Corrected exercises: related rates (201-103)

The coefficients are constant: $\frac{dz}{dt}=4\frac{dx}{dt}+7\frac{dy}{dt}=4\times 2.5+7\times(-1.5)=10-10.5=-0.5$. $z$ decreases even though $x$ grows.

$\frac{dy}{dt}=(6x^{2}-5)\frac{dx}{dt}$. At $x=2$: $6\times 4-5=19$, so $\frac{dy}{dt}=19\times 4=76$. Pitfall: substituting $x=2$ before differentiating, which gives $y=6$, a constant, and $\frac{dy}{dt}=0$.

$y=12x^{-1}$, so $\frac{dy}{dt}=-\frac{12}{x^{2}}\frac{dx}{dt}$. At $x=4$: $\frac{dy}{dt}=-\frac{12}{16}\times(-3)=2.25$. The two minus signs cancel: $x$ decreases, so $y$ increases.

Two-link chain rule: $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}=(3u^{2}+2)\cdot 2t$. At $t=1$, $u=2$: $\frac{dy}{dt}=(12+2)\times 2=28$. Pitfall: forgetting to compute $u$ at $t=1$ and replacing $u$ with $1$.

Product: $\frac{dz}{dt}=2x\frac{dx}{dt}\cdot y+x^{2}\frac{dy}{dt}$. So $\frac{dz}{dt}=2\times 3\times 2\times 5+9\times(-1)=60-9=51$. Pitfall: writing $2x\frac{dx}{dt}\cdot\frac{dy}{dt}$, that is, differentiating both factors at once.

$z=xy^{-2}$, a product: $\frac{dz}{dt}=\frac{1}{y^{2}}\frac{dx}{dt}-\frac{2x}{y^{3}}\frac{dy}{dt}$. So $\frac{dz}{dt}=\frac{3}{4}-\frac{2\times 8}{8}\times 0.5=0.75-1=-0.25$. The numerator grows and yet the quotient falls: the square of the denominator weighs more.

Differentiate both sides: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$, so $\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$. At the point $(5, 12)$: $\frac{dy}{dt}=-\frac{5}{12}\times 6=-2.5$. Pitfall: differentiating $y^{2}$ as $2y$, without the factor $\frac{dy}{dt}$, and having nothing left to solve for.

$\frac{dz}{dt}=\frac{1}{2\sqrt{x^{2}+y^{2}}}\left(2x\frac{dx}{dt}+2y\frac{dy}{dt}\right)=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{\sqrt{x^{2}+y^{2}}}$. Here $\sqrt{36+64}=10$, so $\frac{dz}{dt}=\frac{6\times 3+8\times(-2)}{10}=\frac{2}{10}=0.2$.

$\frac{dy}{dt}=\frac{2x}{x^{2}+1}\frac{dx}{dt}$: the derivative of the logarithm, then that of the inside, then the rate of $x$. At $x=3$: $\frac{dy}{dt}=\frac{6}{10}\times(-2)=-1.2$.

Take the logarithm: $\ln z=2\ln x+3\ln y$, then differentiate with respect to $t$: $\frac{1}{z}\frac{dz}{dt}=2\cdot\frac{1}{x}\frac{dx}{dt}+3\cdot\frac{1}{y}\frac{dy}{dt}$. Relative rates are weighted by the exponents: $2\times 1.5+3\times(-0.8)=3-2.4=0.6$% per year. Pitfall: adding the rates without the exponents, $1.5-0.8=0.7$%.

First step, NAME the variables and say which ones depend on time: here $T$ and $n$ both depend on $t$. Second step, WRITE the relation that links them, $T=\dfrac{n^{2}}{20{,}000}+3$. Third step, DIFFERENTIATE both sides with respect to TIME, which brings out the chain rule. Fourth step, SUBSTITUTE the numerical values, and only now. Fifth step, CONCLUDE with the units and a sentence. The order of steps 3 and 4 is the only real difficulty of the chapter, and exercise 2 is devoted to it.

