Q.The length $L$ (in centimetre) of a copper rod is a linear function of its Celsius temperature $C$. In an experiment, if $L = 124.942$ when $C = 20$ and $L = 125.134$ when $C = 110$, express $L$ in terms of $C$.

Linear Interpolation: The Intuition

Imagine you're walking along a straight road between two towns. You know the distance to Town A (at the start) and the distance to Town B (at the end). If someone asks, "How far have you walked when you're one-third of the way there?" — you don't need a GPS. You just take one-third of the total distance and add it to the starting point.

That's linear interpolation in a nutshell: filling in the gaps between two known points by assuming the change happens at a constant rate.

The key assumption is linearity — the quantity changes by the same amount for each unit of the input. If the temperature at 2 PM was 30°C and at 4 PM it was 40°C, linear interpolation says at 3 PM it was exactly 35°C. The real temperature might have been 34°C or 36°C, but if you have no other data, the straight-line guess is the most natural one.

The Precise Statement

Let's formalise this. You have two known points:

• $(x_0, y_0)$ — the first point
• $(x_1, y_1)$ — the second point, with $x_1 \neq x_0$

You want to estimate the value $y$ at some $x$ that lies between $x_0$ and $x_1$.

Here's what each part means:

• $\frac{y_1 - y_0}{x_1 - x_0}$ is the slope — the rate of change of $y$ per unit change in $x$. It tells you how much $y$ increases (or decreases) when $x$ increases by 1.
• $(x - x_0)$ is how far you've moved from the starting point.
• Multiplying them gives the total change in $y$ from $x_0$ to $x$.
• Adding $y_0$ gives the final estimate.

A Worked Example

Suppose a car's speed is recorded at two times:

• At $t = 10$ seconds, speed $= 20$ m/s
• At $t = 20$ seconds, speed $= 50$ m/s

What was the speed at $t = 14$ seconds?

Here $x_0 = 10$, $y_0 = 20$, $x_1 = 20$, $y_1 = 50$, and $x = 14$.

Slope: $\frac{50 - 20}{20 - 10} = \frac{30}{10} = 3$ m/s per second.

Distance from start: $14 - 10 = 4$ seconds.

Change in speed: $3 \times 4 = 12$ m/s.

Estimated speed: $20 + 12 = 32$ m/s.

Why "Linear"?

The formula gives a straight line between the two points. If you plot $(x_0, y_0)$ and $(x_1, y_1)$ on a graph and draw a line through them, the interpolated value for any $x$ is simply the $y$-coordinate of that line at that $x$. The word "linear" refers to this straight-line relationship — not to the fact that the original data might be linear. …