Exponential Growth and Decay: Free Response

Plan Fixed adds the same 3000 dollars every year on top of the starting 50000, so after $t$ years the budget is the starting amount plus $t$ copies of 3000:

$F(t) = 50000 + 3000t.$

Plan Percent multiplies the budget by the growth factor $b = 1 + 0.06 = 1.06$ every year, so after $t$ years it has been multiplied by $1.06$ a total of $t$ times:

$E(t) = 50000(1.06)^{t}.$

Substitute $t=4$ into each model.

$F(4) = 50000 + 3000(4) = 50000 + 12000 = 62000.$

For Plan Percent,

$E(4) = 50000(1.06)^{4} = 50000(1.26247696) \approx 63123.85.$

Comparing the two, Plan Percent's budget of about 63123.85 dollars is larger than Plan Fixed's 62000 dollars.

Plan Fixed's yearly increase is fixed by definition: it is 3000 dollars whether the budget is small or large, because a flat dollar amount does not reference the current budget at all.

Plan Percent's yearly increase, by contrast, is $6\%$ of whatever the budget already is that year, which is

$E(t) - E(t-1) = 0.06\,E(t-1).$

Since $E(t-1)$ itself keeps growing year over year, the quantity $0.06\,E(t-1)$, the dollar amount Plan Percent adds, grows right along with it. Plan Fixed's increase stays flat because it never looks at the current budget in the first place.

With one compounding period a year, $n=1$, so the model is

$A = P(1+r)^{t} = 5000(1.05)^{10}.$

Since $1.05^{10} \approx 1.628895$,

$A \approx 5000(1.628895) \approx 8144.47.$

The balance compounded annually is about 8144.47 dollars after 10 years.

Quarterly compounding means $n=4$ periods a year, so the period rate is $r/n = 0.05/4 = 0.0125$ and the number of periods is $nt = 4(10) = 40$:

$A = 5000\left(1+\frac{0.05}{4}\right)^{40} = 5000(1.0125)^{40}.$

Since $1.0125^{40}\approx1.643620$,

$A \approx 5000(1.643620) \approx 8218.10.$

That is about $8218.10 - 8144.47 = 73.63$ dollars more than compounding annually.

Monthly compounding uses $n=12$ periods a year rather than $n=4$, so within the same 10 years there are $12(10)=120$ periods instead of $40$, each crediting a smaller amount of interest, $r/12$ rather than $r/4$, but more often:

$\frac{r}{12} < \frac{r}{4}, \qquad 12t > 4t.$

Because interest that has already been credited starts earning interest of its own on the very next period, crediting it sooner and more often can only help the balance, never hurt it. So for a fixed principal, rate, and time, a larger number of compounding periods per year produces a balance that is at least as large as one with fewer periods, and here strictly larger.

Each patch doubles every 8 days, so its model has the doubling-time form $P\cdot 2^{t/d}$ with $d=8$. Patch A starts at 15 square meters:

$A_A(t) = 15\cdot 2^{t/8}.$

Patch B starts at 30 square meters, twice as large, with the same doubling time:

$A_B(t) = 30\cdot 2^{t/8}.$

Set each model equal to 480 and solve for $t$. For Patch A:

$15\cdot 2^{t/8} = 480 \;\Longrightarrow\; 2^{t/8} = 32 = 2^{5} \;\Longrightarrow\; \frac{t}{8}=5 \;\Longrightarrow\; t=40.$

For Patch B:

$30\cdot 2^{t/8} = 480 \;\Longrightarrow\; 2^{t/8} = 16 = 2^{4} \;\Longrightarrow\; \frac{t}{8}=4 \;\Longrightarrow\; t=32.$

Since both ratios came out as exact powers of 2, no decimal logarithm was even needed here.

