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
- 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
- \(\displaystyle \frac{r}{h}=\frac{R}{H}=\frac{8}{12}=\frac{2}{3},\quad r=\frac{2h}{3}\)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
Difficulty levels
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