11.20 Small increments and approximations
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 |
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:
$\delta y\approx\frac{dy}{dx}\times\delta x$δy≈dydx×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:
$\delta y\approx f'\left(a\right)\times\delta x$δy≈f′(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}$3×43πr3−1 |
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 $\delta y\approx\frac{dy}{dx}\times\delta x$δy≈dydx×δx.
Think: The given formula for $\delta y$δy can be adapted for the volume function:
$\delta V\approx V'\left(r\right)\times\delta r$δV≈V′(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 $\delta y\approx\frac{dy}{dx}\times\delta x$δy≈dydx×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:
$\delta V\approx V'\left(r\right)\times\delta r$δV≈V′(r)×δr
And together we have: $\text{Percentage change}\approx\frac{V'\left(r\right)\times\delta r}{V\left(r\right)}\times100%$Percentage change≈V′(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%$4πr2×0.03r43πr3×100% |
|
|
| $=$= | $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
Determine $f'\left(x\right)$f′(x)
Express your answer in positive index form.
Use the formula $\delta y$δy$\approx$≈$f'\left(x\right)\times\delta x$f′(x)×δx to calculate the approximate change in $y$y when $x$x changes from $1$1 to $1.001$1.001
Express your answer as an exact value.
Question 2
Consider the surface area $S$S of a spherical balloon that has a radius measuring $r$r cm.
Determine $S'\left(r\right)$S′(r)
Using the formula $\delta y$δy$\approx$≈$f'\left(x\right)\times\delta x$f′(x)×δx, form an expression for the percentage error in the surface area of a sphere that corresponds to an error of $2%$2% in the radius.
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(t−2)6), where $t$t is the time in hours since midnight.
Determine $H'\left(t\right)$H′(t).
Use the increments formula to calculate the approximate change in height when $t$t changes from $4$4 to $4.2$4.2.
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.