A painter in Pune leans a \(5\,\text{m}\) ladder against a smooth wall. On the polished floor its foot begins to slide away from the wall at a steady \(0.8\,\text{m/s}\). At first the top of the ladder creeps down the wall. Then it comes down faster and faster. The foot never changes its speed, so why does the top?
Differentiate the relation, not the numbers
Let the foot of the ladder be a distance \(x\) from the wall and the top a height \(y\) up it. Both change with time. But at every instant they are tied by the length of the ladder, because the wall, the floor and the ladder form a right-angled triangle. So \(x^2 + y^2 = 25\) at every instant, with lengths in metres.
An equation that holds at every instant can be differentiated with respect to time. The derivative of the constant 25 is zero. By the chain rule, \(x^2\) changes at the rate \(2x\,dx/dt\), and \(y^2\) at the rate \(2y\,dy/dt\). So \(2x\,\dot x + 2y\,\dot y = 0\), which gives \(\dot y = -x\,\dot x/y\). When the foot is \(3\,\text{m}\) out, the top is \(4\,\text{m}\) up. Then \(\dot y = -3 \times 0.8/4 = -0.6\,\text{m/s}\): the top falls at \(0.6\,\text{m/s}\).
The formula \(\dot y = -x\,\dot x/y\) answers the painter's question. The foot speed \(\dot x\) stays the same, but the factor \(x/y\) grows. At first \(x\) is small and \(y\) large, so the top barely moves. Late in the slide \(x\) is large and \(y\) small, so the top falls fast. The rates are tied, but not in a fixed proportion.
The steps are always the same. Name the changing quantities. Write the relation that holds between them at every instant. Differentiate the whole relation with respect to time, using the chain rule on each term. Only then put in the numbers of the instant you want, and read the sign of the answer.
Let two quantities that change with time obey an equation at every instant. Then the equation obtained by differentiating it with respect to time also holds at every instant. It ties their rates of change.
A slope without solving for y
The same idea finds a slope on a curve that is not written as \(y = f(x)\). The circle \(x^2 + y^2 = 25\) is such a curve. Differentiate with respect to \(x\), treating \(y\) as a function of \(x\): \(2x + 2y\,dy/dx = 0\), so \(dy/dx = -x/y\). At the point \((3, -4)\) the slope is \(-3/(-4) = 3/4\). This way of working is called implicit differentiation.
A patch of oil
Oil leaking from a ship spreads on still water as a circle. Suppose its area grows at a steady \(2\pi \approx 6.28\) square metres every second. The area is \(A = \pi r^2\), so \(dA/dt = 2\pi r\,dr/dt\). The edge therefore moves out at \(dr/dt = (dA/dt)/(2\pi r) = 1/r\) metres per second. At \(r = 1\,\text{m}\) that is \(1\,\text{m/s}\); at \(r = 4\,\text{m}\) only \(0.25\,\text{m/s}\).
A ladder \(5\,\text{m}\) long slides with its foot on a smooth floor and its top against a smooth wall. At one instant the foot is \(4\,\text{m}\) from the wall and moves away from it at \(1.5\,\text{m/s}\). How fast does the top slide down, in m/s?
A student puts \(x = 3\) into \(x^2 + y^2 = 25\) first, getting \(9 + y^2 = 25\). Differentiating this gives \(2y\,\dot y = 0\), so the top of the ladder seems not to move. What went wrong?
A spherical balloon is filled with air at a steady rate \(dV/dt\). Since \(V = \tfrac43\pi r^3\), \(dV/dt = 4\pi r^2\,dr/dt\). When the radius has doubled, how fast does the radius grow compared with before?
For the circle \(x^2 + y^2 = 25\), implicit differentiation gives \(dy/dx = -x/y\). Find the slope of the tangent at the point \((3, -4)\).
Back to physics
Reflection: Plane and Spherical Mirrors finds how fast an image moves by differentiating the mirror formula with respect to time. That is a related-rates problem, and the steps are the ones above.
This is the worked example of Reflection: Plane and Spherical Mirrors, lesson 6, done with the tool. An object is \(30\,\text{cm}\) in front of a concave mirror of focal length \(20\,\text{cm}\). It moves towards the mirror along the axis at \(2\,\text{cm/s}\). Find the velocity of the image.
