Related rates

When two quantities that change with time are tied together by an equation, their rates are tied together too. Here a ladder, a filling tank, a lamp’s shadow and a ripple each drive one quantity at a steady rate and show how the other follows. If y depends on x through $y = f(x)$ and x changes with time, the chain rule gives

The graph on the right draws the relation itself, the follower against the driver, and the current state is a point on that curve. The rate triangle starts at the point. Its red and yellow sides are the two rates times the same time, a minute on the tank and a second elsewhere. The yellow change divided by the red one, signs included, is the slope, so the third side lies along the tangent. Turning the rate knob scales the triangle. Moving along a curved relation changes the slope, so the same driver rate gives a different follower rate; the shadow’s straight line keeps its slope.

Differentiate first, then substitute. On the ladder, at the instant x = 3 the top is at y = 4. Substituting first turns $x^2 + y^2 = 25$ into $9 + 16 = 25$, an equation between constants whose derivative says nothing. Differentiating the general equation gives $2x\,x' + 2y\,y' = 0$, and only then do x = 3, y = 4 and $x' = 1$ m/s give $y' = -3/4$ m/s.

The four situations each tie a driven quantity to one that follows:

• Ladder (5 m): $x^2 + y^2 = 5^2 \Rightarrow y' = -\frac{x}{y}\,x'$. The farther the foot is out, the faster the top falls.
• Cone tank (point down): $V' = A\,h'$, where A is the area of the water’s surface. A steady inflow raises the level quickly when the surface is small and slowly when it is large.
• Lamp and shadow: similar triangles put the tip of the shadow at $T = 1.5x$, so $T' = 1.5\,x'$. The relation is a straight line, and the rates never change.
• Ripple: $A = \pi r^2 \Rightarrow A' = 2\pi r\,r'$. At a steady radial speed, a longer circumference means faster growth of the area.

A rate is not a change over one unit of time. The tip of the triangle is where the state would be after that time if both rates stayed as they are now. The hollow point on the same vertical is where it really is. The gap between them is the error of the tangent-line approximation: on the ladder from x = 3 with $x' = 1$ m/s, the top really drops 1 m in the next second, not the predicted 0.75 m. Only on the straight line of the shadow do the two points coincide.

New layer compares what the next tenth of a time unit adds with surface area × thickness on the tank, and with circumference × width on the ripple. For a thin enough layer the two nearly agree, which is where $V' = A\,h'$ and $A' = 2\pi r\,r'$ come from.

The ladder is an idealization. Its top would fall faster and faster without limit as y approaches 0, because the problem keeps the foot moving at a steady speed whatever happens. The page stops at x = 4.8.