A Consistent Strategy for Related Rates
Related-rates problems describe two or more quantities that change with time and are connected by a relationship. The goal is to use the rate of change of one quantity to find the rate of another at a particular instant. The geometry or physical setup supplies the relationship; differentiation with respect to time turns it into a relationship between rates.
The main challenge is often not the differentiation itself. It is deciding which quantities to name, expressing the constraint accurately, keeping track of signs and units, and waiting until the constraint has been differentiated before inserting instantaneous values.
A dependable solution has four central parts: define time-dependent variables, write a constraint connecting them, differentiate that constraint with respect to time, and evaluate the resulting equation at the specified instant. Then interpret the sign, units, and direction in context.
Why Differentiating the Constraint Works
Suppose changing quantities \(x(t)\) and \(y(t)\) satisfy a constraint \(F(x(t),y(t))=0\). Since this equation holds as time passes, its left side is a constant function of time. The derivative of a constant is zero. Applying the chain rule gives the rate equation.
For a constraint written \(y=g(x)\), the same reasoning gives \(dy/dt=g'(x)\,dx/dt\). This is the chain rule: \(y\) changes because \(x\) changes, and the rate of \(y\) is the rate of change of \(g\) with respect to \(x\), multiplied by the rate of \(x\) with respect to time.
A Four-Step Workflow
Use variables that match the diagram or physical setup. State what each variable measures and use consistent units.
Use geometry, a volume formula, or a length balance to relate the variables. Include any fixed quantities and explain why they are fixed.
Apply the chain rule to every changing quantity. Keep the rate symbols visible so their signs and units can be tracked.
Insert values for the instant only after differentiating. Solve for the requested rate, then state its units and what its sign means.
Worked Example: A Person’s Shadow
Worked Example: Finding the Speed of a Shadow Tip
A 6-meter lamp post casts a shadow of a person who is 2 meters tall. The person walks directly away from the post at \(1.5\) meters per second. How quickly is the tip of the shadow moving when the person is 8 meters from the post?
Let \(x(t)\) be the person’s distance from the post, and let \(s(t)\) be the length of the person’s shadow. Both are measured in meters. The person walks away, so \(dx/dt=1.5\) meters per second. The shadow tip is \(x+s\) meters from the post.
The two right triangles formed by the lamp and the person’s shadow are similar. Their height-to-base ratios give
Cross-multiplying gives \(6s=2x+2s\), so \(4s=2x\), and therefore \(s=x/2\). Differentiate this relationship with respect to time:
The tip’s distance from the post is \(x+s\), so its rate is
At the stated instant, \(x=8\) meters, so \(s=8/2=4\) meters. The person-to-tip distance is 4 meters, and the tip is \(8+4=12\) meters from the post. The answer is \(2.25\) meters per second, away from the post. As a check, the similar-triangle equation gives \(6/(8+4)=6/12=1/2\) and \(2/4=1/2\); the geometry and the calculated shadow length agree.
Worked Example: Filling a Cylindrical Container
Worked Example: Water Level in a Cylinder
Water enters a vertical cylindrical tank of radius 2 meters at \(0.6\) cubic meters per minute. How quickly is the water depth increasing?
Let \(h(t)\) be the water depth in meters and \(V(t)\) the water volume in cubic meters. The radius is fixed at 2 meters, while the volume changes at \(dV/dt=0.6\) cubic meters per minute. The cylinder’s water volume is
Since \(r=2\) meters is constant, substitute it into the relationship: \(V=4\pi h\). Differentiate with respect to time:
Insert the known volume rate and solve:
The level is rising because the result is positive. The units also check: \((\text{m}^3/\text{min})/\text{m}^2=\text{m}/\text{min}\). To verify the rate, substitute it into the differentiated volume equation: \(4\pi(3/(20\pi))=12/20=0.6\) cubic meters per minute, as given. No particular value of \(h\) is needed because the tank’s cross-sectional area stays constant.
Worked Example: Two Moving Objects
Worked Example: Rate of Separation
One object moves east from an intersection at 3 meters per second. At the same time, another moves north from the same intersection at 4 meters per second. How quickly is the distance between them increasing 5 seconds after they start?
