1.07t - Forming differential equation models

1.07t - Forming differential equation models

Turn information about changing quantities into first-order differential equations. You will distinguish a value from a rate, calibrate proportionality constants, combine competing rates and explain the assumptions that make a model useful.

Expressing a rate as a differential equation

Can a tank keep the same volume while liquid is still flowing through it? Suppose liquid enters at 66 litres per minute and leaves at 0.4Q0.4Q litres per minute, where QQ is the volume in litres. At Q=10Q=10, the net increase is 6−4=26-4=2 litres per minute. At Q=15Q=15, both rates are 66: the volume stays constant. We will build that equation, and see what it predicts either side of 1515.

You already know

A derivative gives an instantaneous rate or a curve's gradient. A point condition, such as y(3)=4y(3)=4, gives a value of the function. Proportional quantities satisfy a relationship with a constant multiplier.

The derivative belongs on the left because the information describes change, rather than the quantity itself.

First-order differential equation

A differential equation relates an unknown function to its derivatives. It is first-order when its highest derivative is a first derivative. For example, dy/dx=2y−x\mathrm dy/\mathrm dx=2y-x is first-order; y=2x+3y=2x+3 is a function rule with no derivative.

Forming an equation does not require solving it. You can often find a gradient immediately, even when the whole curve is still unknown.

Worked example 1: translating a gradient rule

Question. A curve's gradient is twice its yy-coordinate minus its xx-coordinate. It passes through (3,4)(3,4). State the differential equation and point condition, then find the gradient there.

Plan. Translate the rule for every point, then substitute the one given point.

dydx=2y−xgradient ruley(3)=4point conditiondydx∣(3,4)=2(4)−3=5substitute the coordinates\begin{aligned} \frac{\mathrm dy}{\mathrm dx}&=2y-x && \text{gradient rule} \\ y(3)&=4 && \text{point condition} \\ \left.\frac{\mathrm dy}{\mathrm dx}\right|_{(3,4)}&=2(4)-3=5 && \text{substitute the coordinates} \end{aligned}

Why this step? The derivative is the gradient; the point condition tells us where to evaluate it.

Answer. dy/dx=2y−x\mathrm dy/\mathrm dx=2y-x, y(3)=4y(3)=4, and the gradient is 55.

Check. The yy-contribution is 88, reduced by the xx-contribution 33, giving 55.

Now change only “minus” to “plus”. A small change in the words changes the rate rule everywhere.

Worked example 2: changing one sign

Question. A curve's gradient is twice its yy-coordinate plus its xx-coordinate. It passes through (3,4)(3,4). State the equation and point condition, then find the gradient there.

Plan. Keep the same variables and point, but use addition in the rate rule.

dydx=2y+xtranslate “plus”y(3)=4the same pointdydx∣(3,4)=2(4)+3=11evaluate the rate\begin{aligned} \frac{\mathrm dy}{\mathrm dx}&=2y+x && \text{translate “plus”} \\ y(3)&=4 && \text{the same point} \\ \left.\frac{\mathrm dy}{\mathrm dx}\right|_{(3,4)}&=2(4)+3=11 && \text{evaluate the rate} \end{aligned}

Why this step? The point has not changed, but the law assigning its gradient has.

Answer. dy/dx=2y+x\mathrm dy/\mathrm dx=2y+x, y(3)=4y(3)=4, and the gradient is 1111.

Check. Changing −3-3 to +3+3 increases the previous gradient by 66.

In a population model, dN/dt=20−0.1N\mathrm dN/\mathrm dt=20-0.1N gives a rate of 1212 individuals per year at N=80N=80, but −5-5 at N=250N=250. A larger positive population can have a negative rate. A smooth model also treats discrete individuals as a continuous quantity; that approximation is usually more useful for a large population.

Watch out

Wrong: N=20−0.1tN=20-0.1t for a rate of 20−0.1N20-0.1N. Right: dN/dt=20−0.1N\mathrm dN/\mathrm dt=20-0.1N. Keep the current quantity in the stated rate rule, and distinguish N(0)=80N(0)=80 from N′(0)=12N'(0)=12.

