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Calculus 1

Exponential growth and decay models

If a quantity changes at a rate proportional to its own size, what must the quantity be, and how do a growing population, a decaying sample, a bank balance and a cooling dish all follow from that one sentence?

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Separable differential equationsCalculus 1Solving logarithmic equationsAlgebra 2
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Logistic growthCalculus 2
Where you are 7 in Differential equations
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▴ 2 aboveSlope fieldsEuler's methodSeparable differential equations
Exponential growth and decay models
Logistic growth

Why this exists

Many quantities change at a rate set by how much of them there is. Each cell in a culture divides on its own schedule, so twice the cells make twice the new cells per hour. Each atom of carbon-14 has the same chance of decaying in the next year, so twice the atoms give twice the decays. A bank paying interest continuously adds money in proportion to the balance. Writing a differential equation from a description turned sentences like these into \dfrac{dy}{dt} = ky, and a cooling object into \dfrac{dT}{dt} = k(T - A), without solving them. This lesson solves both once and for all. The solutions are exponential functions and nothing else, which is why one family of functions serves populations, radioactive dating, savings and cooling alike: the single assumption about the rate forces it.

Definition

Day 1: rate proportional to amount

A quantity y(t) changes at a rate proportional to its size if there is a constant k with

\frac{dy}{dt} = ky

for every t in an interval. Where y \neq 0 the constant is

k = \frac{1}{y}\,\frac{dy}{dt},

the rate of change per unit of amount, called the relative growth rate (or continuous growth rate). Its units are one over time: per hour, per year. The model is exponential growth when k > 0 and exponential decay when k < 0; when k = 0 the quantity is constant. A relative rate of 0.05 per year is often stated as “5\% a year, compounded continuously”.

For a positive quantity, the doubling time (when k > 0) is the time it takes to double, and the half-life (when k < 0) the time it takes to halve. Doubling time and half life worked with these from formulas in base 2 and \tfrac{1}{2}; here they come out of the differential equation.

Day 2: Newton’s law of cooling

An object at temperature T(t) sits in surroundings held at a constant temperature A, the ambient temperature. Newton’s law of cooling says the rate of change of T is proportional to the difference between the object and its surroundings:

\frac{dT}{dt} = k(T - A), \qquad k < 0.

With k < 0, an object hotter than its surroundings (T > A) has \dfrac{dT}{dt} < 0 and cools, and a colder one (T < A) has \dfrac{dT}{dt} > 0 and warms. Despite its name, the law covers both. The constant k depends on the object and on how it meets the air or water around it, and is found from a measurement.

Intuition

Under \dfrac{dy}{dt} = ky with k > 0, the larger y is, the faster it grows, which makes it larger still. Over every stretch of time of a fixed length the amount is multiplied by the same factor. In Figure 1 the carp population of Example 1 doubles every 3.85 years, from 150 to 300 to 600 to 1200, whichever moment is taken as the start. Decay is the same picture with a factor less than 1: the amount falls fast while it is large, slows as it shrinks, and never reaches 0.

2 4 6 8 10 12 300 600 900 1200 t P
Figure 1The carp population P = 150e^{0.18t} of Example 1. Each doubling takes the same time, \dfrac{\ln 2}{0.18} \approx 3.85 years: from 150 to 300, to 600, to 1200.

Newton’s law is that picture seen from the ambient temperature. Since A is constant, the gap T - A obeys \dfrac{d}{dt}(T - A) = k(T - A), so the gap decays exponentially. It shrinks by the same factor in each equal stretch of time, whether the object starts above A or below it (Figure 2), and the temperature approaches A along a horizontal asymptote without ever reaching it.

15 30 45 60 75 90 68 100 150 200 t T
Figure 2Solutions of \dfrac{dT}{dt} = k(T - 68) with e^{15k} = \tfrac{2}{3}: the cider of Example 7 cooling from 194 degrees, and an object warming from 20 degrees. Both gaps to 68 shrink by the factor \tfrac{2}{3} every 15 minutes.

Results

Exponential models. The solutions of \dfrac{dy}{dt} = ky are exactly the functions

y = y_0 e^{kt}, \qquad y_0 = y(0),

defined for every t. No other function solves the equation on any interval.

The constant from two values. If y(t_1) = y_1 and y(t_2) = y_2 are both positive, then k = \dfrac{\ln(y_2 / y_1)}{t_2 - t_1}.

