How to use small changes and approximation

In Add Math, $\frac{dy}{dx}$ usually stands for an instantaneous rate of change. Small changes puts the same derivative to a different use: for a very small change $\delta x$ in $x$, the curve looks almost exactly like its own tangent line over that tiny stretch, so the resulting small change in $y$, written $\delta y$, is well approximated by $\delta y \approx \frac{dy}{dx}\,\delta x$.

This lets us do two useful things without a full recalculation: estimate the value of an expression that sits close to a value we can already work out exactly, such as $(2.03)^{3}$ sitting close to $2^{3}$; and find how much one quantity is affected when a shape's dimension changes by a small amount, such as the area of a circle when its radius increases slightly. Both are the same idea, trade an exact but awkward calculation for a fast, reliable estimate built from the gradient at a point you already know.

Watch for a question that specifically asks you to estimate or find the approximate value of an expression such as $(2.03)^{3}$ or $\sqrt{4.02}$, a number that sits close to a value you can compute exactly and easily. This is different from finding an exact value, and avoiding that longer exact calculation is the whole point of the method.

It also appears whenever one quantity depends on another through a formula, area, volume, perimeter, and a dimension changes by a small amount. A circle of radius $r$ has area $A=\pi r^{2}$; differentiating gives $\frac{dA}{dr}=2\pi r$, so a small increase $\delta r$ in the radius produces a small change $\delta A \approx 2\pi r\,\delta r$ in the area.

For example, if $r=5$ cm and $\delta r=0.1$ cm, then $\delta A \approx 2\pi(5)(0.1)=\pi \approx 3.14$ cm², no need to work out two full areas and subtract.

1. 1

   Identify y and the known point

   Write the quantity as $y=f(x)$ and note a value of $x$, call it $a$, where $y$ is easy to work out exactly.

2. 2

   Differentiate

   Find $\frac{dy}{dx}$ as a function of $x$.

3. 3

   Evaluate the gradient at the known point

   Substitute $x=a$ into $\frac{dy}{dx}$, never the new, unknown value of $x$.

4. 4

   State the small change δx

   Work out $\delta x$, the small change in $x$ from $a$ to the new value, keeping the sign: positive for an increase, negative for a decrease.

5. 5

   Apply the formula

   Substitute into $\delta y \approx \frac{dy}{dx}\,\delta x$ to get the approximate small change in $y$.

6. 6

   Give the value asked for

   If the question wants the change, state $\delta y$. If it wants the new value of $y$, add it on: $y_{\text{new}} \approx y+\delta y$.

• Evaluating $\frac{dy}{dx}$ at the new, unknown value of $x$ (such as $2.03$) instead of the known starting point $x=2$, the gradient must come from the point you already know exactly.
• Getting the sign of $\delta x$ wrong, a decrease, such as from $5$ cm to $4.98$ cm, gives $\delta x=-0.02$, not $+0.02$.
• Stopping at $\delta y$ when the question asks for the approximate new value, remember to add it back on: $y_{\text{new}} \approx y+\delta y$.
• Applying the method to a change that is not actually small, which makes the straight-line approximation unreliable; it only works well when $\delta x$ is small.
• In area or volume questions, forgetting to convert a percentage change into an actual small number for $\delta x$ before substituting it into the formula.

The step students most often lose marks on is evaluating $\frac{dy}{dx}$ at the wrong point, substituting the new, approximate value of $x$ instead of the known one. In a one-to-one lesson our teachers watch you set out $y$, $\frac{dy}{dx}$ and $\delta x$ line by line and catch that slip immediately, so choosing the right point to substitute becomes automatic.

Because your working is shown step by step, it is easy to see exactly where $\delta y$ is added back to $y$ to answer the question that was actually asked. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

If you would like to see how we teach small changes and approximation, message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.