From breaking down life goals into smaller, more manageable tasks, to lines on a page, if you zoom in close enough, pretty much anything becomes much simpler. This is the concept behind linear approximations and differentials – if you zoom in close enough, your function looks like a line and can be manipulated like one.
In this article, you will learn how to approximate functions at a given point, or locally, by using linear functions to make these approximations. Because linear functions are the easiest type of function with which to work, they are a powerful approximation tool. You will also learn about a related concept, differentials, so that you will be able to use linear approximations to estimate the amount a function changes as a result of a change to its input value.
Linear Approximation and Differentials Definitions
What are linear approximation and differentials?
Linear Approximation Definition and Equation
Linear approximation is a method that uses the tangent line to a curve to approximate another point on that curve. It is a great method to estimate values of a function, \( f(x) \), as long as \( x \) is near \( x = a \).
Given a differentiable function, \( f(x) \), you can find its tangent line at \( x = a \). The equation of this tangent line, \( L(x) \) is
\[ L(x) = f(a) + f'(a) (x - a) \]
and it is called the linear approximation, or tangent line approximation of \( f(x) \) at \( x = a \). It is also known as the linearization of the function \( f(x) \) at \( x = a \).
The graph below visualizes the concept of nearness. You will notice that when you zoom in, the further you zoom, the closer the function looks like its tangent line.
Figure 1. The linear approximation of the function, \( f(x) \), at \( x = a \).
The idea behind this is that while it could be difficult to calculate nearby values of the function, it might be much simpler to perform the linear approximation.
Steps to Calculate a Linear Approximation
The following example outlines the basic steps used to perform linear approximations and demonstrates a practical use for them.
Approximate \( f(\theta) = \sin(\theta) \) at \( \theta = 0 \).
Solution:
- Use the given value for \( a \) in place of \( x \) to find \( f(a) \). This gives you the ordered pair \( (a, f(a)) \).
- In the context of this problem, you need to solve for \( f(0) \).\[ f(0) = \sin(0) = 0 \]
- The ordered pair is \( (0, 0) \).
- Take the derivative of the given function, \( f(x) \), to find the slope of the tangent line.
- In the context of this problem, you need to find the derivative of \( f(\theta) \).\[ f'(\theta) = \cos(\theta) \]
- Solve for the derivative at the point \( x = a \).
- In the context of this problem, you need to plug in the value for \( \theta \) to solve for the derivative at that point.\[ \begin{align}f'(\theta) &= \cos(\theta) \\f'(0) &= \cos(0) \\f'(0) &= 1\end{align} \]
- Plug the values for \( f(a) \), \( f'(a) \), and \( a \) into the equation: \( L(x) = f(a) + f'(a) (x - a) \).
- In the context of this problem, you need to plug the value for \( \theta \) and the values you got in steps \( 1–3 \) into the linear approximation equation.\[ \begin{align}L(x) &= f(a) + f'(a) (x - a) \\L(\theta)&= f(0) + f'(0) (\theta - 0) \\&= 0 + 1 (\theta - 0) \\\mathbf{L(\theta)} &= \mathbf{\theta}\end{align} \]
This linear approximation is regularly used in the field of optics to simplify formulas. It is also used to describe the motion of a pendulum and the vibrations of a string.
Differential Definition and Equation
To be able to use linear approximations to estimate the amount a function changes, you must understand the concept of differentials; they provide you with a method of estimating the amount a function changes as a result of a small change in its input values.
Given a differentiable function, \( y = f(x) \), let \( \mathrm{d}x \) be an independent variable that can be any nonzero real number. Define the dependent variable \( \mathrm{d}y \) by the equation:
\[ \mathrm{d}y = f'(x) ~\mathrm{d}x. \]
You call the expressions \( \mathrm{d}y \) and \( \mathrm{d}x \) differentials.
- Notice that \( \mathrm{d}y \) is a function of both \( x \) and \( \mathrm{d}x \).