Doubling time and half-life. For y_0 > 0, the doubling time when k > 0 is \dfrac{\ln 2}{k}, and the half-life when k < 0 is \dfrac{\ln 2}{|k|}. Neither depends on y_0 or on when the clock starts.

Continuous compounding. A deposit of P dollars earning an annual rate r compounded continuously obeys \dfrac{dB}{dt} = rB, so its balance after t years is B = Pe^{rt}.

Newton’s law of cooling. The solution of \dfrac{dT}{dt} = k(T - A) with T(0) = T_0 is

T = A + (T_0 - A)e^{kt}.
Why it is true

Finding the solutions by separating. Where y \neq 0, separate as in Separable differential equations:

\begin{aligned} \int \frac{1}{y}\,dy &= \int k\,dt \\ \ln|y| &= kt + C. \end{aligned}

So |y| = e^C e^{kt}, and y = Be^{kt} with B = \pm e^C, any nonzero number. The equilibrium y = 0, lost when dividing by y, is B = 0. Setting t = 0 gives B = y_0.

No other solutions. Separating assumed y \neq 0 along the way, so it leaves open whether some solution could reach 0 and do something else. A direct argument closes that gap. Let y be any differentiable solution on an interval containing 0, and put g(t) = y(t)e^{-kt}. By the product rule,

\begin{aligned} g'(t) &= y'(t)e^{-kt} - k\,y(t)e^{-kt} \\ &= \big(ky(t) - ky(t)\big)e^{-kt} = 0. \end{aligned}

A function whose derivative is 0 on an interval is constant there (a consequence of the mean value theorem), so g(t) = g(0) = y_0 and y(t) = y_0 e^{kt}. In particular a solution that is 0 at one moment is 0 at every moment, and a positive solution stays positive. Conversely, y = y_0e^{kt} has y' = ky_0e^{kt} = ky.

Two values. Divide y_2 = y_0 e^{kt_2} by y_1 = y_0 e^{kt_1}: \dfrac{y_2}{y_1} = e^{k(t_2 - t_1)}. The ratio is positive, so its logarithm exists, and k(t_2 - t_1) = \ln\dfrac{y_2}{y_1}. A logarithm needs a positive input, the condition Solving logarithmic equations uses to reject answers.

Doubling time. For any t and any D, y(t + D) = y_0e^{kt}e^{kD} = y(t)\,e^{kD}. This equals 2y(t) exactly when e^{kD} = 2, that is, D = \dfrac{\ln 2}{k}, and neither t nor y_0 appears. For the half-life, e^{kH} = \tfrac{1}{2} gives H = -\dfrac{\ln 2}{k} = \dfrac{\ln 2}{|k|}.

Continuous compounding. Interest paid at every instant at annual rate r adds rB dollars per year to a balance of B dollars, which is \dfrac{dB}{dt} = rB. The same formula Pe^{rt} is the limit of compounding n times a year, P\left(1 + \dfrac{r}{n}\right)^{nt}, as n \to \infty (Compound interest and compounding periods); that limit is stated here without proof.

Newton’s law. Put u = T - A. Since A is constant,

\frac{du}{dt} = \frac{dT}{dt} = k(T - A) = ku.

By the first result, u = u(0)e^{kt} = (T_0 - A)e^{kt}, so T = A + (T_0 - A)e^{kt}, and every solution has this form.

Worked examples

Example 1

Day 1: Rawan’s carp

Rawan stocks a pond with 150 carp. While food and space are plentiful, the population grows at a relative rate of 0.18 per year. Write the model, find the population after 5 years, find when it reaches 600, and find the doubling time.

Solution

Let P be the number of carp t years after stocking. A relative rate of 0.18 per year means \dfrac{1}{P}\,\dfrac{dP}{dt} = 0.18, so \dfrac{dP}{dt} = 0.18P with P(0) = 150, and by the Results

P = 150e^{0.18t}.

After 5 years, P(5) = 150e^{0.9} \approx 368.9, about 369 carp. A count of fish is an integer; the model is a smooth approximation to it.

For 600 carp, divide by 150 and take the natural logarithm: e^{0.18t} = 4, so 0.18t = \ln 4 and t = \dfrac{\ln 4}{0.18} \approx 7.70 years.

The doubling time is \dfrac{\ln 2}{0.18} \approx 3.85 years. As a check, 600 is two doublings of 150, and 2 \times 3.85 = 7.70 (Figure 1).

