Unit 8Net Change vs. Total Change

Concept 4.147Built for May 2027 AP Exam

Net Change vs. Total Change

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

If a quantity $Q$ changes at a known rate $Q'(t)$ , then the net change in $Q$ over $[a, b]$ is $\int_a^b Q'(t)\,dt = Q(b) - Q(a)$ . That is the net change theorem, and it is just the fundamental theorem read backwards. Net change is a signed quantity: gains and losses cancel. Total change, by contrast, is $\int_a^b |Q'(t)|\,dt$ , which adds every gain and every loss at full size. Population is the classic example. If births run at $B(t)$ and deaths at $D(t)$ , the net change is $\int (B - D)$ , while the total change is $\int B + \int D$ . Two numbers, one situation, no ambiguity once you know which noun the question used.

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: A rate with units, a time interval, and a question phrased as “what is the net change in …” or “by how much did … change.” Occasionally the rate is split into two competing rates, one adding and one removing.
What you’ll do: Integrate the net rate for net change; integrate each rate separately and add magnitudes for total change. State units, since the net rate times time returns the original quantity.
Memory hook: Net cancels, total counts. The integral is the same operation; the absolute value is the difference.
Use it on the exam: For MCQs: If two rates are given, check whether the question wants the difference integrated or the two integrals added. The wording “net” decides instantly. For FRQs: Write the integral with limits and a rate before evaluating. The setup line carries a point, and the units on the answer carry another.

Concept Review

The Net Change Theorem

Figure 1 — Net accumulation can fall back to zero while total accumulation keeps climbing.

For a differentiable $Q$ with continuous derivative,

$\int_a^b Q'(t)\,dt = Q(b) - Q(a)$

Read left to right, it says: accumulate the rate and you recover the change in the quantity. Read right to left, it says: subtract two function values and you have the accumulated rate. The theorem is not new machinery. It is the fundamental theorem with the two sides swapped in emphasis, and it is the reason rate problems can be answered without ever building an antiderivative by hand.

Left panel: a falling rate curve with blue area above the axis and amber area below. Right panel: two accumulation curves that coincide until the rate reaches zero at t equals 2.4, then split, with the amber band between them

Where the Two Numbers Part Company

Net change keeps the sign of the rate. Total change discards it.

Rate behaviour	Net change	Total change
rate always positive	$\int_a^b r\,dt$	equal to net change
rate always negative	$\int_a^b r\,dt$ , negative	$\left\lvert \int_a^b r\,dt \right\rvert$
rate changes sign once	signed sum	signed sum plus twice the backward part
two competing rates $B$ and $D$	$\int (B - D)\,dt$	$\int B\,dt + \int D\,dt$

Row four is the one that appears in context. A reservoir gains water at one rate and loses it at another; the net change is the difference of the two integrals, while the total water that moved through the system is their sum.

Units Are the Check

A rate always carries “quantity per time.” Multiplying by time returns the quantity, so the answer to a net change question carries the units of the quantity itself, not the rate.

Rate and units	Interval	Net change carries
gallons per minute	minutes	gallons
people per year	years	people
cubic feet per second	seconds	cubic feet
degrees per hour	hours	degrees

If your answer comes out in gallons per minute, you forgot to integrate. If it comes out in gallons per minute squared, you integrated something twice.

Recovering the Quantity Itself

Net change gives you a difference, not a level. To find $Q(b)$ you still need one known value:

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

Released parts often state the initial amount in the stem and then ask for the amount at a later time. That sentence exists so you can add it. A question that gives no starting value can only be asking for the change, and answering with a level is unscoreable.

Two Rates, One Quantity

When a quantity has an inflow and an outflow, the net rate is the difference and the total movement is the sum. A tank with inflow $5$ and outflow $2$ gains $3$ per minute net, but $7$ gallons per minute move through it.

$\text{net change} = \int_a^b \big(\text{in} - \text{out}\big)\,dt \qquad\qquad \text{total moved} = \int_a^b \text{in}\,dt + \int_a^b \text{out}\,dt$

The distinction is physical, not cosmetic. A problem about the tank’s level wants the first. A problem about wear on the pump wants the second.

Common Misconception Check

Myth 1: “Net change and total change are the same whenever the rate is positive.” Truth: They are equal in that case, which is exactly why the distinction gets forgotten until a rate turns negative.
Myth 2: “Net change gives the final amount.” Truth: It gives the difference. Adding the starting amount is a separate step that the stem usually supplies.
Myth 3: “A net change of zero means nothing happened.” Truth: It means the gains and losses balanced. The total change can be large while the net change is zero.

