Quickly find missing values between two known points (x1, y1) and (x2, y2) with our free Linear Interpolation Calculator. Just enter the coordinates, and it will compute the interpolated value for x or y instantly.
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What is Linear Interpolation?
Linear interpolation is a method used to calculate a value that lies between two known values on a straight line. Imagine you have two points on a graph, and you draw a straight line connecting them. If you want to find a value between these points, you assume that the change between them is consistent, and you use the straight line to estimate the unknown value.
Formula to Find Linear Interpolation
The formula for linear interpolation is:
Here:
- (x1, y1) and (x2, y2) are the two known points.
- x is the value at which you want to estimate y.
- y is the estimated value at x.
You can also find the value of x when you have a known value of y using linear interpolation. The formula for linear interpolation can be rearranged to solve for x when you know y.
How to Use the Calculator
- Enter x1 and y1 for the first known point.
- Enter x2 and y2 for the second known point.
- If you know x3 and want to find y3, enter the value of x3. If you know y3 and want to find x3, enter the value of y3.
- Here, x3 and y3 represent a point between the two known points. You’ll need to provide one of these values to find the other.
- Once you’ve entered these values, the linear interpolation tool will calculate and display the estimated value of the unknown coordinate (y3 or x3) based on the straight-line relationship between the two known points.
Step-by-Step Examples: Calculating Values Using Linear Interpolation
Example:
1. Find y3 when x3 = 6, given the points (4, 10) and (8, 20).
Solution:
– Let x1 = 4
– Let y1 = 10
– Let x2 = 8
– Let y2 = 20
– Let x3 = 6
– Let y3 = ?
Using the linear interpolation formula:
\[ y_3 = y_1 + \frac{(x_3 – x_1) (y_2 – y_1)}{x_2 – x_1} \]
Substituting values:
\[ y_3 = 10 + \left( \frac{(6 – 4) \times (20 – 10)}{8 – 4} \right) \]
\[ y_3 = 10 + \left( \frac{2 \times 10}{4} \right) \]
\[ y_3 = 10 + 5 \]
\[ y_3 = 15 \]
The value of y3 is 15
Example:
2. Find x3 when y3 = 18, given the points (3, 12) and (9, 24).
Solution:
– Let x1 = 3
– Let y1 = 12
– Let x2 = 9
– Let y2 = 24
– Let x3 = ?
– Let y3 = 18
Using the linear interpolation formula:
\[ x_3 = x_1 + \frac{(y_3 – y_1) (x_2 – x_1)}{y_2 – y_1} \]
Substituting values:
\[ x_3 = 3 + \frac{(18 – 12) (9 – 3)}{24 – 12} \]
\[ x_3 = 3 + \frac{(6) (6)}{12} \]
\[ x_3 = 3 + \frac{36}{12} \]
\[ x_3 = 3 + 3 \]
\[ x_3 = 6 \]
The value of x3 is 6
Example:
3. Estimating Temperature at a Specific Time.
Scenario:
At 2 PM, the temperature was 20℃.
At 5 PM, the temperature rose to 29℃.
We want to estimate the temperature at 3:30 PM.
Using linear interpolation, find the estimated temperature at 3:30 PM.
Solution:
– Let x1 = 2 (since 2 PM is 2 hours after noon).
– Let y1 = 20 (temperature at 2 PM).
– Let x2 = 5 (since 5 PM is 5 hours after noon).
– Let y2 = 29 (temperature at 5 PM).
– Let x3 = 3.5 (since 3:30 PM is 3.5 hours after noon).
– We need to find y3 (the estimated temperature at 3:30 PM).
Use the linear interpolation formula.
$$ y_3 = y_1 + \frac{(x_3 – x_1)(y_2 – y_1)}{(x_2 – x_1)} $$
Substituting values:
$$ y_3 = 20 + \frac{(3.5 – 2)(29 – 20)}{5 – 2} $$ $$ y_3 = 20 + \frac{1.5 \times 9}{3} $$ $$ y_3 = 20 + \frac{13.5}{3} $$ $$ y_3 = 20 + 4.5 $$ $$ y_3 = 24.5 $$
The estimated temperature at 3:30 PM is 24.5℃.
Can Linear Interpolation Be Used if the Data Points Are Not Linear?
If the data points are not on a straight line, using linear interpolation may not give accurate results. Linear interpolation assumes a constant rate of change between two points, forming a straight-line segment connecting them.
If the data points are very close to each other and the function is smooth, linear interpolation can give a rough estimate.
If the data points follows a nonlinear pattern, linear interpolation will lead to errors.
What Can Be Use Instead?
- Polynomial Interpolation (like Lagrange or Newton methods): Fits a curve through multiple points to model the data more accurately.
- Spline Interpolation: Uses piecewise polynomials to create a smooth curve that passes through all the data points.
- Cubic Interpolation: A better option for smooth transitions, giving more precise results for curved data.