G4 - Connected rates and inverse derivatives

G4 - Connected rates and inverse derivatives

Connect rates through a derivative, keep their signs and units meaningful, and find inverse derivatives at corresponding inputs. Check that the inverse and a finite reciprocal slope are valid before reversing a rate.

Linking rates through a changing quantity

A circular patch has radius 5 m5\,\mathrm m, growing at 0.4 m s−10.4\,\mathrm{m\,s^{-1}}. Is its area growing at 0.4 m2 s−10.4\,\mathrm{m^2\,s^{-1}}? A metre added to a large radius produces more area than a metre added to a small one, so we need the link between area and radius.

You already know

The chain rule multiplies the rate of output with respect to an intermediate quantity by that quantity's input rate. It is not in the formula booklet: learn it.

For the patch, A=πr2A=\pi r^2, so dA/dr=2πrdA/dr=2\pi r. This measures area change per metre of radius, not area change per second. The circle-area formula and its derivative are recall results, not supplied.

Connected rates

If Q=F(r)Q=F(r) and r=r(t)r=r(t), then

dQdt=dQdrdrdt\frac{dQ}{dt}=\frac{dQ}{dr}\frac{dr}{dt}

Evaluate dQ/drdQ/dr at the requested instant. Not in the formula booklet: learn it.

The units travel through the same link: area per length multiplied by length per time gives area per time. Write a decreasing quantity's derivative as negative before you calculate.

Worked example 1: an expanding circular patch

Question. A radius rr metres increases at 0.4 m s−10.4\,\mathrm{m\,s^{-1}}. Find dA/dtdA/dt when r=5 mr=5\,\mathrm m, given A=πr2A=\pi r^2.

Plan. Differentiate area with respect to radius, then connect to time.

dAdr=2πrdifferentiate the areadAdt=2πrdrdtchain rule=2π(5)(0.4)=4πsubstitute the instant\begin{aligned} \frac{dA}{dr}&=2\pi r && \text{differentiate the area}\\ \frac{dA}{dt}&=2\pi r\frac{dr}{dt} && \text{chain rule}\\ &=2\pi(5)(0.4)=4\pi && \text{substitute the instant} \end{aligned}

Why multiply by dr/dtdr/dt? It converts area per metre into area per second.

Answer. dA/dt=4π m2 s−1dA/dt=4\pi\,\mathrm{m^2\,s^{-1}}.

Check. The sign is positive, and m×m s−1=m2 s−1\mathrm m\times\mathrm{m\,s^{-1}}=\mathrm{m^2\,s^{-1}}. The area itself is 25π m225\pi\,\mathrm{m^2}, a different quantity.

Change only the direction of the radius rate. The area-radius relation stays the same; the time rate changes sign.

Worked example 2: a shrinking circular patch

Question. The radius now decreases at 0.4 m s−10.4\,\mathrm{m\,s^{-1}}. Find dA/dtdA/dt when r=5 mr=5\,\mathrm m, given A=πr2A=\pi r^2.

Plan. Use dr/dt=−0.4dr/dt=-0.4 in the same connected-rate relation.

drdt=−0.4decrease means negativedAdt=2π(5)(−0.4)chain rule at r=5=−4πevaluate\begin{aligned} \frac{dr}{dt}&=-0.4 && \text{decrease means negative}\\ \frac{dA}{dt}&=2\pi(5)(-0.4) && \text{chain rule at }r=5\\ &=-4\pi && \text{evaluate} \end{aligned}

Why is rr still positive? The radius is a length; its rate can be negative while the length remains positive.

Answer. dA/dt=−4π m2 s−1dA/dt=-4\pi\,\mathrm{m^2\,s^{-1}}; the area decreases at magnitude 4π m2 s−14\pi\,\mathrm{m^2\,s^{-1}}.

Check. A smaller positive radius gives a smaller area, consistent with the negative derivative.

A related geometric connection is hidden in the sphere formula: V=4πr3/3V=4\pi r^3/3 gives dV/dr=4πr2dV/dr=4\pi r^2, exactly the surface area. A very thin shell has volume approximately surface area times thickness. These sphere formulae and the derivative connection are not supplied in the booklet.

Watch out

Wrong: dA/dt=2πr=10πdA/dt=2\pi r=10\pi. That is dA/drdA/dr, with units of length, so the time rate is missing. Right: multiply by dr/dtdr/dt, including its sign.