Topic-Specific Study Method

Read the Rate’s Sign Structure First

Borrowed idea: Net change and total change both come from a rate, yet they answer different questions about the same interval. The net change is simply the definite integral of the rate, and the fundamental theorem turns it into an endpoint difference that ignores everything in between. Total change adds the absolute value first, so a negative rate contributes its full magnitude instead of subtracting. Which one does the question want? Read whether the word “net” appears. When a rate switches sign, the two numbers split apart by exactly twice the backward accumulation, and no amount of careful arithmetic can repair a wrong choice. Keep both.

Exam disguise	What you see on the page	What your brain must translate	Required execution
“net change in the amount of water”	inflow and outflow rates	difference of rates	integrate $(\text{in} - \text{out})$
“total amount of water that flowed in”	one rate only	plain accumulation	integrate the inflow alone
“by how much did the population change”	birth and death rates	net change in disguise	integrate the difference, state people
“total change in the population”	the same two rates	both integrals, magnitudes added	integrate each, then add
“how many people are in the town at $t = 10$ ”	an initial count plus a rate	level, not a change	add the initial count to the integral

① Underline the word “net” or its absence before writing anything. That single word selects the formula.

② Write the rate with its units on the integral sign, so the answer’s units are forced rather than remembered.

③ For level questions, copy the initial value onto the page as a separate line. It is easy to lose inside a long computation.

Training & Error Diagnosis

Core Archetype (Simplest Form): Water flows into a tank at $R(t) = 4 + t$ gallons per minute and out at $D(t) = 1 + 0.5t$ gallons per minute, for $0 \le t \le 6$ minutes. Find the net change in the volume of water.

Standard Execution:

Net rate. $R - D = 3 + 0.5t$ , which is positive throughout, so no splitting is needed.

Integrate. $\int_0^6 (3 + 0.5t)\,dt = \left[3t + 0.25t^{2}\right]_0^6 = 18 + 9 = 27$ .

Units. Gallons per minute times minutes gives gallons.

Check. Both rates are positive and the inflow dominates, so a positive net change of $27$ gallons is plausible.

Exam Variation (Sign Change / Context Shift): A town’s population changes at a rate of $P'(t) = 40 - 8t$ people per year for $0 \le t \le 10$ . Find the net change and the total change in the population over the decade.

Solution: The rate is zero at $t = 5$ . Net change is $\int_0^{10}(40 - 8t)\,dt = \left[40t - 4t^{2}\right]_0^{10} = 400 - 400 = 0$ . Total change requires the split: $\int_0^5 (40 - 8t)\,dt = 200 - 100 = 100$ and $\int_5^{10}(40 - 8t)\,dt = (400 - 400) - 100 = -100$ . Adding absolute values gives $200$ people. The population ends where it began after gaining and losing $100$ people.

Common Traps & Failure Signals:

Absolute value outside (Signal: $\left|\int r\right|$ written for a total change): the absolute value belongs on each piece.
Level for a change (Signal: an answer of “ $500$ people” when the stem gave no starting count): a change needs no initial value; a level does.
Units dropped (Signal: “ $27$ ” with no gallons): context points are awarded for units.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Net Only: You used the signed integral for a total change.
- Level Confused: You reported the accumulated change as the final amount and never added the starting value the stem provided, so the arithmetic was right and the answer described the wrong quantity entirely.
- Sign Ignored: You assumed the rate stayed positive because the first term looked large, and the single reversal at $t = 5$ silently erased half of the total change.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; The net change theorem appears often in multiple choice, as a rate integrated over a span, but no released FRQ part since 2017 asks for it under that name
10-Year Total Count: 0 of 9 Exams · 0 Parts; Every released FRQ that touches accumulation phrases the task as an amount or an average instead, and those are scored under topics 8.1 and 8.6
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

Net Change From a Linear Rate

A quantity $Q$ changes at a rate of $Q'(t) = 6 - 2t$ units per hour. The net change in $Q$ on $0 \le t \le 4$ is

MCQ 2

Total Change With a Reversal

Using the rate $Q'(t) = 6 - 2t$ again, the total change in $Q$ on $0 \le t \le 4$ is

MCQ 3

Two Rates, One Quantity

A reservoir receives water at $5$ cubic meters per hour and drains at $2$ cubic meters per hour. Over $6$ hours, the net change in stored water and the total water moved are, respectively,

FRQ 4

A Tank, a Net Change, and a Level

Water flows into a tank at $R(t) = 4 + t$ gallons per minute and drains out at $D(t) = 1 + 0.5t$ gallons per minute, for $0 \le t \le 6$ . At $t = 0$ the tank holds $30$ gallons.

(a) Find the net change in the volume of water in the tank on $0 \le t \le 6$ and state its units.

(b) Find the volume of water in the tank at $t = 6$ .