Differentiating with respect to $t$: $\dfrac{dT}{dt}=\dfrac{2n}{20{,}000}\times\dfrac{dn}{dt}=\dfrac{n}{10{,}000}\times\dfrac{dn}{dt}$. The factor $\dfrac{dn}{dt}$ comes from the chain rule, because $n$ is not the variable of differentiation: it is a function of time. Leaving it out would amount to computing $\dfrac{dT}{dn}$, a completely different number, whose unit would be days per case and not days per week.

For $n=400$: $\dfrac{dT}{dt}=\dfrac{400}{10{,}000}\times 12=0.04\times 12=0.48$ day per week. For $n=800$: $\dfrac{dT}{dt}=0.08\times 12=0.96$ day per week. The same inflow of twelve extra cases a week degrades the delay twice as fast when the service is already loaded.

First find the corresponding $n$: $\dfrac{n^{2}}{20{,}000}+3=20$ gives $n^{2}=17\times 20{,}000=340{,}000$, so $n=\sqrt{340{,}000}\approx 583.1$ cases. Since $n(t)=400+12t$, solve $400+12t=583.1$, which gives $t\approx 15.26$ weeks. In fifteen weeks, the delay will have almost doubled.

Because the relation between $T$ and $n$ is QUADRATIC, while the growth of $n$ is linear. Formally, $\dfrac{dT}{dt}=\dfrac{n}{10{,}000}\times 12$ is proportional to $n$, so it grows with it. In concrete terms, a lightly loaded service absorbs the extra cases in its idle time, and a loaded one stacks them up. It is the same phenomenon as the asymptote of the waiting line seen in chapter 2, in polynomial form: the degradation of a saturated service is never proportional to the extra demand.

$R=\dfrac{4{,}800{,}000}{120}=40{,}000$ dollars per employee. If you substitute first, you write $R=40{,}000$, which is a CONSTANT, and the derivative of a constant is zero. The calculation is perfectly correct and the answer is perfectly wrong: you have differentiated a snapshot of one instant, not the film. The values $4{,}800{,}000$ and $120$ are true at one date only, and it is precisely their variation that is asked for.

Differentiate the relation $R=\dfrac{M}{n}$ with respect to $t$, with the quotient rule and the chain rule: $\dfrac{dR}{dt}=\dfrac{\dfrac{dM}{dt}\,n-M\,\dfrac{dn}{dt}}{n^{2}}$. It is the WHOLE relation that is differentiated, before any substitution.

$\dfrac{dR}{dt}=\dfrac{180{,}000\times 120-4{,}800{,}000\times 6}{120^{2}}=\dfrac{21{,}600{,}000-28{,}800{,}000}{14{,}400}=\dfrac{-7{,}200{,}000}{14{,}400}=-500$ dollars per year. The AVERAGE salary falls by 500 dollars a year, while the payroll rises by 180,000 dollars. Both facts are true at the same time: the organization spends more on salaries and pays less well, because it hires faster than it grows its budget. A report that quoted only the payroll would give exactly the opposite impression of the reality experienced.

Set the numerator to zero: $\dfrac{dM}{dt}\times 120=4{,}800{,}000\times 6$, so $\dfrac{dM}{dt}=\dfrac{28{,}800{,}000}{120}=240{,}000$ dollars per year. In other words, it would take 240,000 dollars more per year, not 180,000, for the average salary to hold. It can be read another way: $\dfrac{dM}{dt}=R\times\dfrac{dn}{dt}$, that is $40{,}000\times 6$, which amounts to funding each new employee at the current average salary.

Differentiate the RELATION, then substitute: never the reverse. What the rule protects is the distinction between a value and a function. Every quantity that varies must remain a letter until the last line, and only the quantities that are TRULY constant, a contractual price, a fixed area, a number of weeks per year, may be substituted from the start.

$R=pq=24\times 540=12{,}960$ dollars per month.

