Linear Interpolation

What Is Linear Interpolation?

Linear interpolation is a method used to find a new value that lies on a straight line between two known values. The name itself gives us clues about what it does. "Linear" tells us it deals with straight lines. The root word "inter" means "between," so we know we are working in the space between our known points. It's a way of making a smart, calculated guess for a point we don't have, assuming the trend between our known points is a steady, straight line.

Imagine you are on a long, straight road. You see a sign that says you are at mile marker 10. A while later, you see another sign for mile marker 30. If you wanted to know the exact spot of mile marker 20, you would naturally assume it's halfway between the two signs you saw. In essence, you just performed linear interpolation! You used two known points (marker 10 and marker 30) to estimate the location of an unknown point (marker 20) in between them.

In algebra, we do the same thing but with coordinates on a graph. If we know the coordinates of two points, we can connect them with a line segment. Any point that falls on that line segment can be found using linear interpolation.

Why Is Linear Interpolation Useful?

Linear interpolation isn't just a classroom exercise; it's used constantly in the real world to fill in missing information. Here are a few places where it comes in handy:

• Weather Forecasting: A weather station might record the temperature every hour. If it was ${62}^{\circ }F$ at 2:00 PM and ${68}^{\circ }F$ at 3:00 PM, a meteorologist can use linear interpolation to estimate the temperature at 2:30 PM was likely around ${65}^{\circ }F$.
• Computer Graphics and Animation: When animators create a character, they might define keyframes—for example, the character's arm is up at frame $10$ and down at frame $20$. The computer uses linear interpolation to automatically draw all the frames in between, creating smooth motion.
• Finance: Stock prices are recorded at the end of each day, but what if you want to estimate a stock's value at mid-day? Interpolation can provide a reasonable guess based on the previous and next day's closing prices.
• Science and Engineering: When conducting experiments, scientists collect data at specific intervals. If a measurement was missed or is needed at a point between two measurements, interpolation can be used to estimate the missing data point, assuming the process being measured is continuous and behaves linearly over short intervals.

In all these cases, linear interpolation provides a fast and simple way to make a very reasonable estimate without needing to perform a new measurement.

How Can We Visualize Linear Interpolation?

The easiest way to understand linear interpolation is to see it on a graph. Let's imagine we have two points, which we'll call $P_1$ and $P_2$. In algebra, we write these points with coordinates.

Let our first point be $P_1=(x_1,y_1)$ and our second point be $P_2=(x_2,y_2)$.

Now, picture these two points on a standard Cartesian coordinate plane. We can draw a straight line segment that connects them. Linear interpolation is all about finding another point, let's call it $P=(x,y)$, that lies somewhere on that line segment. We will know the $x$-coordinate of this new point, and our goal is to find its $y$-coordinate.

Because the point $(x,y)$ is on the same line as $(x_1,y_1)$ and $(x_2,y_2)$, the slope between $(x_1,y_1)$ and $(x,y)$ must be the same as the slope between $(x_1,y_1)$ and $(x_2,y_2)$. This is the fundamental concept behind the formula. You're essentially using the property of constant slope to find the missing $y$-value. The ratio of the vertical change to the horizontal change is consistent all along the line.

What Is the Linear Interpolation Formula?

To perform linear interpolation without drawing a graph every time, we use a standard formula. This formula is derived from the concept of similar triangles on the graph, or more simply, from the slope equation.

Given two known points $(x_1,y_1)$ and $(x_2,y_2)$, we can find the $y$-value for a given $x$ that lies between $x_1$ and $x_2$ using the following formula:

Let's break down each piece of the formula:

• $(x_1,y_1)$ is your first known coordinate pair (the starting point).
• $(x_2,y_2)$ is your second known coordinate pair (the ending point).
• $x$ is the input value for which you want to find the corresponding output value. It must be between $x_1$ and $x_2$.
• $y$ is the unknown output value you are solving for.

You might notice that the fraction part of the formula, $\frac{y_2-y_1}{x_2-x_1}$, is simply the formula for the slope of the line between the two known points! Let's call the slope $m$. The formula can then be rewritten as $y=y_1+m(x-x_1)$, which is just a rearranged version of the point-slope form of a linear equation, $y-y_1=m(x-x_1)$. So, if you know point-slope form, you already understand the foundation of linear interpolation.

How Do You Solve a Linear Interpolation Problem?

Solving a linear interpolation problem is a straightforward process. By following these steps, you can ensure you get the correct answer every time.

1. Identify Your Knowns: Read the problem carefully and identify your two known points, $(x_1,y_1)$ and $(x_2,y_2)$. Also, identify the $x$-value for which you need to find the corresponding $y$.
2. Assign Variables: Assign the smaller $x$-value to be $x_1$ and its corresponding $y$-value to be $y_1$. Assign the larger $x$-value to be $x_2$ and its $y$-value to be $y_2$. This helps keep your work organized.
3. Write Down the Formula: Write the linear interpolation formula on your paper. This helps you remember it and reduces the chance of errors.
   $y=y_1+(x-x_1)\frac{y_2-y_1}{x_2-x_1}$
4. Substitute the Values: Carefully substitute your known values for $x_1,y_1,x_2,y_2,$ and $x$ into the formula.
5. Calculate the Result: Follow the order of operations (PEMDAS/BODMAS) to solve for $y$.
   • First, solve the subtractions inside the parentheses: $(x-x_1)$, $(y_2-y_1)$, and $(x_2-x_1)$.
   • Next, perform the division to calculate the slope.
   • Then, multiply that result by $(x-x_1)$.
   • Finally, add $y_1$ to get your final answer for $y$.
6. Check Your Answer: Does your answer make sense? The resulting $y$-value should be between $y_1$ and $y_2$. If it's not, you may have made a calculation error.

Worked Examples of Linear Interpolation

Let's walk through a few examples to see how the formula works in practice.

What Are Common Mistakes to Avoid?

Linear interpolation is powerful, but small mistakes can lead to the wrong answer. Be on the lookout for these common errors:

• Mixing up Coordinates: A very common mistake is mixing up $x_1,y_1,x_2,$ and $y_2$. Always make sure that $y_1$ is the y-coordinate that goes with $x_1$, and $y_2$ goes with $x_2$. Writing them down as pairs, like $(x_1,y_1)$, can help prevent this.
• Forgetting Order of Operations (PEMDAS): The formula has addition, subtraction, multiplication, and division. You must follow the correct order. The most common error is adding $y_1$ before performing the multiplication. Remember to calculate the slope, multiply it by $(x-x_1)$, and then add $y_1$.
• Using Interpolation for Non-Linear Data: Linear interpolation works perfectly if the data follows a straight line. If the data actually follows a curve (like the path of a thrown ball), the linear estimate will be inaccurate. It's an approximation that is only as good as the assumption of linearity. For a slight curve, it might be a good estimate; for a sharp curve, it will be a poor one.
• Extrapolating Instead of Interpolating: Interpolation is estimating a value between known points. If you try to use the same formula to estimate a value outside the range of your known points (e.g., using data from day 5 and day 10 to predict day 20), it's called extrapolation. Extrapolation is much less reliable because you have no guarantee the linear trend continues.

Quick Summary

Here are the key takeaways for linear interpolation:

• Purpose: To estimate an unknown value that lies on a straight line between two known values.
• Core Idea: It assumes a constant rate of change (slope) between the two known points.
• The Formula:
  $y=y_1+(x-x_1)\frac{y_2-y_1}{x_2-x_1}$
• Key Check: Your final answer for $y$ should always fall between the values of $y_1$ and $y_2$.