Example 2

Day 1: Ghufran finds the constant from two counts

Ghufran’s bacterial culture holds 2{,}000 cells per milliliter at 3 hours and 4{,}500 at 7 hours, growing at a rate proportional to its size. Find k, the count at time 0, the doubling time, and the count predicted at 10 hours.

Solution

From the two values,

k = \frac{\ln(4500/2000)}{7 - 3} = \frac{\ln 2.25}{4} \approx 0.2027 \text{ per hour}.

The starting count satisfies y_0e^{3k} = 2000. Since 3k = \tfrac{3}{4}\ln 2.25, e^{3k} = 2.25^{3/4}, so y_0 = \dfrac{2000}{2.25^{3/4}} \approx 1089 cells per milliliter.

The doubling time is \dfrac{\ln 2}{k} = \dfrac{4\ln 2}{\ln 2.25} \approx 3.42 hours. The 10-hour count is 3 hours past the 7-hour count, so it is that count times e^{3k}: 4500 \cdot 2.25^{3/4} \approx 8267 cells per milliliter. Any measured value can serve as the starting point; y_0 was not needed for the prediction.

Example 3

Day 1: Shatha dates a charcoal sample

Carbon-14 has a half-life of 5{,}730 years. A living thing keeps the same share of carbon-14 as the air; after it dies the carbon-14 decays and is not replaced. Shatha’s charcoal from an old hearth holds 38\% of the carbon-14 that the same mass of living wood holds. About how old is it?

Solution

From the half-life, k = -\dfrac{\ln 2}{5730} \approx -0.000121 per year. With y = y_0e^{kt}, the charcoal’s amount is 0.38y_0:

0.38\,y_0 = y_0 e^{kt}.

The unknown y_0 cancels, leaving e^{kt} = 0.38, so

t = \frac{\ln 0.38}{k} = \frac{5730\,\ln(1/0.38)}{\ln 2} \approx 7{,}999,

about 8{,}000 years. As a check, 38\% lies between \tfrac{1}{2} and \tfrac{1}{4}, so the age lies between one and two half-lives, 5{,}730 and 11{,}460 years.

Example 4

Day 1: Rawan’s savings account

Rawan deposits 2{,}500 dollars in an account paying 4.2\% a year compounded continuously. Find the balance after 8 years and compare it with once-a-year compounding at the same rate. Find the doubling time, and the continuous rate that would double the deposit in 12 years.

Solution

The balance obeys \dfrac{dB}{dt} = 0.042B, so B = 2500e^{0.042t} and

B(8) = 2500e^{0.336} \approx 3498.35 \text{ dollars}.

Once a year, the balance is multiplied by 1.042 each year: 2500(1.042)^8 \approx 3474.42 dollars. Continuous compounding pays 23.93 dollars more, because interest starts earning interest the moment it is paid.

The doubling time is \dfrac{\ln 2}{0.042} \approx 16.50 years. To double in 12 years, e^{12r} = 2, so r = \dfrac{\ln 2}{12} \approx 0.05776, about 5.78\% a year.

Example 5

Day 2: Ghufran’s casserole

Ghufran takes a casserole out of the oven at 350 degrees Fahrenheit and sets it in a kitchen at 72 degrees. Ten minutes later it is at 260 degrees. Find k, the temperature after 30 minutes, when it reaches 140 degrees, and how fast it cools at the start.

Solution

With T_0 = 350 and A = 72, Newton’s law gives T = 72 + 278e^{kt}, t in minutes. The reading at t = 10:

\begin{aligned} 260 &= 72 + 278e^{10k} \\ e^{10k} &= \tfrac{188}{278}, \qquad k = \tfrac{1}{10}\ln\tfrac{188}{278} \approx -0.03912. \end{aligned}

After 30 minutes, e^{30k} = \left(e^{10k}\right)^3, so T(30) = 72 + 278\left(\tfrac{188}{278}\right)^3 \approx 158.0 degrees.

For 140 degrees, 278e^{kt} = 68, so t = \dfrac{\ln(68/278)}{k} \approx 36.0 minutes.

At the start, \dfrac{dT}{dt} = k(350 - 72) = 278k \approx -10.87 degrees per minute. In the first 10 minutes it fell 90 degrees, an average of 9 degrees per minute, less than the starting rate because the cooling slows as the gap closes.