(c) Explain why the answer to (a) is not the answer to (b), and identify the quantity the stem supplied to connect them.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	locate the word “net” or its absence	the wrong of two formulas gets used
2	write the rate with units	the answer’s units become a guess
3	solve $r(t) = 0$ if a total change is wanted	the reversal is missed and the pieces cancel
4	integrate the net rate for net change	the answer reports total movement
5	add absolute pieces for total change	the answer can come out smaller than the net change
6	add the initial value for a level question	a difference is reported where a level was asked

Sentences That Score

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

Ready to check your understanding?

Strengthen this topic with AP-style practice, or continue to next topic.

Start Topic Practice Next Topic: 8.6 The integral of a rate of change (e.g., rate of flow) gives the total net change (e.g., total quantity).

Unit 8Net Change vs. Total Change

Concept 4.147Built for May 2027 AP Exam

Net Change vs. Total Change

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: A rate with units, a time interval, and a question phrased as “what is the net change in …” or “by how much did … change.” Occasionally the rate is split into two competing rates, one adding and one removing.
What you’ll do: Integrate the net rate for net change; integrate each rate separately and add magnitudes for total change. State units, since the net rate times time returns the original quantity.
Memory hook: Net cancels, total counts. The integral is the same operation; the absolute value is the difference.
Use it on the exam: For MCQs: If two rates are given, check whether the question wants the difference integrated or the two integrals added. The wording “net” decides instantly. For FRQs: Write the integral with limits and a rate before evaluating. The setup line carries a point, and the units on the answer carry another.

Concept Review

The Net Change Theorem

Figure 1 — Net accumulation can fall back to zero while total accumulation keeps climbing.

For a differentiable $Q$ with continuous derivative,

$\int_a^b Q'(t)\,dt = Q(b) - Q(a)$

Where the Two Numbers Part Company

Net change keeps the sign of the rate. Total change discards it.

Rate behaviour	Net change	Total change
rate always positive	$\int_a^b r\,dt$	equal to net change
rate always negative	$\int_a^b r\,dt$ , negative	$\left\lvert \int_a^b r\,dt \right\rvert$
rate changes sign once	signed sum	signed sum plus twice the backward part
two competing rates $B$ and $D$	$\int (B - D)\,dt$	$\int B\,dt + \int D\,dt$

Units Are the Check

A rate always carries “quantity per time.” Multiplying by time returns the quantity, so the answer to a net change question carries the units of the quantity itself, not the rate.

Rate and units	Interval	Net change carries
gallons per minute	minutes	gallons
people per year	years	people
cubic feet per second	seconds	cubic feet
degrees per hour	hours	degrees

If your answer comes out in gallons per minute, you forgot to integrate. If it comes out in gallons per minute squared, you integrated something twice.

Recovering the Quantity Itself

Net change gives you a difference, not a level. To find $Q(b)$ you still need one known value:

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

Two Rates, One Quantity

$\text{net change} = \int_a^b \big(\text{in} - \text{out}\big)\,dt \qquad\qquad \text{total moved} = \int_a^b \text{in}\,dt + \int_a^b \text{out}\,dt$

The distinction is physical, not cosmetic. A problem about the tank’s level wants the first. A problem about wear on the pump wants the second.

Common Misconception Check

Myth 1: “Net change and total change are the same whenever the rate is positive.” Truth: They are equal in that case, which is exactly why the distinction gets forgotten until a rate turns negative.
Myth 2: “Net change gives the final amount.” Truth: It gives the difference. Adding the starting amount is a separate step that the stem usually supplies.
Myth 3: “A net change of zero means nothing happened.” Truth: It means the gains and losses balanced. The total change can be large while the net change is zero.

Topic-Specific Study Method

Read the Rate’s Sign Structure First

Exam disguise	What you see on the page	What your brain must translate	Required execution
“net change in the amount of water”	inflow and outflow rates	difference of rates	integrate $(\text{in} - \text{out})$
“total amount of water that flowed in”	one rate only	plain accumulation	integrate the inflow alone
“by how much did the population change”	birth and death rates	net change in disguise	integrate the difference, state people
“total change in the population”	the same two rates	both integrals, magnitudes added	integrate each, then add
“how many people are in the town at $t = 10$ ”	an initial count plus a rate	level, not a change	add the initial count to the integral

① Underline the word “net” or its absence before writing anything. That single word selects the formula.

② Write the rate with its units on the integral sign, so the answer’s units are forced rather than remembered.

③ For level questions, copy the initial value onto the page as a separate line. It is easy to lose inside a long computation.

Training & Error Diagnosis

Standard Execution:

Net rate. $R - D = 3 + 0.5t$ , which is positive throughout, so no splitting is needed.

