14.7 Rates of Change and Approximations

Part 1: Small Increments and Approximations

The Big Idea: Connecting δy and δx Using the Derivative

Imagine you have a curve y = f(x). Pick a point P(x, y) on the curve. Now move to a nearby point Q(x + δx, y + δy) — also on the curve, but very close to P.

• The gradient of the chord PQ (the straight line connecting P and Q) is: $\dfrac{\delta y}{\delta x}$
• The gradient of the tangent at P is: $\dfrac{dy}{dx}$

When P and Q are very close together (i.e., δx is very small), the chord PQ almost lines up with the tangent. This means:

$\frac{\delta y}{\delta x} \approx \frac{dy}{dx}$

Rearranging this gives the key formula for small increments:

$\boxed{\delta y \approx \frac{dy}{dx} \times \delta x}$

In plain English: To find the approximate change in y (called δy), multiply the derivative $\dfrac{dy}{dx}$ by the small change in x (called δx).

Step-by-Step Method for Small Increments

Given: an equation connecting x and y, and a small change δx.

To find δy (the approximate change in y):

1. Differentiate y with respect to x to get $\dfrac{dy}{dx}$.
2. Substitute the given value of x into $\dfrac{dy}{dx}$.
3. Multiply by δx: $\delta y \approx \frac{dy}{dx} \times \delta x$

✏️ Worked Example 1

Step 1: Identify δx. $\delta x = 2.05 - 2 = 0.05$

Step 2: Differentiate. $\frac{dy}{dx} = 3x^2 + 2x$

Step 3: Substitute x = 2. $\frac{dy}{dx} = 3(2)^2 + 2(2) = 12 + 4 = 16$

Step 4: Use the formula. $\delta y \approx 16 \times 0.05 = 0.8$

✅ The approximate increase in y is 0.8.