Rates Of Change
Learn to apply rates of change for Queensland Year 11 Mathematical Methods (QCAA). Here the derivative becomes a real-world rate — how fast a quantity such as volume, temperature or population is changing.
You will learn to build a rate rule from a quantity, read off the instantaneous rate of change at an instant, find when a rate reaches a set value, and locate maximum or minimum values with a sign test.
Every question with a fully worked solution.
- Rates Of Change - Video - Rates of change Watch
Theory
In Year 11 Mathematical Methods, the derivative gives the instantaneous rate of change of a quantity: if \(Q\) depends on \(t\), then \(\dfrac{dQ}{dt}\) is the rate at which \(Q\) changes at that instant. This page shows how to find a rate rule by differentiating a polynomial, evaluate the rate at an instant, find when a rate takes a given value, and use \(\dfrac{dQ}{dt}=0\) with a first-derivative sign test to find a maximum or minimum value.
The instantaneous rate of change of a quantity \(Q(t)\) is the value of its derivative \(\dfrac{dQ}{dt}\) at a particular instant. It is the gradient of the tangent to the graph of \(Q\) against \(t\) at that point, so it measures how fast \(Q\) is changing right then.
A positive rate means \(Q\) is increasing; a negative rate means \(Q\) is decreasing; a zero rate (\(\dfrac{dQ}{dt}=0\)) means \(Q\) is momentarily neither rising nor falling, which is where a maximum or minimum value can occur.
For a quantity \(Q\) depending on \(t\), the instantaneous rate of change is the derivative:
The rate at a particular instant \(t=a\) is found by substitution:
How to work with a rate of change
- Differentiate: apply the power rule to \(Q(t)\) to build the rate rule \(\dfrac{dQ}{dt}\).
- Substitute or solve: for a rate at an instant, substitute the time; to find when a rate is reached, set \(\dfrac{dQ}{dt}\) equal to that value and solve.
- Interpret: attach the correct unit (quantity per unit time) and read the sign — positive increasing, negative decreasing, zero stationary.
Rate rule — differentiate \(h(t)\):
| \(\dfrac{dh}{dt}\) | \(=\) | \(t+2\) |
Value at \(t=6\) — substitute:
| \(\left.\dfrac{dh}{dt}\right|_{t=6}\) | \(=\) | \((6)+2\) |
| \(=\) | \(8\) |
The height is increasing at \(8\) cm/week.
Rate rule — differentiate:
| \(\dfrac{dV}{dt}\) | \(=\) | \(60-6t\) |
Value at \(t=4\):
| \(\left.\dfrac{dV}{dt}\right|_{t=4}\) | \(=\) | \(60-6(4)\) |
| \(=\) | \(60-24\) | |
| \(=\) | \(36\) |
The rate is positive, so water is still flowing in.
At \(t=4\) min the volume is increasing at \(36\) L/min.
Rate rule:
| \(\dfrac{dQ}{dt}\) | \(=\) | \(3t^2-18t+15\) |
Set the rate to zero and solve:
| \(3t^2-18t+15\) | \(=\) | \(0\) |
| \(3(t^2-6t+5)\) | \(=\) | \(0\) |
| \(3(t-1)(t-5)\) | \(=\) | \(0\) |
| \(t\) | \(=\) | \(1 \ \text{or}\ 5\) |
The rate of change is zero at \(t=1\) and \(t=5\).
Rate rule — differentiate, then solve \(\dfrac{dP}{dx}=0\):
| \(\dfrac{dP}{dx}\) | \(=\) | \(-4x+40\) |
| \(-4x+40\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(10\) |
First-derivative sign test around \(x=10\):
| \(\left.\dfrac{dP}{dx}\right|_{x=9}\) | \(=\) | \(-4(9)+40=4>0\) |
| \(\left.\dfrac{dP}{dx}\right|_{x=11}\) | \(=\) | \(-4(11)+40=-4<0\) |
The rate goes from positive to negative, so \(x=10\) gives a maximum.
Maximum profit — substitute \(x=10\) into \(P\):
| \(P(10)\) | \(=\) | \(-2(10)^2+40(10)-50\) |
| \(=\) | \(-200+400-50\) | |
| \(=\) | \(150\) |
Profit is greatest at \(x=10\) thousand items, giving \(\$150\,000\).
Common pitfalls
Frequently asked questions
What is an instantaneous rate of change?
It is the derivative \(\dfrac{dQ}{dt}\) evaluated at an instant — the gradient of the tangent to the graph of \(Q\) against \(t\) at that point.
How do I find the rate of change at a particular time?
Differentiate \(Q(t)\) to get \(\dfrac{dQ}{dt}\), then substitute the time. The result is a signed value with a per-unit-time unit.
How do I find when a rate of change equals a given value?
Set \(\dfrac{dQ}{dt}\) equal to that value and solve for \(t\). For a stationary value, set \(\dfrac{dQ}{dt}=0\).
How do I find a maximum or minimum value of a quantity?
Solve \(\dfrac{dQ}{dt}=0\), then use a first-derivative sign test: if the rate goes \(+\) to \(-\) it is a maximum, and \(-\) to \(+\) a minimum.
What does the sign of the rate tell me?
A positive rate means the quantity is increasing, a negative rate means it is decreasing, and a zero rate means it is momentarily stationary.
Do I need the second derivative for these problems?
No. In Year 11 Methods the nature of a maximum or minimum is confirmed with the first-derivative sign test, not the second derivative.