Why Time Variables Matter
In a related-rates problem, the quantities of interest may change while time passes. A water level rises, a vehicle changes position, or a rotating arm changes its angle. To describe these changes precisely, represent each changing quantity as a function of time.
This choice of variables is an important first step. It makes clear which values can vary, what each symbol measures, and which instantaneous rates may be relevant. A fixed dimension, by contrast, is a parameter: it helps describe the physical situation but does not change with time.
For example, the water depth in a tank might be written \(h(t)\), while the tank's fixed radius is written \(R\). The notation \(h(t)\) emphasizes that the depth can depend on time. Writing \(R\), without a time argument, emphasizes that the radius is fixed in the situation being modelled.
Choosing the Time-Dependent Variables
Begin by identifying the physical quantities that may change. Give each one a symbol and attach \(t\) as its argument. Common choices include \(x(t)\) for a position, \(r(t)\) for a radius, \(h(t)\) for a height, and \(\theta(t)\) for an angle. The symbol should match what the quantity measures, and its units should be stated.
Next, identify the quantities that are fixed in the model. These may include a rigid object's length, a tank's radius, or the height of a stationary structure. Give these quantities symbols too, but do not treat them as functions of time unless the situation explicitly says they change.
It is useful to separate three kinds of information:
- Changing values: for example, \(h(t)\), the water depth at time \(t\).
- Instantaneous rates: for example, \(\frac{dh}{dt}\), the rate at which that depth is changing at time \(t\).
- Fixed parameters: for example, \(R\), the radius of a tank that is not changing shape.
A rate is not the same thing as a quantity. If \(h(t)\) is measured in metres and \(t\) in minutes, then \(\frac{dh}{dt}\) is measured in metres per minute. The numerical value of \(h(t)\) tells you the depth at an instant; the numerical value of \(\frac{dh}{dt}\) tells you how quickly the depth is changing at that instant.
What an Instantaneous Rate Means
The derivative of a time-dependent quantity describes its instantaneous change. For a quantity \(q(t)\), its derivative at time \(t\) is defined by the limit of average rates over shorter and shorter time intervals, provided that limit exists:
Here, \(\Delta t\) is a change in time, and \(q(t+\Delta t)-q(t)\) is the corresponding change in the quantity. The quotient is an average rate over that time interval. The limit gives the instantaneous rate.
This definition explains why assigning a time function matters: the derivative describes how that function changes as its input, time, changes. It also explains the units. If \(q\) has units \(U\) and time has units \(T\), then the quotient has units \(U/T\), and so does the derivative.
Worked Example: Water Depth in a Tank
Worked Example: Water Depth in a Tank
A tank has a fixed radius \(R=0.75\) m. Its water depth, measured in metres, is modelled for \(t\) minutes by \(h(t)=0.30+0.05t\). Find the depth and its instantaneous rate of change at \(t=6\) min, and identify the fixed parameter.
The water depth changes with time, so represent it by \(h(t)\). The tank radius does not change, so \(R=0.75\) m is a fixed parameter.
First calculate the depth at \(t=6\):
Differentiate the time function to find its rate:
The rate is constant, so at \(t=6\) min it is still \(0.05\) m/min. Check the derivative directly from the definition. For any nonzero \(\Delta t\),
Thus the depth at that instant is \(0.60\) m, and it is increasing at \(0.05\) m/min. The radius \(R\) is a parameter, not a changing water-level variable.
Worked Example: Position of an Elevator
Worked Example: Position of an Elevator
An elevator's vertical position relative to a fixed ground-level reference is \(y(t)=2+1.5t-0.25t^2\) metres, where \(t\) is measured in seconds. Find its position and instantaneous rate of change at \(t=2\) s. Treat the ground-level reference as fixed.
Position changes over time, so it is represented by \(y(t)\). The reference level does not move; it is not another time-dependent position in this model.
Substitute \(t=2\) into the position function:
Differentiate to obtain the instantaneous rate of change of position:
Check the rate using the difference quotient. For a time increment \(\Delta t\), first expand the position at \(2+\Delta t\):
Therefore, for \(\Delta t\ne 0\),
Taking the limit as \(\Delta t\to0\) gives \(0.5\) m/s, matching the derivative. At \(t=2\) s, the elevator is \(4\) m from the reference level, and its position is changing at \(0.5\) m/s.
Worked Example: Angle of a Rotating Arm
Worked Example: Angle of a Rotating Arm
A rigid arm of fixed length \(L=8\) m rotates about a fixed point. Its angle from a reference direction is \(\theta(t)=\frac{\pi}{6}+\frac{\pi}{60}t\) radians, where \(t\) is measured in minutes. Find the angle and its instantaneous rate of change at \(t=10\) min, and distinguish the changing quantity from the fixed one.
The angle changes, so write it as \(\theta(t)\). The arm's length remains \(L=8\) m, so \(L\) is a fixed parameter.
Evaluate the angle at \(t=10\):
Differentiate the angle function:
For a check, the difference quotient is
It agrees with the derivative. The angle at \(10\) min is \(\frac{\pi}{3}\) rad, its rate of change is \(\frac{\pi}{60}\) rad/min, and the arm length remains the fixed value \(8\) m.
A Common Mistake: Giving a Fixed Parameter a Changing Rate
Suppose a rigid ladder has length \(L=5\) m throughout a situation. A common error is to write \(\frac{dL}{dt}=5\) m/s, as though the length were changing at a rate equal to its value. This confuses a quantity with its rate and gives the rate the wrong units: \(L\) is measured in metres, while \(\frac{dL}{dt}\) would be measured in metres per second.
In this model the ladder's length is constant in time. If written as a function at all, it would be \(L(t)=5\) m, and therefore
The changing position of one end of the ladder would need its own time-dependent variable. Naming that position does not make the ladder's fixed length change.
A Reliable Naming Process
Before using rates, make a short inventory of the situation. This prevents fixed dimensions from being mixed up with changing quantities.
Look for quantities described as moving, growing, shrinking, filling, draining, or rotating.
Represent each changing quantity by a function such as \(x(t)\), \(h(t)\), or \(\theta(t)\), and state its units.
Record fixed lengths, radii, heights, and reference values without treating them as changing quantities.
For a quantity \(q(t)\), use \(q'(t)\) or \(\frac{dq}{dt}\), with units of quantity per unit time.
A value answers “how much at this time?” A rate answers “how quickly is it changing at this time?”
This inventory establishes the roles of the symbols before any relationship among the changing quantities is used. Once the quantities have been named, the next task in a related-rates problem is to express how they are connected. Keeping those two tasks distinct makes it easier to interpret a later rate calculation correctly.
Check Your Understanding
For each question, pay attention to which symbols represent changing quantities and which represent fixed parameters.
- A circular pond's radius is fixed, while its water depth changes. Which quantity should be written as a function of time, and which is a fixed parameter?
- If \(x(t)\) is a position measured in feet and \(t\) is in seconds, what are the units of \(x'(t)\)?
- A rigid rod has length \(7\) m throughout a model. What is its rate of change with respect to time?
- For \(q(t)=4+3t\) centimetres, find \(q(2)\) and \(q'(2)\), including units if time is measured in seconds.
- In your own words, distinguish the value \(h(5)\) from the rate \(h'(5)\).