Example 6

Day 2: Shatha’s orange juice warms

Shatha takes a pitcher of orange juice from the refrigerator at 38 degrees Fahrenheit and sets it on a porch at 86 degrees. After 25 minutes it is at 50 degrees. When does it reach 70 degrees?

Solution

Here T_0 - A = 38 - 86 = -48, so T = 86 - 48e^{kt}. The reading gives 50 = 86 - 48e^{25k}, so e^{25k} = \tfrac{36}{48} = \tfrac{3}{4} and k = \tfrac{1}{25}\ln\tfrac{3}{4} \approx -0.01151 per minute. The constant is negative, exactly as for cooling: the gap shrinks either way.

At 70 degrees, 48e^{kt} = 16, so e^{kt} = \tfrac{1}{3} and

t = \frac{\ln(1/3)}{k} = \frac{25\ln 3}{\ln(4/3)} \approx 95.5 \text{ minutes}.

As a check, the gap is multiplied by \tfrac{3}{4} every 25 minutes: 48, 36, 27, 20.25, 15.19. It passes 16 between 75 and 100 minutes.

Example 7

Day 2: Rawan’s cider, porch temperature unknown

Rawan carries a mug of hot cider onto a porch whose temperature she does not know. Her thermometer reads 194, 152 and 124 degrees Fahrenheit at 0, 15 and 30 minutes. Find the porch temperature and when the cider reaches 100 degrees.

Solution

Over each 15 minutes the gap T - A is multiplied by the same factor e^{15k}, so

\frac{152 - A}{194 - A} = \frac{124 - A}{152 - A}.

Cross-multiplying, (152 - A)^2 = (194 - A)(124 - A), that is, 23104 - 304A + A^2 = 24056 - 318A + A^2. The A^2 terms cancel, leaving 14A = 952, so A = 68 degrees. The factor is then \tfrac{84}{126} = \tfrac{2}{3}, and \tfrac{56}{84} = \tfrac{2}{3} agrees.

So e^{15k} = \tfrac{2}{3} and T = 68 + 126\left(\tfrac{2}{3}\right)^{t/15}. At 100 degrees, 126\left(\tfrac{2}{3}\right)^{t/15} = 32:

t = \frac{15\ln(32/126)}{\ln(2/3)} \approx 50.7 \text{ minutes}.

Figure 2 shows this curve, with k = \tfrac{1}{15}\ln\tfrac{2}{3} \approx -0.02703 per minute.

Boundaries

  • The rate must stay proportional. A real population slows as food or space runs short, so exponential growth describes only its early stage; a ceiling is the subject of logistic growth (and of Logistic growth: when growth levels off).
  • The constants must stay constant. If the interest rate changes, or the ambient temperature does (an oven heating up, the air outdoors over a day), the formulas here no longer apply. The equation becomes \dfrac{dT}{dt} = k\big(T - A(t)\big), which this lesson does not solve.
  • Doubling time and half-life need y_0 > 0 and k \neq 0. The two-value formula needs y_1 and y_2 of the same sign, so that the ratio has a logarithm.
  • Newton’s law fits only some objects. It assumes the object has one temperature throughout. A large roast warms from the outside in, so a reading at its center lags and does not follow a single exponential. During a change of state, ice melting or water boiling, the temperature stays fixed while heat flows, and the law fails. At large temperature differences, radiation and evaporation (a very hot pan, a steaming drink) add cooling the law does not include, so a k found from one reading may not hold later.
  • Carbon dating assumes the starting share is known. The share of carbon-14 in the air has varied over time, and laboratories correct for that; the computation above is the model before correction.

Where this goes next

Logistic growth replaces ky with a rate that slows as the population nears a ceiling, and it is solved by the same separation. Equations of the form \dfrac{dy}{dt} = ky + b (Exercise 24) are Newton’s law in disguise and the first of the linear differential equations met in later courses. Euler’s method is tested against y_0e^{kt}, since the exact answer is known. The facts proved here, that the rate decides the function and that one family of functions answers a whole class of questions, are the reason differential equations are the language of models in science.

Exercises

Day 1: solving and reading the model

  1. Solve \dfrac{dy}{dt} = -0.08y with y(0) = 650. Is this growth or decay, and what is y(5)?
  2. A quantity starts at 300 and grows at a relative rate of 0.065 per hour. When does it reach 1{,}000?
  3. A quantity grows at a rate proportional to its size, with y(0) = 80 and y(4) = 200. Find k and the doubling time.
  4. A decaying quantity has y(2) = 500 and y(6) = 320. Find k, y(0) and the half-life.