Let $L$ be any target area both patches could reach. Solving each model for the exponent gives

$2^{t_A/8} = \frac{L}{15}, \qquad 2^{t_B/8} = \frac{L}{30}.$

Since Patch B's starting area is exactly double Patch A's, $\dfrac{L}{30} = \dfrac12\cdot\dfrac{L}{15}$, so

$2^{t_B/8} = \frac12\cdot 2^{t_A/8} = 2^{t_A/8 - 1}.$

Two powers of the same base are equal exactly when their exponents are equal, so

$\frac{t_B}{8} = \frac{t_A}{8} - 1 \;\Longrightarrow\; t_B = t_A - 8.$

This used no particular value of $L$, only that Patch B always starts twice as large, so Patch B reaches EVERY shared target exactly 8 days, one full doubling time, before Patch A does.

Each isotope follows the half-life form $P\left(\tfrac12\right)^{t/h}$ with the same starting dose $P=50$ but a different half-life $h$. Isotope X has $h=5$:

$X(t) = 50\left(\tfrac12\right)^{t/5}.$

Isotope Y has $h=8$:

$Y(t) = 50\left(\tfrac12\right)^{t/8}.$

Substitute $t=8$ into each model. For Isotope X,

$X(8) = 50\left(\tfrac12\right)^{8/5} = 50\left(\tfrac12\right)^{1.6} \approx 50(0.32988) \approx 16.49.$

For Isotope Y, the exponent comes out to exactly $1$ since 8 hours is exactly one half-life:

$Y(8) = 50\left(\tfrac12\right)^{8/8} = 50\left(\tfrac12\right)^{1} = 25.$

So about 16.49 millicuries of Isotope X remain, against exactly 25 millicuries of Isotope Y.

Fix any elapsed time $t>0$. Dividing the same $t$ by a larger half-life produces a smaller exponent:

$h=8 > h=5 \;\Longrightarrow\; \frac{t}{8} < \frac{t}{5}.$

The function $\left(\tfrac12\right)^{x}$ is decreasing in $x$, since each additional unit of exponent halves the value again, so a smaller exponent always produces a LARGER value of $\left(\tfrac12\right)^{x}$:

$\frac{t}{8}<\frac{t}{5} \;\Longrightarrow\; \left(\frac12\right)^{t/8} > \left(\frac12\right)^{t/5}.$

Since both isotopes start from the same dose, multiplying this inequality by 40 shows Isotope Y always has strictly more activity remaining than Isotope X, at every positive elapsed time, not only at 8 hours.

Both readings fit $A(t) = P\cdot b^{t}$ for the same unknown $P$ and $b$, so dividing the later reading by the earlier one cancels $P$ and leaves a pure power of $b$:

$\frac{A(5)}{A(2)} = \frac{P b^{5}}{P b^{2}} = b^{3} = \frac{2430}{720} = 3.375.$

Since $1.5^{3} = 3.375$,

$b = 3.375^{1/3} = 1.5.$

The day-2 reading is $P b^{2} = 720$. Using $b=1.5$ from part A,

$b^{2} = 1.5^{2} = 2.25, \qquad P = \frac{720}{2.25} = 320.$

So the video had 320 views on day 0, before the tracker began recording it.

With $P=320$ and $b=1.5$, the day-0-anchored model is $A(t) = 320(1.5)^{t}$. Checking whole days,

$A(14) = 320(1.5)^{14} \approx 320(291.929) \approx 93417,$

still under 100000, while

$A(15) = 320(1.5)^{15} \approx 320(437.894) \approx 140126,$

which is over 100000. So day 15 is the first whole day the count exceeds 100000.

Building the day-0-anchored form specifically needed both earlier parts: part A's ratio step only ever isolates $b$, since $P$ cancels out of it by construction, and part B's back-solving step only ever isolates $P$, and only once $b$ is already known. Neither part alone produces the pair $(P,b)$ that this particular form requires; a model anchored at a DIFFERENT known day, such as day 2 or day 5, could reach the same numeric answer using $b$ alone, but writing it in the day-0 form specifically needs both results.