Related Rates

01 / UNDERSTAND & APPLY

1. A snapshot contains values, not a whole motion

A balloon expands, a ladder slides, or a shadow stretches. Several quantities change together because geometry or another physical constraint ties them together. Related rates asks how a known instantaneous rate of one quantity determines the instantaneous rate of another. You usually do not need a complete formula describing the entire motion as a function of time.

Imagine a circular ripple with radius r(t) and area A(t). At one instant its radius may be 5 metres. That does not make r a constant during the motion. The identity A(t) = πr(t)² holds through the motion, whereas r = 5 describes only a snapshot. Differentiating the snapshot instead of the identity would erase precisely the change we are trying to study.

Use t for time and choose a positive direction for every signed coordinate. A distance such as a radius is nonnegative, but its derivative can be negative if it shrinks. A height can be positive while its rate is negative. A phrase such as “falls at 2 cm/s” means dh/dt = −2 cm/s when upward height is positive.

Every relationship is a model with assumptions. A ladder of fixed length must stay in contact with both surfaces; a spherical balloon is assumed spherical; the water surface in a cone is assumed horizontal. State these assumptions before using the formulas. A calculation can be algebraically flawless and still fail because its geometric model stops applying.

02 / UNDERSTAND & APPLY

2. The chain rule supplies the rate factors

If A = πr² and r depends differentiably on t, then dA/dt = 2πr dr/dt. The factor dr/dt is essential: differentiating with respect to time is not the same operation as differentiating with respect to radius. More generally, if y = f(x(t)), then dy/dt = f′(x) dx/dt. The first factor converts an x-change into a y-change, and the second tells how quickly x changes in time.

Implicit relationships work the same way. For x² + y² = L² with constant L, differentiating gives 2x dx/dt + 2y dy/dt = 0. If L also changes, the right side becomes 2L dL/dt. The equation itself tells you which quantities are connected; the wording of the problem tells you which of those quantities are constant.

A helpful units check follows directly. In dA/dt = 2πr dr/dt, the right side has length multiplied by length/time, hence area/time. For a volume derivative, expect volume/time. If your answer has length² instead of length²/time, you have probably reported an area or forgotten a rate factor. Converting centimetres and metres before substitution prevents less obvious errors.

Rates do not need to be constant unless the problem says so. Differentiating a geometric identity at one instant remains valid for varying rates. Conversely, a constant input rate need not produce a constant output rate: a sphere supplied with constant volume per second has radius increasing more slowly as it grows because its surface area is larger.

03 / UNDERSTAND & APPLY

3. A diagram is an equation-finding device

Begin by sketching the geometry and labelling the changing lengths with letters. Fixed lengths may be labelled with their numerical values. Draw a small arrow next to each moving point to show the sign convention. Then list the given instantaneous rates and the desired rate in derivative notation. This separates information about where the system is from information about how it is moving.

Next find a constraint that holds for more than the single pictured configuration. It may be Pythagoras, similar triangles, a volume identity, or a trigonometric ratio. If the equation includes a rate you do not know, look for another geometric relationship that eliminates that variable. In a cone, for example, similarity can express the surface radius in terms of the water depth before differentiation.

The usual sequence is: define variables; record constants and directions; write the persistent constraint; reduce variables where useful; differentiate; substitute the snapshot; solve; interpret units and sign. There is no benefit in inserting a special value such as y = 0 earlier. In fact, a zero value can hide a singular point where division by y is invalid.

For a ladder diagram, label the wall-foot corner O, the ground contact B, and the upper contact A. OB = x(t), OA = y(t), and AB = L. For a cone diagram, mark the full fixed height H and radius R as well as the current water height h(t) and surface radius r(t). These labels reveal x² + y² = L² and r/h = R/H before any calculus starts.

04 / UNDERSTAND & APPLY

4. One changing length: circles and spheres

When a shape is controlled by one size variable, the related-rate equation is especially transparent. A sphere has V = 4πr³/3. Differentiation gives dV/dt = 4πr² dr/dt. The multiplier 4πr² is its surface area: adding a very thin layer of thickness dr adds approximately surface area times thickness to the volume. The derivative is the exact limiting version of that geometric picture.

If the surface area S = 4πr² is requested instead, differentiate that identity too. You can first find dr/dt from the volume rate and then obtain dS/dt = 8πr dr/dt. Alternatively, eliminate r between the formulas, but the two-step route often keeps units and assumptions clearer.

The formulas apply at positive radius. A request at r = 0 needs separate thought because division by r² is impossible. Under a positive constant volume input, the idealized solution r(t) is proportional to t^(1/3) from an initially empty sphere, so the initial radius rate is unbounded in that model. That warns you about a limiting idealization rather than authorizing division by zero.