Day 1: half-life and doubling time

  1. Iodine-131, used in medicine, has a half-life of 8.02 days. Find k and the fraction of a dose left after 30 days.
  2. Cesium-137 has a half-life of 30.2 years. How long until 90\% of a sample has decayed?
  3. A wooden tool handle holds 61\% of the carbon-14 of living wood. About how old is it?
  4. A bacterial population doubles every 9 hours. Find its relative growth rate and the factor by which it grows in 24 hours.

Day 1: money and populations

  1. Ghufran deposits 4{,}000 dollars at 3.5\% a year compounded continuously. Find the balance after 15 years.
  2. What annual rate, compounded continuously, triples a deposit in 20 years?
  3. A town of 18{,}000 people is growing by 540 people per year, at a rate proportional to its size. Find k and when the town reaches 25{,}000.

Day 2: Newton’s law of cooling

  1. Solve \dfrac{dT}{dt} = -0.12(T - 20) with T(0) = 90, temperature in degrees Celsius and time in minutes, and find T(10).
  2. A loaf of bread leaves the oven at 210 degrees Fahrenheit into a kitchen at 70 degrees. Fifteen minutes later it is at 150 degrees. When does it reach 90 degrees?
  3. Shatha leaves a bottle of water at 40 degrees Fahrenheit on a beach at 85 degrees. After 30 minutes it is at 55 degrees. Without logarithms, find its temperature after one hour.
  4. An object’s temperature is T = 72 + 130e^{-0.05t} (degrees Fahrenheit, minutes). Read off the ambient temperature, the starting temperature and k; find how fast it cools at t = 0; and find when it is cooling at 2 degrees per minute.
  5. An egg at 95 degrees Celsius is dropped into ice water held at 0 degrees. After 4 minutes it is at 40 degrees. Find k, and when the egg reaches 15 degrees.
  6. A pan in a garage reads 160, 124 and 100 degrees Fahrenheit at 0, 10 and 20 minutes. Find the garage temperature and the pan’s temperature at 30 minutes.
  7. Ghufran finds a bowl of oatmeal at 100 degrees Fahrenheit at 8:00 in a room at 70 degrees; at 8:10 it is at 90 degrees. If it was 180 degrees when served, about when was it served?