With the Cartesian signs, \(u = -30\,\text{cm}\) and \(f = -20\,\text{cm}\). Moving towards the mirror, \(u\) increases towards zero: \(du/dt = +2\,\text{cm/s}\).
The relation \(1/v + 1/u = 1/f\) holds at every instant. Differentiate it with respect to time: \(-(1/v^2)\,dv/dt - (1/u^2)\,du/dt = 0\), since \(f\) is constant.
The mirror formula gives \(1/v = -1/20 + 1/30 = -1/60\), so \(v = -60\,\text{cm}\). Then \(dv/dt = -(v^2/u^2)\,du/dt = -(3600/900) \times 2 = -8\,\text{cm/s}\).
\(v\) is becoming more negative, so the image moves away from the mirror at \(8\,\text{cm/s}\), four times as fast as the object.
A rod slides on two rails \(0.1\,\text{m}\) apart, closing a circuit in a field of \(2\,\text{T}\). The flux is \(\Phi = Blx\), where \(x\) is the rod's distance from the closed end. The rod moves at \(dx/dt = 0.02\,\text{m/s}\). Find the emf \(d\Phi/dt\), in millivolts.
Small changes by the tangent
Lesson 1 showed that a curve and its tangent cannot be told apart close to a point. So for a small step \(\Delta x\), the curve rises almost exactly as much as the tangent does. That gives the linear approximation: \(\Delta y \approx (dy/dx)\,\Delta x\), or \(f(x_0 + \Delta x) \approx f(x_0) + f'(x_0)\,\Delta x\). The derivative is taken at the known point \(x_0\).
For example, estimate \((2.01)^3\). Take \(f(x) = x^3\) and \(x_0 = 2\), where \(f = 8\) and \(f' = 3x^2 = 12\). With \(\Delta x = 0.01\), \(f(2.01) \approx 8 + 12 \times 0.01 = 8.12\). The exact value is \(8.120601\). The error, about \(0.0006\), is close to \(\tfrac12 f''\,\Delta x^2 = \tfrac12 \times 12 \times 0.0001\). Halve the step and the error falls four times.
Physics uses this estimate all the time. The period of a pendulum is \(T = 2\pi\sqrt{L/g}\), which goes as \(L^{1/2}\). The power rule gives \(dT/T = \tfrac12\,dL/L\). So a string \(1\,\%\) too long makes the period only about \(0.5\,\%\) too long. The same step estimates the error in a measured result from the errors in the quantities it depends on.
Use the tangent at \(x = 2\) to estimate \((2.01)^3\). Here \(f(2) = 8\) and \(f'(2) = 12\).
Put in the numbers only after differentiating. A relation frozen at one instant, such as \(9 + y^2 = 25\), has lost the rates it was meant to relate. And the small-change estimate works only for small steps. Its error grows as \(\tfrac12 f''\,\Delta x^2\). A step that is large compared with the scale on which the curve bends gives a wrong answer.
Back to the painter
The foot of the painter's ladder moves at a steady \(0.8\,\text{m/s}\), and the top falls at \(0.8\,x/y\). When the foot is \(1\,\text{m}\) out, the top falls at only \(0.16\,\text{m/s}\). When it is \(3\,\text{m}\) out, \(0.6\,\text{m/s}\). When it is \(4.6\,\text{m}\) out, \(1.88\,\text{m/s}\). The rates are tied by the triangle, and the factor \(x/y\) is what makes the top rush down at the end.
Related rates tie the moving parts of many physics situations. The equation of continuity and the time to drain a tank in Fluid Flow come from a volume that changes as a level falls. Faraday's law meets the motional emf of a sliding rod through the flux \(Blx\) in Faraday's Law, Lenz's Law and Motional EMF. The image velocity \(v_I = -m^2v_O\) comes from differentiating the mirror formula in Reflection: Plane and Spherical Mirrors. The velocities of the two ends of a sliding ladder are tied by its length in Rigid Bodies and Moment of Inertia.
- Substituting the numbers of one instant before differentiating.Differentiate the relation while every quantity is still a variable. Put in the numbers afterwards.
- Dropping the rate factor when differentiating a variable.With respect to time, \(y^2\) gives \(2y\,dy/dt\), not \(2y\). Every changing quantity brings its own rate.
- Losing the sign of a rate.A quantity that is decreasing has a negative rate. The top of a sliding ladder has \(\dot y < 0\).