Average and instantaneous rates
An average rate uses two input values, while an instantaneous rate describes the change at one input. The derivative supplies the instantaneous rate for a differentiable quantity function
\[\overline{r}=\frac{Q(b)-Q(a)}{b-a},\quad r(t)=Q'(t)\]
- Identify the quantities State which quantity changes and which variable measures the change, then keep their units in that order \(\displaystyle Q=\text{quantity},\quad t=\text{input time}\)
- Find an average rate Use the change in the quantity divided by the change in the input over the stated interval \(\displaystyle \overline{r}=\frac{Q(b)-Q(a)}{b-a}\)
- Find an instantaneous rate Differentiate the quantity function and substitute the required input value \(\displaystyle r(t)=Q'(t)\)
- Interpret the sign and units A negative instantaneous rate describes a decrease at that input. A negative average rate means a net decrease over the interval, not necessarily a decrease throughout it. Units are quantity units per input unit \(\displaystyle \frac{\text{quantity units}}{\text{input units}}\)
Worked example
Water fills a right circular cone with its vertex at the bottom, height 12 cm and radius 8 cm. At the instant when the water depth is 4 cm, the depth is increasing at 2 cm per minute. Find the rate at which the water volume is increasing
\[H=12,\quad R=8,\quad h=4,\quad \frac{dh}{dt}=2\]
- \(\displaystyle \frac{r}{h}=\frac{R}{H}=\frac{8}{12}=\frac{2}{3},\quad r=\frac{2h}{3}\)Similar cross-sections give the water radius r in terms of its depth h.
- \(\displaystyle V=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi\left(\frac{2h}{3}\right)^2h=\frac{4}{27}\pi h^3\)Substitute the radius relation into the cone-volume formula
- \(\displaystyle \frac{dV}{dh}=\frac{4}{9}\pi h^2,\quad \left.\frac{dV}{dh}\right|_{h=4}=\frac{64}{9}\pi\)Differentiate with respect to depth and substitute the stated depth
- \(\displaystyle \frac{dV}{dt}=\frac{dV}{dh}\frac{dh}{dt}=\frac{64}{9}\pi\times2=\frac{128}{9}\pi\text{ cm}^3\text{ per minute}\)Multiply the volume change per centimetre of depth by the depth rate
Common mistakes
Forgetting the interval width
✕\(\displaystyle \overline{r}=Q(b)-Q(a)\)✓\(\displaystyle \overline{r}=\frac{Q(b)-Q(a)}{b-a}\)
An average rate divides the total change by the interval width, so an endpoint difference alone is not a rateUsing a quantity value as a rate
✕\(\displaystyle Q(t)=t^2+4t+10,\quad r(2)=Q(2)=22\)✓\(\displaystyle r(2)=Q'(2)=8\)
Q(2) is the quantity value, whereas Q'(2) is its instantaneous rate of changeInverting the units
✕\(\displaystyle \frac{\text{minutes}}{\text{unit}}\)✓\(\displaystyle \frac{\text{unit}}{\text{minute}}\)
A rate of quantity with respect to time measures quantity units for each time unitDifficulty levels
introFind an average or instantaneous rate for a polynomial quantity function
standardUse the stated geometric constraint for a fixed cone or an explicitly similar family of cylinders
challengeCompare rates, interpret signs and report units for a geometric or algebraic model
examChoose a suitable rate method for a geometric or interpretation question and state the result with units
Questions about this topic
What is the difference between an average and an instantaneous rate?+
An average rate uses a finite interval, while an instantaneous rate is the derivative at one input value
Why are similar triangles used in a related-rates problem?+
In a fixed cone, similar cross-sections give a fixed radius-to-depth ratio. For a family of cylinders, a fixed height-to-radius ratio must be stated; an arbitrary changing cylinder need not remain similar
What does a negative rate of change mean?+
A negative average rate means the final quantity is smaller than the initial quantity. A negative derivative describes the instantaneous rate at one input; to conclude that the quantity decreases throughout an interval, establish that the derivative is negative throughout it
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 6: Applications of differentiation
In this chapter
Differentiate polynomialsChallenge enrichment: rational powers and derivative domainsDifferentiate natural exponential functions and solve simple derivative conditionsDifferentiate sine, cosine and tangent functions and solve simple derivative conditionsDifferentiate a composite function by the chain ruleDifferentiate a product or quotient of two functionsEvaluate and determine function rules6B Differentiation rules: Find the maximal domain of an algebraic functionExpand and factor polynomial expressionsSimplify and combine algebraic fractions6B Differentiation rules: Find the equation of a line from its graph or from its intercept and slopeAnalyse polynomial curves, stationary points and extremaUse a derivative to analyse and recognise the graph of a function6C Tangents, normals and curve sketching: Evaluate a polynomial or recover a coefficient parameter6C Tangents, normals and curve sketching: Move between roots, factors, polynomial rules and equationsRecognise and classify a polynomialConnect a factorised higher-degree rule and its graphAnalyse a cubic graph or determine its ruleWork with sets and interval notation6C Tangents, normals and curve sketching: Find the maximal domain of an algebraic function6C Tangents, normals and curve sketching: Find the equation of a line from its graph or from its intercept and slopeForm an equation of a parallel or perpendicular lineSolve a maximum or minimum problem by differentiationSet up and solve a practical linear equationRearrange literal equations and recognise equivalent line forms6D Optimisation: Find the maximal domain of an algebraic function6D Optimisation: Evaluate a polynomial or recover a coefficient parameter6D Optimisation: Move between roots, factors, polynomial rules and equationsCalculate and interpret average and instantaneous rates from a modelVolume of cones, pyramids and spheresVolume of prisms and cylindersPerimeter and area of standard shapesSimilar figures and scale factorsFind a side of a right-angled triangleAnalyse straight-line motion through displacement, velocity and acceleration6F Kinematics: Evaluate a polynomial or recover a coefficient parameter