 iGCSE (2021 Edition)

# 11.20 Small increments and approximations

Lesson

## Estimating change

We can estimate the change in the dependent variable of a function. The tangent at a point gives a linear approximation to the curve at that point. As such, provided we select a value of the independent variable close to the point of contact, the tangent can calculate a reasonable approximation for the change in the value of the function. Which we can also relate as follows: For a small change, $\delta x$δx, in the independent variable $x$x, the gradient of the secant between $A$A and $B$B will be approximately the gradient of the tangent at $A$A.

 Hence, $\frac{\delta y}{\delta x}$δyδx​ $\approx$≈ $\frac{dy}{dx}$dydx​ Therefore, $\delta y$δy $\approx$≈ $\frac{dy}{dx}\times\delta x$dydx​×δx

Estimation change

Given a small change in the independent variable, $\delta x$δx, the corresponding change in the dependent variable, $\delta y$δy, can be estimated by:

$\delta y\approx\frac{dy}{dx}\times\delta x$δydydx×δx

More specifically, the change in $y=f(x)$y=f(x), from $x=a$x=a to $x=a+\delta x$x=a+δx, can be estimated by:

$\delta y\approx f'\left(a\right)\times\delta x$δyf(a)×δx

And the percentage change (or percentage error) can be calculated by:

 Percentage change $=$= $\frac{\text{Change}}{\text{Original}}\times100%$ChangeOriginal​×100% $=$= $\frac{\delta y}{y}\times100%$δyy​×100% $\approx$≈ $\frac{\frac{dy}{dx}\times\delta x}{y}\times100%$dydx​×δxy​×100%

#### Worked example

##### Example 2

Consider the volume $V$V of a balloon that is in the shape of a sphere of radius $r$r cm.

a) Determine $V'\left(r\right)$V(r).

Think: The volume of a sphere is given by $V=\frac{4}{3}\pi r^3$V=43πr3, differentiate this in terms of $r$r.

Do:

 $V'\left(r\right)$V′(r) $=$= $3\times\frac{4}{3}\pi r^{3-1}$3×43​πr3−1 Using the power rule $=$= $4\pi r^2$4πr2

b) Find the approximate change in the volume of a spherical balloon when the radius changes from $5$5 cm to $5.01$5.01 cm, using $\delta y\approx\frac{dy}{dx}\times\delta x$δydydx×δx.

Think: The given formula for $\delta y$δy can be adapted for the volume function:

$\delta V\approx V'\left(r\right)\times\delta r$δVV(r)×δr

We are estimating the change in $V$V from $r=5$r=5 to $r=5.01$r=5.01, for our formula we require $V'\left(5\right)$V(5) and the change in $r$r$\delta r=0.01$δr=0.01.

Do:

 $\delta V$δV $\approx$≈ $V'\left(r\right)\times\delta r$V′(r)×δr $=$= $V'\left(5\right)\times0.01$V′(5)×0.01 $=$= $4\pi\left(5\right)^2\times0.01$4π(5)2×0.01 $=$= $100\pi\times0.01$100π×0.01 $=$= $\pi$π

Thus, if the radius of the balloon changed from $5$5 cm to $5.01$5.01 cm the volume would change by approximately $\pi$π cm3.

Reflect: We can check the actual change by calculating $V\left(5.01\right)-V\left(5\right)$V(5.01)V(5) to see if our answer is a reasonable approximation.

c) Find the approximate percentage change in the volume of the balloon that corresponds to a $3%$3% increase in its radius, using $\delta y\approx\frac{dy}{dx}\times\delta x$δydydx×δx.

Think: To get the percentage change, we need to divide the change in $V$V by $V$V, and multiply this by $100%$100%. That is:

$\text{Percentage change}=\frac{\delta V}{V}\times100%$Percentage change=δVV×100%

The given formula for $\delta y$δy can be adapted for the volume function:

$\delta V\approx V'\left(r\right)\times\delta r$δVV(r)×δr

And together we have: $\text{Percentage change}\approx\frac{V'\left(r\right)\times\delta r}{V\left(r\right)}\times100%$Percentage changeV(r)×δrV(r)×100%, with the change in radius being $3%$3% of $r$r or $0.03r$0.03r.

Do:

 Percentage change $=$= $\frac{V'\left(r\right)\times\delta x}{V\left(r\right)}\times100%$V′(r)×δxV(r)​×100% $=$= $\frac{4\pi r^2\times0.03r}{\frac{4}{3}\pi r^3}\times100%$4πr2×0.03r43​πr3​×100% $=$= $3\times0.03\times100%$3×0.03×100% $=$= $9%$9%

Hence, a $3%$3% change in radius will lead to approximately a $9%$9% change in volume.

#### Practice questions

##### Question 1

Consider the function $y=$y=$f\left(x\right)=\frac{4}{x^2}+\sqrt{x}$f(x)=4x2+x

1. Determine $f'\left(x\right)$f(x)

##### Question 3

In a particular bay, the height in metres of the tide above the mean sea level is given by $H\left(t\right)=4\sin\left(\frac{\pi\left(t-2\right)}{6}\right)$H(t)=4sin(π(t2)6), where $t$t is the time in hours since midnight.

1. Determine $H'\left(t\right)$H(t).

2. Use the increments formula to calculate the approximate change in height when $t$t changes from $4$4 to $4.2$4.2.

3. Hence, estimate the percentage change in height when $t$t changes from $4$4 to $4.2$4.2.

Round your percentage to two decimal places.

### Outcomes

#### 0606C14.5D

Apply differentiation to small increments and approximations.