Let \(x(t)\) and \(y(t)\) be the eastward and northward distances, respectively, and let \(r(t)\) be the distance between the objects. The rates are \(dx/dt=3\) meters per second and \(dy/dt=4\) meters per second. By the Pythagorean theorem, the constraint is
Differentiate every changing quantity with respect to time:
After 5 seconds, \(x=3(5)=15\) meters and \(y=4(5)=20\) meters. The distance is
Substitute these values and the rates into the differentiated equation:
The separation is increasing at 5 meters per second. Check the geometry: \(15^2+20^2=625=25^2\). Check the rate equation: \(2(25)(5)=250\), while \(2(15)(3)+2(20)(4)=90+160=250\). Both sides agree.
Worked Example: A Movable Pulley
Worked Example: Relating Rope and Load Speeds
A load hangs from a movable pulley supported by two vertical segments of one rope. The free end of the rope is pulled downward at \(0.8\) meters per second. How quickly does the load move?
Let \(y(t)\) be the downward distance of the load from a fixed reference point, and let \(x(t)\) be the downward distance of the free end. The two rope segments supporting the movable pulley each lengthen by the load’s downward displacement. The free-end segment changes by \(x\). Any other parts of the rope have fixed length, denoted by \(C\). Thus the rope-length constraint is
where the total rope length \(L\) and \(C\) are constant. Differentiating with respect to time gives
The free end moves downward, so \(dx/dt=0.8\) meters per second. Therefore,
Because \(y\) is measured downward, the negative sign means that the load moves upward. Its speed is \(0.4\) meters per second upward. Substitution into the differentiated constraint checks the result: \(2(-0.4)+0.8=-0.8+0.8=0\), so the total rope length stays constant.
A Common Pitfall: Substituting Before Differentiating
In the moving-objects example, the distance at 5 seconds is \(r=25\) meters. A tempting but incorrect approach is to substitute the instant’s values into the original equation and then treat them as constants. For example, writing \(25^2=15^2+20^2\) and differentiating those numerical values gives \(0=0\). This says nothing about the rates: those numbers describe one instant, not the quantities throughout the motion.
The correction is to differentiate the equation while \(r\), \(x\), and \(y\) still represent changing quantities:
Only after obtaining this rate equation should the instantaneous values \(r=25\), \(x=15\), and \(y=20\) be substituted. This order preserves the chain-rule factors that describe how the quantities are changing.
Interpreting and Checking a Result
A numerical rate is not a complete answer unless it is tied to the question. A useful final statement identifies the quantity, whether it is increasing or decreasing, its direction where relevant, and its units. For example, “the load moves upward at \(0.4\) meters per second” is clearer than giving only \(-0.4\).
Before finishing, check the model and calculation:
- Does the constraint reflect the stated geometry or physical setup?
- Were all changing variables differentiated with respect to time?
- Were instantaneous values substituted only after differentiation?
- Do the signs match the chosen positive directions?
- Do the units reduce to the requested distance, area, volume, or angle per unit time?
- Does substituting the answer into the differentiated constraint reproduce the known rates?
These checks are especially useful when a diagram contains several lengths or when a pulley’s rope segments move in opposite directions. A correct derivative of an incorrect constraint still gives an incorrect model, so the relationship itself deserves careful attention.
Check Your Understanding
For each problem, identify the changing variables, write a constraint, and decide what rate information is needed before calculating.
- A lamp post is 5 meters tall, and a person 1.5 meters tall walks away from it. What geometric relationship connects the person’s distance and shadow length?
- Water enters a cylindrical tank at a constant volume rate. Which quantity is fixed, and why does that simplify the rate equation?
- Two objects move at right angles. What constraint relates their separation to their distances along the two paths?
- In a pulley constraint \(L=2y+x+C\), what does a negative value of \(dy/dt\) mean if downward is the positive direction for \(y\)?
- Why must you differentiate a changing-quantity constraint before substituting values that apply only at one instant?
