

f’(0.5) = -1.1547

In getting the approximate value of y = arcos (0.52), apply the approximate formula or the tangent line equation.

L(x) = f(a) + f’(a) (x - a)

L(0.52) = f(0.5) + f’(0.5) (0.52-0.5)

L(0.52) = arcos (0.5) – 1.1547 (0.52-0.5)

L(0.52) = 1.0472 – 0.0231

L(0.52) = 1.0241

**Final Answer**

The approximate value of the function y = arcos (0.52) is 1.0241.

## Example 14: Approximate Value of a Linear Function

Find the linear approximation of the function g(x) = 51 + 3x at x = 0.

**Solution**

Solve for the value of the given linear function for g(0) and g'(0). Solve for the value of the function and its derivative at a = 0.

g(a) = 51 + 3(0)

g(0) = 51 + 0

g(0) = 51

g’(a) = 51 + 3x

g’(0) = 0 + 3

g’(0) = 3

In getting the approximate value of g(x) = 51 + x, apply the approximate formula or the tangent line equation.

L(x) = f(a) + f’(a) (x - a)

L(x) = g(a) + g’(a) (x - a)

L(x) = g(0) + g’(0) (x - 0)

L(x) = 51 + 1(0 – 0)

L(x) = 51

**Final Answer**

Therefore, the approximate value of the given function g(x) = 51 + 3x at x = 0 is 51.

## Example 15: Finding the Approximate Value of a Quadratic Function

Given the function f(x) = 5 / x^{2}, use linear approximation at a = 1 to estimate the value of f(0.99).

**Solution**

Solve for the value of the given linear function for f(x) and f'(x) when x = 0.99. Solve for the value of the function and its derivative at a = 1.

f(x) = 5 / x^{2}

f(a) = 5 / a^{2}

f(1) = 5 / 1^{2}

f(1) = 5

f(x) = 5 / x^{2}

f’(x) = 5x^{-2}

f’(1) = 5(-2) a^{-2-1}

f’(1) = -10a^{-3}

f’(1) = -10/a^{3}

f’(1) = -10/(1)^{3}

f’(1) = -10

In getting the approximate value of f(x) = 5 / x^{2}, apply the approximate formula or the tangent line equation.

L(x) = f(a) + f’(a) (x - a)

L(0.99) = f(1) + f’(1) (0.99 – 1)

L(0.99) = 5 - 10 (0.99 - 1)

L(0.99) = 5 + 0.1

L(0.99) = 5.10

**Final Answer**

Therefore, using the linear approximation formula, the approximate value of the function f(x) = 5 / x^{2} is 5.10.

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*This content is accurate and true to the best of the author’s knowledge and is not meant to substitute for formal and individualized advice from a qualified professional.*

**© 2021 Ray**