05 / UNDERSTAND & APPLY

5. Two changing lengths and the sign of motion

A right-triangle constraint is common when objects move along perpendicular lines. The moving lengths may both increase, both decrease, or move in opposite directions. Differentiate with signed rates rather than inserting positive speeds everywhere. A negative result then has a direct geometric interpretation instead of looking like an algebraic error.

For a fixed-length ladder, x dx/dt + y dy/dt = 0 implies dy/dt = −(x/y) dx/dt when y > 0. As the foot moves out, the top moves down. The factor x/y changes with the configuration, so equal time intervals need not produce equal vertical drops. Near the horizontal position, this ideal formula predicts a large downward speed for a prescribed outward speed; real contact and mechanical constraints may fail before the ideal limit.

For two vehicles leaving an intersection along perpendicular roads, their separation s satisfies s² = x² + y². If both move away, ds/dt is positive. The separation rate is generally not the sum of their speeds because the motion directions differ. The sum is appropriate for directly opposite collinear motion, which is a different geometry.

06 / UNDERSTAND & APPLY

6. Eliminate a changing radius in a conical vessel

A common trap is to substitute the full cone radius into the current water volume. The water occupies a smaller similar cone, so its surface radius changes with its depth. If the full cone has radius R and height H, similarity gives r/h = R/H as long as the cone has its vertex at the bottom and circular opening at the top, with the water surface parallel to that opening. The constants R and H describe the vessel; r and h describe the water.

Substitution into V = πr²h/3 gives V = πR²h³/(3H²). Now only h changes, and dV/dt = πR²h²/H² · dh/dt. The coefficient is the current water-surface area πr². A fixed inflow makes the level rise faster in a narrow part and more slowly in a wide part. This agrees with pouring intuition and offers a strong reasonableness check.

If there is both inflow and outflow, use the net volume rate. A 5 L/min inlet and a 2 L/min outlet give dV/dt = 3 L/min, not either individual rate. Convert litres to the same cubic-length units used in the geometric formula. If the cone is inverted in a different orientation or only partly described, draw the water region carefully instead of automatically reusing the smaller-cone formula.

07 / UNDERSTAND & APPLY

7. Angles, shadows and reference points

Trigonometric constraints are useful when the requested rate is angular. Use radians for derivative formulas such as d(tan θ)/dt = sec²θ · dθ/dt. Degrees per second can be reported afterward by multiplying a radian rate by 180/π. Mixing degree-valued angles directly into the standard derivative formula produces a missing conversion factor.

For a person walking away from a lamp, distinguish the shadow’s length from the ground position of its tip. If x is the lamp-to-person distance and s is the shadow length, the lamp-to-tip distance is x + s. Similar triangles connect those distances to the lamp and person heights. Differentiating x + s gives dx/dt + ds/dt: the tip moves both because the person moves and because the shadow length changes.

An angular observation has a similar reference issue. A point on the ground directly below a moving object may itself move. Decide whether the horizontal distance is fixed before differentiating tan θ = height/distance. If both numerator and denominator vary, use the quotient rule or an implicit product form rather than silently setting the distance derivative to zero.

08 / UNDERSTAND & APPLY

8. Original practice: values, rates and interpretation

For each question, write the constraint before reading the solution. These original exercises intentionally vary the requested quantity so that recognition of a familiar picture does not replace modelling.

09 / UNDERSTAND & APPLY

9. Audit the answer before putting down the pen

Check three things independently: dimensions, sign, and limiting behavior. Dimensions ask whether the result has the requested type of quantity per unit time. Sign asks whether the result agrees with the chosen coordinate direction. Limiting behavior asks whether the rate should grow or shrink as the geometry changes. For example, constant volume input into an expanding sphere should produce a smaller radius rate at a larger radius.

The most frequent algebra mistake is omitting a chain-rule rate, such as writing d(r²)/dt = 2r instead of 2r dr/dt. The most frequent modelling mistake is treating a changing quantity as constant because its current value is known. The most frequent interpretation mistake is reporting the rate of a shadow’s length when the question asks for the motion of its tip.

Another subtle failure is taking a magnitude too early. Keep signed derivatives through the equation and convert to a positive speed only when the question asks “how fast” with a direction stated separately. If the result contradicts the assumed direction, inspect your assumptions; it may reveal that the described motion cannot occur with the stated constraints.

Finally, remember that instantaneous does not mean average. A rate at one configuration need not describe the next ten seconds. Unless a constant rate is given or proved, multiplying it by a long time interval is only a local approximation. Related-rate equations can sometimes be integrated to find the full motion, but that is an additional differential-equation problem, not a required step here.