If you divide both sides of the equation in the definition by \( \mathrm{d}x \), you get:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = f'(x) \]
which is how you are familiar with denoting a derivative. Thus, this equation is called differential form.
Note that if you are just given a differentiable function \( f(x) \), then the differentials become \( \mathrm{d}f \) and \( \mathrm{d}x \). Essentially, \( \mathrm{d}y \) is replaced by \( \mathrm{d}f \) in the equation:
\[ \mathrm{d}f = f'(x) ~\mathrm{d}x \]
When you started learning derivatives, you used Leibniz notation \( \frac{\mathrm{d}y}{\mathrm{d}x} \) to represent the derivative of \( y \) with respect to \( x \). While you used the expressions \( \mathrm{d}y \) and \( \mathrm{d}x \) in this notation, they did not yet have any meaning of their own.
Given the definition of differentials above, now you have a meaning behind the expressions \( \mathrm{d}y \) and \( \mathrm{d}x \).
Tangent Line Approximation
Recall from the tangent lines article that the tangent line to the curve of a function, \( f \), that is differentiable at a point, \( a \), is given by the equation:
\[ y = f(a) + f'(a) (x - a). \]
For instance, consider the differentiable function:
\[ f(x) = \frac{1}{x}, \]
where \( x = a = 3 \).
- What is the tangent line to the curve at \( a = 3 \), and
- Can it be used to approximate the value of \( f(x) \) when \( a = 3.1 \)?
Solution:
- What is the tangent line to the curve at \( a = 3 \)?
- Find \( f(a) \).\[ f(a) = f(3) = \frac{1}{3} \]
- Find the derivative of \( f(x) \).\[ f'(x) = -\frac{1}{x^{2}} \]
- Plug in the value for \( a \) to solve for the derivative at that point.\[ \begin{align}f'(a) &= -\frac{1}{a^{2}} \\f'(3) &= -\frac{1}{(3)^{2}} \\&= -\frac{1}{9}\end{align} \]
- Plug the value for \( a \) and the values you got in steps \( 1–3 \) into the tangent line equation.\[ \begin{align}y &= f(a) + f'(a) (x - a) \\\mathbf{y} &= \mathbf{\frac{1}{3} - \frac{1}{9} (x - 3)}\end{align} \]
- Can the tangent line be used to approximate the value of \( f(x) \) when \( a = 3.1 \)?
- Use the tangent line you found in part A to estimate the value of the function at \( a = 3.1 \).\[ \begin{align}y &= \frac{1}{3} - \frac{1}{9} (3.1 - 3) \\&\approx 0.3222\end{align} \]
- Calculate the value of \( f(3.1) \) using the given function.\[ f(3.1) = \frac{1}{3.1} \approx 0.3226\]
- Compare the two values from steps \(1-2\). Use the graphs below to help visualize the comparison.Since the difference between the approximation and the actual values are minimal, \[ y \approx 0.3222 \text{ vs. } f(x) \approx 0.3226 \] you can say that the tangent line can be used to approximate the value of \( f(x) \) when \( a = 3.1 \).
Figure 2. The tangent line to the curve of \( f(x) = \frac{1}{x} \) at \( x = 3 \) gives you a good approximation of \( f(x) \) when \( x \) is near \( 3 \).
Figure 3. When \( x = 3.1 \), the value of \( y \) on the tangent line to the curve of \( f(x) = \frac{1}{x} \) is approximately \( 0.3222 \); the actual value of \( f(3.1) \) is \( \frac{1}{3.1} \approx 0.3226 \).
What this example shows is that in general, for a differentiable function \( f(x) \), the equation of the tangent line to \( f(x) \) where \( x = a \) can be used to approximate \( f(x) \) for values of \( x \) near \( a \).
Therefore,
\[ f(x) \approx f(a) + f'(a) (x - a), \text{ for } x \text{ near } a, \]
and the linear function:
\[ L(x) = f(a) + f'(a) (x - a) \]
is the linear approximation of \( f(x) \) at \( x = a \).
But what is considered “near” \( a \)?