$\dfrac{dR}{dt}=\dfrac{dp}{dt}\,q+p\,\dfrac{dq}{dt}=(-0.50)\times 540+24\times 18=-270+432=+162$ dollars per month. The revenue INCREASES by 162 dollars per month: the volume effect, 432 dollars, outweighs the price effect, $-270$ dollars. The two terms of the product rule each have a name in management, and their sum is what you observe at the end of the month.

We want $\dfrac{dp}{dt}$ such that $\dfrac{dp}{dt}\times 540+24\times 18=0$, so $\dfrac{dp}{dt}=-\dfrac{432}{540}=-0.80$ dollar per month. A drop of 80 cents per month would exactly cancel the volume gain. The current drop, 50 cents, therefore stays below the break-even point, and that is the only question that matters in deciding whether to continue the discount.

$\dfrac{p'}{p}=\dfrac{-0.50}{24}\approx -0.02083$, that is $-2.08$% per month. $\dfrac{q'}{q}=\dfrac{18}{540}=0.03333$, that is $+3.33$% per month. The sum is $0.0125$, that is $+1.25$% per month. We do recover $\dfrac{dR}{dt}=0.0125\times 12{,}960=162$ dollars per month. The two routes agree, and the second can be done in your head.

The revenue grows by about 1.25% per month, the difference of the two percentages. The business conclusion fits in one sentence: the discount is PROFITABLE as long as the relative rise in volume exceeds the relative drop in price. It is a decision rule that can be used without a calculator, and it explains why a company tracks the percentage change in its sales rather than their number: only the percentage compares directly with the discount granted.

$\dfrac{dP}{dt}=B-D+M=1{,}120-860-140=120$ inhabitants per year. The relative rate is $\dfrac{120}{48{,}000}\times 100=0.25$% per year. The municipality is growing, but slowly: the positive natural balance, 260 people, is cut by more than half by the migration deficit.

$\dfrac{dP}{dt}=1{,}120-860+300=560$ inhabitants per year, and the relative rate becomes $\dfrac{560}{48{,}000}\times 100\approx 1.17$% per year. A net migration improved by 440 people multiplies the growth rate by nearly five, without any demographic parameter having changed. It is the quickest lever a municipality has, and this is why attraction policies weigh more in the short run than birth-rate policies.

$N=\dfrac{1{,}000\times 1{,}120}{48{,}000}=\dfrac{1{,}120{,}000}{48{,}000}\approx 23.33$ births per thousand inhabitants.

The numerator is constant and the denominator varies, so $\dfrac{dN}{dt}=1{,}000\,B\times\left(-\dfrac{1}{P^{2}}\right)\dfrac{dP}{dt}=-\dfrac{1{,}000\times 1{,}120\times 120}{48{,}000^{2}}=-\dfrac{134{,}400{,}000}{2{,}304{,}000{,}000}\approx -0.058$ per thousand per year. The crude birth rate FALLS, although the number of births has not moved by a single unit.

Yes, and the previous calculation is the proof. A crude rate is a QUOTIENT: it falls as soon as its denominator grows faster than its numerator, here as soon as the population increases with births constant. The same mechanism applies to the crime rate, the equipment rate, the number of doctors per thousand inhabitants. The methodological consequence is constant: faced with a rate that moves, the first question is which of the two terms has moved, and exercise 10 will ask it about the most quoted figure of all, the unemployment rate.

$c=60+0.5\times 100=110$ cases per day. $\rho=\dfrac{96}{110}\approx 0.8727$, that is 87.3% utilization. $W=\dfrac{0.8727}{1-0.8727}=\dfrac{0.8727}{0.1273}\approx 6.86$ service times: a user waits on average nearly seven times the duration of one processing.

$\dfrac{dc}{dt}=0.5\times\dfrac{dI}{dt}=0.5\times 12=6$ cases per day per year. The capacity grows by six daily cases each year.

$\rho=96\,c^{-1}$, so $\dfrac{d\rho}{dt}=-\dfrac{96}{c^{2}}\times\dfrac{dc}{dt}=-\dfrac{96\times 6}{110^{2}}=-\dfrac{576}{12{,}100}\approx -0.04760$ per year. The utilization rate falls by about 4.76 percentage points per year.

