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Concept 05 / 06Differential Equations

Applications of Differential Equations: Growth, Decay, Cooling and Geometric Problems

Modelling with rates proportional to the quantity (exponential growth and decay, half-life), Newton's law of cooling, mixing problems, and curves defined by conditions on their tangents or normals.

Figure 1
Many physical laws are statements about rates, and translating them into DEs is a standard JEE skill. The translation step is the one that needs care; the solving is usually routine. Exponential growth and decay. If a quantity changes at a rate proportional to its current size, dNdt=kN\frac{dN}{dt} = kN. Separating: dNN=k dt\frac{dN}{N} = k\,dt, so N=N0ektN = N_{0}e^{kt}, where N0=N(0)N_{0} = N(0). For k>0k > 0 this is growth (populations, continuously compounded interest); for decay write dNdt=−kN\frac{dN}{dt} = -kN with k>0k > 0, giving N=N0e−ktN = N_{0}e^{-kt} (radioactivity, drug elimination). The characteristic time is the half-life T1/2=ln⁡2kT_{1/2} = \frac{\ln 2}{k} (or the doubling time for growth): in each such interval the quantity halves, independent of how much is present. The figure shows this: N0→N02→N04N_{0} \to \frac{N_{0}}{2} \to \frac{N_{0}}{4} at equal time steps. Problems usually give two measurements; use their ratio to find kk, since N(t2)N(t1)=ek(t2−t1)\frac{N(t_{2})}{N(t_{1})} = e^{k(t_{2} - t_{1})}. Newton's law of cooling. A body at temperature TT in surroundings at constant temperature TsT_{s} cools at a rate proportional to the temperature difference: dTdt=−k(T−Ts)\frac{dT}{dt} = -k(T - T_{s}). With u=T−Tsu = T - T_{s} this is decay: T−Ts=(T0−Ts)e−ktT - T_{s} = (T_{0} - T_{s})e^{-kt}. The difference, not the temperature itself, decays exponentially. Mixing problems. A tank contains a solution; liquid flows in with a known concentration and out at the same rate, well mixed. If AA is the amount of dissolved substance, then dAdt=(rate in)−(rate out)=rcin−rAV\frac{dA}{dt} = (\text{rate in}) - (\text{rate out}) = r c_{\text{in}} - r\frac{A}{V}, a linear equation. The solution approaches the equilibrium A=cinVA = c_{\text{in}}V exponentially. Geometric problems. Conditions on tangents and normals become DEs, using the formulas from the chapter on derivatives. At a point (x,y)(x, y) with slope y′y': Tangent meets the xx-axis at x−yy′x - \frac{y}{y'} and the yy-axis at y−xy′y - xy'. Normal meets the xx-axis at x+yy′x + yy' and the yy-axis at y+xy′y + \frac{x}{y'}. Length of the subtangent ∣yy′∣\left|\frac{y}{y'}\right|, subnormal ∣yy′∣|yy'|. Write the stated condition in these terms, solve the DE, and use the given point to fix the constant. For example, 'the subnormal is constant' gives yy′=ayy' = a, so y2=2ax+Cy^{2} = 2ax + C, a parabola; 'the subtangent is constant' gives yy′=a\frac{y}{y'} = a, so y=Cex/ay = Ce^{x/a}, an exponential curve. Other physical models: velocity with air resistance proportional to speed, mdvdt=mg−kvm\frac{dv}{dt} = mg - kv, gives terminal velocity mgk\frac{mg}{k}; the charging of a capacitor, Rdqdt+qC=ER\frac{dq}{dt} + \frac{q}{C} = E, is linear with the same exponential approach to equilibrium. All share the form dydt=−k(y−y∞)\frac{dy}{dt} = -k(y - y_{\infty}), whose solution approaches y∞y_{\infty} exponentially with time constant 1k\frac{1}{k}. Key Takeaways: - dNdt=kN⇒N=N0ekt\frac{dN}{dt} = kN \Rightarrow N = N_{0}e^{kt}; half-life or doubling time ln⁡2∣k∣\frac{\ln 2}{|k|}. - Find kk from a ratio of two measurements. - Cooling: T−Ts=(T0−Ts)e−ktT - T_{s} = (T_{0} - T_{s})e^{-kt}. - Mixing: rate of change = rate in −- rate out; concentration out =AV= \frac{A}{V}. - Tangent xx-intercept x−yy′x - \frac{y}{y'}; normal xx-intercept x+yy′x + yy'. - Constant subnormal gives a parabola; constant subtangent gives an exponential. - Pitfall: in cooling problems, applying the exponential to TT instead of T−TsT - T_{s}. Brain-Teaser 1: A radioactive sample loses 10%10\% of its mass in 55 days. How long until only half remains? Solution: N=N0e−ktN = N_{0}e^{-kt} with 0.9=e−5k0.9 = e^{-5k}, so k=ln⁡(10/9)5k = \frac{\ln(10/9)}{5}. Half-life T=ln⁡2k=5ln⁡2ln⁡(10/9)≈5×0.6930.1054≈32.9T = \frac{\ln 2}{k} = \frac{5\ln 2}{\ln(10/9)} \approx \frac{5 \times 0.693}{0.1054} \approx 32.9 days. Note that losing 10%10\% in 55 days does not mean losing 50%50\% in 2525 days: the decay is multiplicative, and each 55-day period removes 10%10\% of what is left. Brain-Teaser 2: A body cools from 80∘80^{\circ}C to 60∘60^{\circ}C in 1010 minutes in a room at 20∘20^{\circ}C. What is its temperature after another 1010 minutes? Solution: The difference from the room decays: 60→4060 \to 40 in 1010 minutes, a factor of 23\frac{2}{3}. After another 1010 minutes the difference is 40×23=80340 \times \frac{2}{3} = \frac{80}{3}, so the temperature is 20+803≈46.7∘20 + \frac{80}{3} \approx 46.7^{\circ}C. There is no need to compute kk: equal time intervals multiply the difference by the same factor. Brain-Teaser 3: Find the curve through (2,3)(2, 3) for which the segment of every tangent between the axes is bisected at the point of contact. Solution: The tangent at (x,y)(x, y) meets the axes at (x−yy′,0)(x - \frac{y}{y'}, 0) and (0,y−xy′)(0, y - xy'). The midpoint of these is the point of contact, so 12(x−yy′)=x\frac{1}{2}(x - \frac{y}{y'}) = x, which gives −yy′=x-\frac{y}{y'} = x, that is, dydx=−yx\frac{dy}{dx} = -\frac{y}{x}. Separating, ln⁡∣y∣=−ln⁡∣x∣+C\ln|y| = -\ln|x| + C, so xy=cxy = c. The point (2,3)(2, 3) gives xy=6xy = 6, a rectangular hyperbola. (The yy-intercept condition gives the same equation, as it must by symmetry.)

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Applications of Differential Equations: Growth, Decay, Cooling and Geometric Problems — DPP

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