The short answer is: it depends.
Approximation in Calculus
Approximation is something you do for a lot of in calculus. You use approximations when you:
just to name a few.
As you might have guessed from the example above, the farther away you get from \( x =a \), the worse your approximation will be.
But again, how far away is too far?
And again, the answer: it depends. It depends on both the function and the value of \( x = a \) that you are using. Ultimately, there is often no easy way to predict how far from \( x = a \) you can get and still have a "good" linear approximation.
Difference between Linear Approximation and Differentials
To get a good idea of the difference between linear approximation and differentials, you need to connect the two concepts.
Suppose you have a function, \( f(x) \), that is differentiable at point \( a \). Say the input, \( x \), changes by some small amount called \( \mathrm{d}x \) (this could also be denoted as \( \Delta x \)).
You are interested in how much the output, \( y \), changes based on this tiny change to \( x \).
- If \( x \) changes from \( a \) to \( a + \mathrm{d}x \), then the total change in \( x \) is \( \mathrm{d}x \), and the change in \( y \) is given by the equation:\[ \Delta y = f(a + \mathrm{d}x) - f(a). \]
- So, if \( \mathrm{d}x \) is small,\[ \begin{align}f(a + \mathrm{d}x) &\approx L(a + \mathrm{d}x) \\&= f(a) + f'(a) (a + \mathrm{d}x - a).\end{align} \]
- Subtracting \( f(a) \) from both sides,\[ \begin{align}f(a + \mathrm{d}x) - f(a) &\approx L(a + \mathrm{d}x) - f(a) \\&= f'(a) \mathrm{d}x.\end{align} \]
In other words,
- the actual change in the function, \( f(x) \), (if \( x \) increases from \( a \) to \( a + \mathrm{d}x \)) is approximately the difference between \( L(a + \mathrm{d}x) \) and \( f(a) \),
- where \( L(x) \) is the linear approximation of \( f(x) \) at \( x = a \).
- By the definition of \( L(x) \), this difference is equal to \( f'(a) \mathrm{d}x \).
To summarize:
\[ \begin{align}\Delta y &= f(a + \mathrm{d}x) - f(a) \\&\approx L(a + \mathrm{d}x) - f(a) \\&= f'(a) \mathrm{d}x \\&= \mathrm{d}y\end{align} \]
This means you can use the differential, \( \mathrm{d}y = f'(a) \mathrm{d}x \), to approximate the change in \( y \) if \( x \) increases from \( x = a \) to \( x = a + \mathrm{d}x \). The graph below shows this in detail.
Figure 4. You use the differential \( \mathrm{d}y = f'(a) \mathrm{d}x \) to approximate the actual change in \( y \) if \( x \) increases from \( a \) to \( a + \mathrm{d}x \).
So, what's the difference between linear approximations and differentials?
Put simply:
Linear approximations give an estimate of the value of a differentiable function at a specific point by using the tangent line to the curve of the function at that point.
Differentials give an approximation for the change in the dependent variable (usually \( y \)) of the function at the same point the linear approximation is calculated.
Let's walk through an example.
Approximating Change with Differentials
Given the function:
\[ y = x^{2} + 2x, \]
calculate
- \( \Delta y \) and
- \( \mathrm{d}y \)
at \( x = 3 \) if \( \mathrm{d}x = 0.1 \).
Solution:
- \( \Delta y \) is the actual change in \( y \) if \( x \) changes from \( 3 \) to \( 3.1 \). To find actual change, use the formula: \( \Delta y = f(a + \mathrm{d}x) - f(a) \).\[ \begin{align}\Delta y &= f(a + \mathrm{d}x) - f(a) \\&= f(3.1) - f(3) \\&= \left( (3.1)^{2} + 2(3.1) \right) - \left( 3^{2} + 2(3) \right) \\&= 15.81 - 15 \\&= 0.81\end{align} \]
- \( \mathrm{d}y \) is the approximate change in \( y \). To find approximate change, use the formula: \( \mathrm{d}y = f'(x) ~\mathrm{d}x \).