The next rule changes the coefficient of the dependent variable.

Translating proportional growth and decline

“Proportional” supplies a constant multiplier; “decreases” supplies the negative sign. Decide these separately.

Proportional change

For a positive quantity QQ, if the stated positive rate of decrease is proportional to QnQ^n, then

dQdt=−kQn,k>0.\frac{\mathrm dQ}{\mathrm dt}=-kQ^n,\qquad k>0.

For growth, use a positive derivative. Retain any domain restrictions, especially when n<0n<0.

Not in the formula sheet: learn it.

Calibrate the constant using a measured rate and its corresponding state. An initial amount alone does not determine that constant.

Worked example 3: a rate proportional to a square

Question. A sample of mass MM grams loses mass at a rate proportional to M2M^2. It loses 22 grams per day when M=5M=5. Form its differential equation, with tt in days.

Plan. Write a negative derivative, then use the measurement to determine kk.

dMdt=−kM2square law and decline−2=−k(5)2use the measured ratek=225divide by 25\begin{aligned} \frac{\mathrm dM}{\mathrm dt}&=-kM^2 && \text{square law and decline} \\ -2&=-k(5)^2 && \text{use the measured rate} \\ k&=\frac2{25} && \text{divide by }25 \end{aligned}

Why this step? “Loses 22” means the positive loss rate is 22, while the signed derivative is −2-2.

Answer. dM/dt=−225M2\mathrm dM/\mathrm dt=-\frac2{25}M^2, for positive MM while the law applies. Here kk has units g−1 day−1\mathrm{g^{-1}\,day^{-1}}.

Check. At M=5M=5, the equation gives −225(25)=−2-\frac2{25}(25)=-2 grams per day.

Change only the dependence on mass: the loss rate is now inversely proportional to MM. The calibration data stay the same, but the constant will have different units.

Worked example 4: an inverse rate law

Question. A positive mass MM loses mass at a rate inversely proportional to MM, losing 22 grams per day when M=5M=5. Form its differential equation.

Plan. Replace the square by a reciprocal, keeping the negative derivative.

dMdt=−kMinverse proportionality−2=−k5calibratek=10multiply by 5\begin{aligned} \frac{\mathrm dM}{\mathrm dt}&=-\frac{k}{M} && \text{inverse proportionality} \\ -2&=-\frac{k}{5} && \text{calibrate} \\ k&=10 && \text{multiply by }5 \end{aligned}

Why this step? The wording identifies the controlling quantity as MM, not time, so the denominator is MM.

Answer. dM/dt=−10/M\mathrm dM/\mathrm dt=-10/M, for M>0M>0. Here kk has units g2 day−1\mathrm{g^2\,day^{-1}}.

Check. At M=5M=5, −10/5=−2-10/5=-2. At M=10M=10, the predicted loss rate is only 11 gram per day.

The controlling quantity can instead be the independent variable. If a gradient is inversely proportional to positive xx, and equals 33 at x=2x=2, then dy/dx=k/x\mathrm dy/\mathrm dx=k/x, 3=k/23=k/2, and k=6k=6. No yy belongs in that denominator.

For dN/dt=0.08N\mathrm dN/\mathrm dt=0.08N, time in years, (1/N)dN/dt=0.08(1/N)\mathrm dN/\mathrm dt=0.08 is a continuous relative rate. It does not mean that a complete year's multiplier is exactly 1.081.08; the rate changes as NN changes.

Watch out

Wrong: use +k/Q+k/Q for a quantity that decreases. Right: dQ/dt=−k/Q\mathrm dQ/\mathrm dt=-k/Q, with k>0k>0. A positive magnitude of loss is different from a positive derivative.

Try the reciprocal law with a different measured state and rate.

Combining rates and preserving equilibrium states

A net change can be zero even when two substantial processes are still happening. The tank from the opening question is a particularly clear example.

Rate balance

When the rates use the same quantity and units,

net rate of increase=input rate−output rate.\text{net rate of increase}=\text{input rate}-\text{output rate}.

For a volume QQ, the net rate is dQ/dt\mathrm dQ/\mathrm dt.

Not in the formula sheet: learn it.

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