Try the same shrinking-circle method with a different set of radius data. The question supplies the whole area relation.

Turning volume flow into a depth rate

A volume flow becomes a depth rate only after the container's geometry links volume and depth.

For a rectangular tank with fixed base area 1200 cm21200\,\mathrm{cm^2}, V=1200hV=1200h. An inflow of 60 cm3 s−160\,\mathrm{cm^3\,s^{-1}} gives 60=1200 dh/dt60=1200\,dh/dt, hence dh/dt=0.05 cm s−1dh/dt=0.05\,\mathrm{cm\,s^{-1}}. In a container that widens upwards, the same flow raises the level more slowly as it gets deeper.

A prism has a constant cross-section along its length. Its volume is cross-sectional area times length. For a triangular cross-section, area is half base times perpendicular height. These geometry results are not in the formula booklet: learn them.

Worked example 3: a widening trough

Question. A horizontal prism trough is 8 cm8\,\mathrm{cm} long. At depth h cmh\,\mathrm{cm}, the water cross-section is a right triangle with perpendicular sides hh and 2h2h. Water enters at 160 cm3 s−1160\,\mathrm{cm^3\,s^{-1}}, with no leakage. Find dh/dtdh/dt at h=5 cmh=5\,\mathrm{cm}. The trough permits this depth.

Plan. Build V(h)V(h), differentiate it and substitute the inflow.

V=8(12h(2h))=8h2area times lengthdVdt=16hdhdtchain rule160=16(5)dhdtuse the instantdhdt=2divide by 80\begin{aligned} V&=8\left(\tfrac12 h(2h)\right)=8h^2 && \text{area times length}\\ \frac{dV}{dt}&=16h\frac{dh}{dt} && \text{chain rule}\\ 160&=16(5)\frac{dh}{dt} && \text{use the instant}\\ \frac{dh}{dt}&=2 && \text{divide by }80 \end{aligned}

Why differentiate V(h)V(h) before substituting h=5h=5? Substituting first would turn the volume into a constant and lose how it changes.

Answer. The depth rises at 2 cm s−12\,\mathrm{cm\,s^{-1}}.

Check. At this depth, dV/dh=80 cm2dV/dh=80\,\mathrm{cm^2}; multiplying by 2 cm s−12\,\mathrm{cm\,s^{-1}} recovers the flow 160 cm3 s−1160\,\mathrm{cm^3\,s^{-1}}.

Change only one assumption: the same trough now leaks at 40 cm3 s−140\,\mathrm{cm^3\,s^{-1}}. The geometry stays fixed, but we must use net flow.

Worked example 4: inflow minus leakage

Question. For the trough in example 3, inflow is 160 cm3 s−1160\,\mathrm{cm^3\,s^{-1}} and leakage is 40 cm3 s−140\,\mathrm{cm^3\,s^{-1}}. Find dh/dtdh/dt when h=5 cmh=5\,\mathrm{cm}.

Plan. Subtract the outflow before using dV/dt=16h dh/dtdV/dt=16h\,dh/dt.

dVdt=160−40=120net flow120=16(5)dhdtsame geometrydhdt=12080=32solve for the depth rate\begin{aligned} \frac{dV}{dt}&=160-40=120 && \text{net flow}\\ 120&=16(5)\frac{dh}{dt} && \text{same geometry}\\ \frac{dh}{dt}&=\frac{120}{80}=\frac32 && \text{solve for the depth rate} \end{aligned}

Why use 120120, not 160160? Only retained water increases the volume in the trough.

Answer. The depth rises at 1.5 cm s−11.5\,\mathrm{cm\,s^{-1}}.

Check. The answer is smaller than 22, as leakage requires; 80(1.5)=12080(1.5)=120.

For positive hh, the rate is net flow divided by 16h16h. Thus constant positive flow gives a decreasing depth rate as the water gets deeper. The formula cannot be divided by h=0h=0; an empty-trough endpoint needs separate treatment.

Watch out

Wrong: 160=16h dh/dt160=16h\,dh/dt when the trough leaks at 4040. This uses incoming flow rather than volume change. Right: 160−40=16h dh/dt160-40=16h\,dh/dt.

Change only the leak rate. All the geometry needed is repeated inside the task.

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