- Find the derivative of \( f(x) \).\[ f'(x) = 2x + 2 \]
- Use the differential formula to solve for the approximate change.\[ \begin{align}\mathrm{d}y &= f'(x) ~\mathrm{d}x \\&= f'(3) ~\mathrm{d}x \\&= (2(3) + 2)(0.1) \\&= 0.8\end{align} \]
As you can see, the calculations using differentials is simpler than calculating the actual values, and the results of both are quite similar, especially as the value of \( \mathrm{d}x \) decreases.
Linear Approximation and Differentials Examples
Try your hand at these examples!
Linear Approximation of a Function
Approximate the function:
\[ f(x) = (1 + x)^{n} \]
at \( x = 0 \). Then use the approximation to estimate \( 1.01^{2} \).
Solution:
- Use the given value for \( a \) in place of \( x \) to find \( f(a) \). This gives you the ordered pair \( (a, f(a)) \).
- In the context of this problem, you need to solve for \( f(0) \).\[ f(0) = (1 + 0)^{n} = 1 \]
- The ordered pair is \( (0, 1) \).
- Take the derivative of the given function, \( f(x) \), to find the slope of the tangent line.
- In the context of this problem, you need to find the derivative of \( f(x) \).\[ f'(x) = n(1 + x)^{n-1} \]
- Solve for the derivative at the point \( x = a \).
- In the context of this problem, you need to plug in the value for \( x \) to solve for the derivative at that point.\[ \begin{align}f'(x) &= n(1 + x)^{n-1} \\f'(0) &= n(1 + 0)^{n-1} \\f'(0) &= n\end{align} \]
- Plug the values for \( f(a) \), \( f'(a) \), and \( a \) into the equation: \( L(x) = f(a) + f'(a) (x - a) \).
- In the context of this problem, you need to plug the value for \( x \) and the values you got in steps \( 1–3 \) into the linear approximation equation.\[ \begin{align}L(x) &= f(a) + f'(a) (x - a) \\&= f(0) + f'(0) (x - 0) \\&= 1 + n (x - 0) \\\mathbf{L(x)} &= \mathbf{1 + nx}\end{align} \]
- Approximate \( 1.01^{2} \) by evaluating \( L(0.01) \) at \( n = 2 \).\[ \begin{align}(1.01)^{2} &= f(1.01) \\&\approx L(1.01) \\&= 1 + n(x - 0) \\&= 1 + 2(0.01) \\\mathbf{(1.01)^{2}} &= \mathbf{1.02}\end{align} \]
Now, an example of differentials.
Calculating Differentials
Find \( \mathrm{d}y \) and evaluate the following function at \( x = 3 \) and \( \mathrm{d}x = 0.1 \).
\[ y = \cos(x) \]
Solution:
- Find the derivative of \( f(x) \).\[ f'(x) = -\sin(x) \]
- Plug the derivative into the differential formula.\[ \begin{align}\mathrm{d}y &= f'(x) ~\mathrm{d}x \\\mathbf{\mathrm{d}y} &= \mathbf{-\sin(x) ~\mathrm{d}x}\end{align} \]
- Plug in the given values of \( x \) and \( \mathrm{d}x \) and simplify.\[ \begin{align}\mathrm{d}y &= -\sin(3)(0.1) \\\mathbf{\mathrm{d}y} &= \mathbf{-0.1\sin(3)}\end{align} \]
Linear Approximations and Differentials – Key takeaways
- The linear approximation of a function is given by the equation:\[ L(x) = f(a) + f'(a) (x - a) \]
- The differential for a function, \( y = f(x) \), is given by the equation:\[ \mathrm{d}y = f'(x) ~\mathrm{d}x \]if \( x \) changes from \( a \) to \( a + \mathrm{d}x \).
- The differential is an approximation for the change in \( y \).
- The actual change in \( y \) is given by the equation:\[ \Delta y = f(a + \mathrm{d}x) - f(a) \]
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