Reasoning

  1. With k > 0 and y_0 > 0, prove that y = y_0 e^{kt} satisfies y(t + D) = 2y(t) for every t, where D = \dfrac{\ln 2}{k}, and explain why the doubling time does not depend on when the clock starts.
  2. Explain how the argument with g(t) = y(t)e^{-kt} shows that a solution of \dfrac{dy}{dt} = ky that is 0 at one moment is 0 at every moment, and why separating the variables alone does not show this.
  3. Show that if T_0 \neq A, the solution T = A + (T_0 - A)e^{kt} never equals A and stays on the same side of A as it starts.
  4. Show that e^r > 1 + r for every r > 0, and conclude that continuous compounding at annual rate r pays more in a year than compounding once a year at rate r. What once-a-year rate matches 5\% compounded continuously?
  5. A block of ice at 20 degrees Fahrenheit is set in a room at 70 degrees. Explain why its temperature does not follow Newton’s law all the way to 70 degrees.
  6. Write \dfrac{dy}{dt} = -0.2y + 6 as \dfrac{dy}{dt} = -0.2(y - 30), substitute u = y - 30, and solve with y(0) = 0. What happens to y in the long run?
Answers
  1. Decay, since k = -0.08 < 0. y = 650e^{-0.08t}, and y(5) = 650e^{-0.4} \approx 435.7.
  2. t \approx 18.5 hours. From 300e^{0.065t} = 1000: t = \dfrac{\ln(10/3)}{0.065}.
  3. k = \dfrac{\ln 2.5}{4} \approx 0.2291, and the doubling time is \dfrac{4\ln 2}{\ln 2.5} \approx 3.03.
  4. k = \dfrac{\ln 0.64}{4} \approx -0.1116. Then \displaystyle e^{2k} = \sqrt{0.64} = 0.8, so y(0) = \dfrac{500}{0.8} = 625; the half-life is \dfrac{\ln 2}{0.1116} \approx 6.21.
  5. k = -\dfrac{\ln 2}{8.02} \approx -0.0864 per day; the fraction left is \left(\tfrac{1}{2}\right)^{30/8.02} \approx 0.0748, about 7.5\%.
  6. About 100.3 years. With 10\% left, e^{kt} = 0.1, so t = \dfrac{30.2\ln 10}{\ln 2}.
  7. About 4{,}086 years: t = \dfrac{5730\ln(1/0.61)}{\ln 2}.
  8. k = \dfrac{\ln 2}{9} \approx 0.0770 per hour; the factor in 24 hours is 2^{24/9} \approx 6.35.
  9. 4000e^{0.525} \approx 6761.84 dollars.
  10. r = \dfrac{\ln 3}{20} \approx 0.0549, about 5.49\%, from e^{20r} = 3.
  11. k = \dfrac{540}{18000} = 0.03 per year, and t = \dfrac{\ln(25/18)}{0.03} \approx 10.95 years.
  12. T = 20 + 70e^{-0.12t}, and T(10) = 20 + 70e^{-1.2} \approx 41.1 degrees Celsius.
  13. About 52.2 minutes. T = 70 + 140e^{kt} with e^{15k} = \tfrac{80}{140} = \tfrac{4}{7}, so k \approx -0.03731; then 140e^{kt} = 20 gives t = \dfrac{15\ln 7}{\ln(7/4)}.
  14. 65 degrees. The gap T - 85 went from -45 to -30 in 30 minutes, a factor of \tfrac{2}{3}; after 30 more minutes it is -20.
  15. Ambient 72 degrees, starting 202 degrees, k = -0.05 per minute. T'(0) = -0.05 \cdot 130 = -6.5 degrees per minute. From 6.5e^{-0.05t} = 2: t = 20\ln 3.25 \approx 23.6 minutes.
  16. T = 95e^{kt} with e^{4k} = \tfrac{8}{19}, so k = \tfrac{1}{4}\ln\tfrac{8}{19} \approx -0.2162 per minute. From 95e^{kt} = 15: t = \dfrac{4\ln(3/19)}{\ln(8/19)} \approx 8.5 minutes.
  17. A = 52 degrees: (124 - A)^2 = (160 - A)(100 - A) gives 15376 - 248A = 16000 - 260A, so 12A = 624. The factor is \tfrac{72}{108} = \tfrac{2}{3}, so T(30) = 52 + 108\left(\tfrac{2}{3}\right)^3 = 84 degrees.
  18. About 7:28. With t in minutes after 8:00, T = 70 + 30e^{kt} and e^{10k} = \tfrac{20}{30} = \tfrac{2}{3}. Served when 30e^{kt} = 110: t = \dfrac{10\ln(11/3)}{\ln(2/3)} \approx -32.0.
  19. y(t + D) = y_0e^{kt}e^{kD} = y(t)\,e^{\ln 2} = 2y(t). The factor e^{kD} does not involve t, so every stretch of length D doubles the amount, wherever it starts.
  20. The argument shows every solution equals y_0e^{kt} with y_0 = y(0). If y(t_1) = 0, then y_0e^{kt_1} = 0, and since e^{kt_1} > 0, y_0 = 0, so y = 0 everywhere. Separating divides by y, so it describes a solution only on stretches where y \neq 0 and says nothing about one that reaches 0.
  21. T - A = (T_0 - A)e^{kt}, and e^{kt} > 0, so T - A is never 0 and always has the sign of T_0 - A.
  22. Let f(r) = e^r - 1 - r. Then f(0) = 0 and f'(r) = e^r - 1 > 0 for r > 0, so f increases on r \ge 0 and f(r) > 0 for r > 0. In one year, continuous compounding multiplies by e^r and once-a-year by 1 + r. For 5\%: e^{0.05} - 1 \approx 0.0513, about 5.13\% once a year.
  23. At 32 degrees the ice melts, and the heat flowing in goes into melting rather than raising the temperature, so it stays at 32 degrees until the ice is gone. Newton’s law would have the temperature keep rising toward 70 the whole time.
  24. y = 30 - 30e^{-0.2t}. With u = y - 30, \dfrac{du}{dt} = \dfrac{dy}{dt} = -0.2u and u(0) = -30, so u = -30e^{-0.2t}. As t \to \infty, y \to 30 from below.
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