P4 5.2 - Connected rates and differential equations

P4 5.2 - Connected rates and differential equations

Connect changing quantities with the chain rule, then translate a physical rate statement into a differential equation. Learn to choose signs, constants, units and initial conditions so that the equation describes the stated model.

Linking rates by the chain rule

If a balloon gains the same volume every second, does its radius also increase at a constant rate? A larger sphere needs more added volume for the same small increase in radius, so the answer should be no.

For a sphere, V=4πr3/3V=4\pi r^3/3, hence dV/dr=4πr2dV/dr=4\pi r^2. The volume rate and radius rate are linked by the chain rule:

Connected rates

If VV depends on rr, which depends on time tt, then

dVdt=dVdrdrdt.\frac{dV}{dt}=\frac{dV}{dr}\frac{dr}{dt}.

For a sphere, this becomes dV/dt=4πr2(dr/dt)dV/dt=4\pi r^2(dr/dt).

The general chain rule is not in the formula booklet: learn it. The sphere's volume formula is assumed knowledge in the specification and is also often supplied in questions.

Use the radius at the specified instant after differentiating. Substituting it into the volume formula first would turn a changing quantity into a fixed number.

[DIAGRAM: asset_name: Equal volume rates and unequal radius rates; asset_slug: sphere-rates; caption: With the same volume input, doubling the radius quarters its instantaneous rate of increase.; description: Two circles representing spherical balloons at radii 2 cm and 4 cm, with labelled equal volume rates 24 pi cubic centimetres per second and respective radius rates 3/2 and 3/8 centimetres per second.]
With the same volume input, doubling the radius quarters its instantaneous rate of increase.

Worked example 1: a radius rate

Question. A spherical balloon gains volume at 24π cm3 s−124\pi\,\mathrm{cm^3\,s^{-1}}. Find dr/dtdr/dt when r=2 cmr=2\,\mathrm{cm}.

Plan. Differentiate volume with respect to radius, then use the chain rule and the instantaneous radius.

24π=4π(2)2drdt,drdt=24π16π=32.24\pi=4\pi(2)^2\frac{dr}{dt}, \qquad\frac{dr}{dt}=\frac{24\pi}{16\pi}=\frac32.

Answer. The radius increases at 3/2 cm s−13/2\,\mathrm{cm\,s^{-1}}.

Check. (16π cm2)(3/2 cm s−1)=24π cm3 s−1(16\pi\,\mathrm{cm^2})(3/2\,\mathrm{cm\,s^{-1}})=24\pi\,\mathrm{cm^3\,s^{-1}}, recovering the supplied volume rate with consistent units.

Change only the instantaneous radius from 22 to 44. The same volume input is spread over a larger spherical surface.

Worked example 2: a larger balloon

Question. With the same volume rate 24π cm3 s−124\pi\,\mathrm{cm^3\,s^{-1}}, find dr/dtdr/dt at r=4 cmr=4\,\mathrm{cm}.

Plan. Keep the chain-rule relationship and substitute the new radius.

24π=4π(4)2drdt,drdt=24π64π=38.24\pi=4\pi(4)^2\frac{dr}{dt}, \qquad\frac{dr}{dt}=\frac{24\pi}{64\pi}=\frac38.

Answer. 3/8 cm s−13/8\,\mathrm{cm\,s^{-1}}.

Check. Doubling rr multiplies 4πr24\pi r^2 by 44, so dr/dtdr/dt must be a quarter of 3/23/2, namely 3/83/8.

In general, for constant positive volume rate cc,

drdt=c4π1r2.\frac{dr}{dt}=\frac{c}{4\pi}\frac1{r^2}.

Since c/(4π)c/(4\pi) is constant, the radius rate is inversely proportional to r2r^2. For a shrinking sphere at volume-loss rate c>0c>0, use dV/dt=−cdV/dt=-c; the signed radius rate is negative and its magnitude is c/(4πr2)c/(4\pi r^2).

Watch out

A rate of decrease given as a positive magnitude needs a minus sign in the derivative. A volume decreasing by 10 cm3 s−110\,\mathrm{cm^3\,s^{-1}} has dV/dt=−10 cm3 s−1dV/dt=-10\,\mathrm{cm^3\,s^{-1}}.

Use the radius at the stated instant in the differentiated relationship.

Translating proportional rates

A differential equation relates an unknown function to one or more of its derivatives. Forming one means describing the rate correctly; solving it to find the whole function is a later task.

“Rate of increase of yy is proportional to yy” translates to dy/dt=kydy/dt=ky, with k>0k>0. A rate of decrease proportional to yy gives dy/dt=−kydy/dt=-ky. “Inversely proportional to y2y^2” means a factor 1/y21/y^2, not −y2-y^2. The independent variable could also be distance xx rather than time.

Constant of proportionality

In dy/dt=kF(y,t)dy/dt=kF(y, t), the constant of proportionality kk does not vary with yy or tt within the model. Find it from a supplied value of the rate and the corresponding quantities. Its units depend on the equation.

The quantity after “proportional to” must be read carefully: current amount and remaining capacity lead to different models.

Worked example 3: rate proportional to the current volume

Question. A controlled tank has volume VV litres at time tt minutes. Its volume increases at a rate proportional to VV. When V=50V=50, it is increasing at 15 L min−115\,\mathrm{L\,min^{-1}}. Form the equation and find the model's rate at V=150V=150.

Plan. Write dV/dt=kVdV/dt=kV, determine kk, then use the same constant at the new volume.

15=50k⇒k=310,dVdt=310V.15=50k\quad\Rightarrow\quad k=\frac3{10}, \qquad\frac{dV}{dt}=\frac3{10}V.

At V=150V=150, the rate is (3/10)150=45(3/10)150=45.

Answer. dV/dt=0.3VdV/dt=0.3V; the rate is 45 L min−145\,\mathrm{L\,min^{-1}}.

Check. Tripling the volume triples the rate in a direct-proportionality model: 15×3=4515\times3=45.

Change only the proportional quantity to the unfilled capacity of a 200200-litre tank. Keep the same calibration point.

Worked example 4: rate proportional to a remaining amount

Question. The tank now follows a model in which its rate of increase is proportional to 200−V200-V. At V=50V=50, the rate is 15 L min−115\,\mathrm{L\,min^{-1}}. Form the equation and find the rate at V=150V=150.

Plan. Use the remaining capacity in the equation before finding kk.

dVdt=k(200−V),15=k(150)⇒k=110.\frac{dV}{dt}=k(200-V), \qquad15=k(150)\quad\Rightarrow\quad k=\frac1{10}.

At V=150V=150, the remaining capacity is 5050, so the rate is 55.

Answer. dV/dt=0.1(200−V)dV/dt=0.1(200-V); the rate at V=150V=150 is 5 L min−15\,\mathrm{L\,min^{-1}}.

Check. The remaining capacity falls from 150150 to 5050, one third as large, so its proportional rate falls from 1515 to 55. At V=200V=200, the model gives rate zero.

A proportionality constant is fixed by the model, not recalculated arbitrarily at each later point.

Watch out

“Proportional to 200−V200-V” does not mean dV/dt=200−kVdV/dt=200-kV. The constant multiplies the entire stated quantity: k(200−V)k(200-V).

Calibrate the model from the given volume and its simultaneous rate.

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