$\dfrac{dW}{d\rho}=\dfrac{1}{(1-\rho)^{2}}=\dfrac{1}{0.1273^{2}}\approx 61.73$. By the chain rule, $\dfrac{dW}{dt}=61.73\times(-0.04760)\approx -2.94$ service times per year. The waiting time therefore decreases by about three service times each year, which is considerable compared with the starting 6.86: at the initial rate, more than 40% of the wait (2.94 out of 6.86) disappears over the first year.

No, and the gain would be much SMALLER. The only factor that changes in the chain is $\dfrac{dW}{d\rho}=\dfrac{1}{(1-\rho)^{2}}$, which equals 61.7 at 87.3% utilization and only 4 at 50%. The other two links, $\dfrac{dc}{dt}=6$ and the form of $\rho$, are almost unchanged. The same investment would therefore produce about fifteen times less reduction in waiting in a lightly loaded service. This is the quantitative argument for investing where saturation is strongest, and it rests only on the square in the denominator.

The product rule gives $z'=x'y+xy'$. Divide both sides by $z=xy$, which is nonzero: $\dfrac{z'}{z}=\dfrac{x'y}{xy}+\dfrac{xy'}{xy}=\dfrac{x'}{x}+\dfrac{y'}{y}$. The simplification is immediate, each term losing the factor that it does not contain in the numerator.

The quotient rule gives $z'=\dfrac{x'y-xy'}{y^{2}}$. Divide by $z=\dfrac{x}{y}$, which amounts to multiplying by $\dfrac{y}{x}$: $\dfrac{z'}{z}=\dfrac{x'y-xy'}{y^{2}}\times\dfrac{y}{x}=\dfrac{x'y-xy'}{xy}=\dfrac{x'}{x}-\dfrac{y'}{y}$. The minus sign comes from the numerator of the quotient rule, and it carries all the economic interest of the result.

Revenue $R=pq$ with $p=24$, $q=540$, $p'=-0.50$ and $q'=18$. Relative rates: $\dfrac{-0.50}{24}\approx -2.08$% and $\dfrac{18}{540}\approx 3.33$% per month. Sum: $1.25$% per month. Multiplied by $R=12{,}960$, it gives back $162$ dollars per month, exactly the value found in exercise 3 with the product rule.

Average salary $R=\dfrac{M}{n}$ with $M=4{,}800{,}000$, $n=120$, $M'=180{,}000$ and $n'=6$. Relative rates: $\dfrac{180{,}000}{4{,}800{,}000}=3.75$% and $\dfrac{6}{120}=5$% per year. Difference: $3.75-5=-1.25$% per year. Multiplied by $R=40{,}000$ dollars, it gives $-500$ dollars per year, the value of exercise 2. The reading is direct: the payroll grows more slowly, in percent, than the headcount, so the average salary falls.

Because it answers, with no calculation, the only question that comes up in a meeting: which of the numerator and the denominator is growing faster in PERCENT? The absolute values, 180,000 dollars and 6 people, cannot be compared; the percentages, 3.75% and 5%, can be compared at once. The rule also explains why so many indicators deteriorate without any of their components deteriorating: gross domestic product per capita, average salary, density of doctors, crude birth rate. In each case, both terms rise, and it is their ORDER in percent that decides the direction of the quotient.

By the quotient rule, $M'(t)=\dfrac{80(t+12)-80t}{(t+12)^{2}}=\dfrac{960}{(t+12)^{2}}$, in percentage points per month. At $t=12$: $M(12)=\dfrac{960}{24}=40$% and $M'(12)=\dfrac{960}{576}\approx 1.667$ points per month.

$T(12)=250+36=286$ million dollars and $A(12)=0.40\times 286=114.4$ million. By the product rule, $\dfrac{dA}{dt}=\dfrac{M'}{100}T+\dfrac{M}{100}T'=\dfrac{1.667}{100}\times 286+0.40\times 3\approx 4.767+1.200=5.97$ million dollars per month.