Accumulation in Context | Math Coach Amy

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Key Ideas Easy Medium Hard

\(A(b)=A(a)+\int_a^b A'(t)\,dt\)

What to Remember

A rate accumulates into an amount

\(A(b)=A(a)+\int_a^b R(t)\,dt\)

Include the starting amount. The integral alone is only the change.

In minus out

If \(A'(t)=I(t)-O(t)\), the amount is increasing when \(I(t)>O(t)\).

A table gives an approximation

A decreasing positive rate makes a left sum too big and a right sum too small. Say so when the problem gives that shape.

Units

Gallons per minute times minutes is gallons. The starting amount uses those same units.

Level 1

Easy

The rate is given by a formula, or the integral is already known.

Example

A tank starts with 100 gallons. Water enters at \(R(t)=6\) gallons per minute for 8 minutes, and no water leaves. How much water is in the tank then?

\(100+\int_0^8 6\,dt\)

\(100+48\)

\(148\)

Answer: 148 gallons

Sand falls into a pile at \(r(t)=2t\) kilograms per minute. The pile starts empty. How much sand is in the pile after 5 minutes?

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\(\int_0^5 2t\,dt=\left[t^2\right]_0^5\)

\(25\)

Answer: 25 kg

\(\int_0^{10} R(t)\,dt=40\) and the tank starts with 15 gallons. Find the amount at \(t=10\) if \(R\) is the only rate.

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\(15+40\)

\(55\)

Answer: 55 gallons

People enter a room at 5 per minute and leave at 2 per minute. Both rates are constant. The room starts with 12 people. How many are there after 6 minutes?

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Net rate: \(5-2=3\) people per minute.

\(12+\int_0^6 3\,dt\)

\(12+18\)

\(30\)

Answer: 30 people

A rate \(R(t)\) is positive on \((0,8)\). Is the accumulated amount increasing there?

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A positive rate of change means the amount is increasing.

Answer: yes

The change in temperature from \(t=1\) to \(t=4\) is \(\int_1^4 T'(t)\,dt=-6\). The temperature at \(t=1\) is 20. Find the temperature at \(t=4\).

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\(20+(-6)\)

\(14\)

Answer: \(14\)

Level 2

Medium

Approximate a table, and keep the starting amount.

Example

A tank contains 100 gallons at \(t=0\). Water enters at a decreasing rate.

t (min)	0	3	6	9
R(t) (gal/min)	10	7	5	2

Use a trapezoidal sum to approximate the amount at \(t=9\). Is the trapezoidal sum guaranteed to be too big?

\(\dfrac{3}{2}(10+2\cdot 7+2\cdot 5+2)\)

\(\dfrac{3}{2}\cdot 36\)

\(54\)

\(100+54=154\)

Decreasing is not the same as concave. The trapezoidal error needs concavity, which is not given.

Answer: about 154 gallons; the direction of the trapezoid error is not known

Using the same tank and table, approximate the gallons added with a right Riemann sum. Because the rate is decreasing, is this too big or too small?

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\(3(7+5+2)\)

\(42\)

A decreasing positive function makes the right sum too small.

Answer: 42 gallons, an underestimate

Water flows in and out.

t (hours)	0	2	4	6
in (gal/h)	8	6	5	4
out (gal/h)	3	4	5	7

At which of these times is the amount decreasing?

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Net rates: \(5\), \(2\), \(0\), and \(-3\).

The amount is decreasing when out exceeds in.

Among the listed times, that happens at \(t=6\).

Answer: \(t=6\)

The tank starts with 40 gallons. Use a left Riemann sum on the net rates from the previous table to approximate the amount at \(t=6\).

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Net: 5, 2, 0, and \(-3\).

Left sum: \(2(5+2+0)=14\).

\(40+14\)

\(54\)

Answer: about 54 gallons

A culture grows at \(r(t)=e^{t/2}\) grams per hour and starts at 10 grams. Find the amount after 2 hours.

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\(10+\int_0^2 e^{t/2}\,dt\)

\(\int e^{t/2}\,dt=2e^{t/2}\)

\(\left[2e^{t/2}\right]_0^2\)

\(2e-2\)

\(10+2e-2\)

\(8+2e\)

Answer: \(8+2e\) grams

Why is \(\int_0^9 R(t)\,dt\) not the amount of water at \(t=9\) when the tank started with 100 gallons?

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The integral is the change during the nine minutes.

The amount is the start plus the change.

Answer: add the starting 100 gallons

Level 3

Hard

Compare two approximations, or decide when the amount is greatest.

Example

For the decreasing rate 10, 7, 5, 2 at \(t=0,3,6,9\), the left sum for the change is \(3(10+7+5)=66\) and the right sum is 42. The tank starts at 100 gallons. Give an interval that contains the amount at \(t=9\) if the rate is decreasing and positive.

A decreasing positive rate traps the integral between the right sum and the left sum.

\(42 < \int_0^9 R(t)\,dt < 66\)

Add 100.

\(142 < ext{amount} < 166\)

Answer: between 142 and 166 gallons

Let \(A'(t)=I(t)-O(t)\). On \([0,5)\), \(I>O\). On \((5,9]\), \(I

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\(A\) increases on \([0,5]\) and decreases on \([5,9]\).

The greatest amount is at \(t=5\).

Answer: \(t=5\)

A pool starts with 200 gallons. The net rate is \(A'(t)=8-2t\) gallons per hour. Find the amount at \(t=6\), and the time when the pool holds the most water.

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\(A(t)=200+8t-t^2\)

\(A(6)=200+48-36=212\)

\(A'(t)=0\) at \(t=4\).

\(A'\) changes from positive to negative, so the maximum is at \(t=4\).

\(A(4)=200+32-16=216\)

Answer: \(A(6)=212\), maximum 216 gal at \(t=4\)

Oil leaks at \(r(t)=10e^{-t}\) liters per hour, starting at \(t=0\). How much leaks out from \(t=0\) until the leak has essentially ended, meaning through \(t\to\infty\)?

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\(\lim_{b\to\infty}\int_0^b 10e^{-t}\,dt\)

\(\lim_{b\to\infty}\left[-10e^{-t}\right]_0^b\)

\(0-(-10)\)

\(10\)

Answer: 10 liters

Two pumps add water. Pump A has added \(\int_0^4 a(t)\,dt=30\) gallons. Pump B runs at a constant 4 gal/min for those same 4 minutes. The tank started with 12 gallons and none leaves. How much is in the tank?

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Pump B adds \(4\cdot 4=16\) gallons.

\(12+30+16\)

\(58\)

Answer: 58 gallons

A student approximates \(\int_0^6 |I(t)-O(t)|\,dt\) and calls it the change in the amount. What did the student actually approximate?

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The absolute value erases the sign of the net rate.

That integral adds the gallons in and the gallons out as positive quantities.

The change in the amount is \(\int (I-O)\) without the absolute value.

Answer: total flow through, not the net change

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