Integrate. $\int_0^6 (3 + 0.5t)\,dt = \left[3t + 0.25t^{2}\right]_0^6 = 18 + 9 = 27$ .

Units. Gallons per minute times minutes gives gallons.

Check. Both rates are positive and the inflow dominates, so a positive net change of $27$ gallons is plausible.

Solution: The rate is zero at $t = 5$ . Net change is $\int_0^{10}(40 - 8t)\,dt = \left[40t - 4t^{2}\right]_0^{10} = 400 - 400 = 0$ . Total change requires the split: $\int_0^5 (40 - 8t)\,dt = 200 - 100 = 100$ and $\int_5^{10}(40 - 8t)\,dt = (400 - 400) - 100 = -100$ . Adding absolute values gives $200$ people. The population ends where it began after gaining and losing $100$ people.

Common Traps & Failure Signals:

Absolute value outside (Signal: $\left|\int r\right|$ written for a total change): the absolute value belongs on each piece.
Level for a change (Signal: an answer of “ $500$ people” when the stem gave no starting count): a change needs no initial value; a level does.
Units dropped (Signal: “ $27$ ” with no gallons): context points are awarded for units.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Net Only: You used the signed integral for a total change.
- Level Confused: You reported the accumulated change as the final amount and never added the starting value the stem provided, so the arithmetic was right and the answer described the wrong quantity entirely.
- Sign Ignored: You assumed the rate stayed positive because the first term looked large, and the single reversal at $t = 5$ silently erased half of the total change.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; The net change theorem appears often in multiple choice, as a rate integrated over a span, but no released FRQ part since 2017 asks for it under that name
10-Year Total Count: 0 of 9 Exams · 0 Parts; Every released FRQ that touches accumulation phrases the task as an amount or an average instead, and those are scored under topics 8.1 and 8.6
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

Net Change From a Linear Rate

A quantity $Q$ changes at a rate of $Q'(t) = 6 - 2t$ units per hour. The net change in $Q$ on $0 \le t \le 4$ is

MCQ 2

Total Change With a Reversal

Using the rate $Q'(t) = 6 - 2t$ again, the total change in $Q$ on $0 \le t \le 4$ is

MCQ 3

Two Rates, One Quantity

A reservoir receives water at $5$ cubic meters per hour and drains at $2$ cubic meters per hour. Over $6$ hours, the net change in stored water and the total water moved are, respectively,

FRQ 4

A Tank, a Net Change, and a Level

Water flows into a tank at $R(t) = 4 + t$ gallons per minute and drains out at $D(t) = 1 + 0.5t$ gallons per minute, for $0 \le t \le 6$ . At $t = 0$ the tank holds $30$ gallons.

(a) Find the net change in the volume of water in the tank on $0 \le t \le 6$ and state its units.

(b) Find the volume of water in the tank at $t = 6$ .

(c) Explain why the answer to (a) is not the answer to (b), and identify the quantity the stem supplied to connect them.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	locate the word “net” or its absence	the wrong of two formulas gets used
2	write the rate with units	the answer’s units become a guess
3	solve $r(t) = 0$ if a total change is wanted	the reversal is missed and the pieces cancel
4	integrate the net rate for net change	the answer reports total movement
5	add absolute pieces for total change	the answer can come out smaller than the net change
6	add the initial value for a level question	a difference is reported where a level was asked

Sentences That Score

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

Ready to check your understanding?

Strengthen this topic with AP-style practice, or continue to next topic.

Start Topic Practice Next Topic: 8.6 The integral of a rate of change (e.g., rate of flow) gives the total net change (e.g., total quantity).

Unit 8Net Change vs. Total Change

Concept 4.147Built for May 2027 AP Exam

Net Change vs. Total Change

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: A rate with units, a time interval, and a question phrased as “what is the net change in …” or “by how much did … change.” Occasionally the rate is split into two competing rates, one adding and one removing.
What you’ll do: Integrate the net rate for net change; integrate each rate separately and add magnitudes for total change. State units, since the net rate times time returns the original quantity.
Memory hook: Net cancels, total counts. The integral is the same operation; the absolute value is the difference.
Use it on the exam: For MCQs: If two rates are given, check whether the question wants the difference integrated or the two integrals added. The wording “net” decides instantly. For FRQs: Write the integral with limits and a rate before evaluating. The setup line carries a point, and the units on the answer carry another.

Concept Review

The Net Change Theorem

Figure 1 — Net accumulation can fall back to zero while total accumulation keeps climbing.

For a differentiable $Q$ with continuous derivative,

$\int_a^b Q'(t)\,dt = Q(b) - Q(a)$

Where the Two Numbers Part Company

Net change keeps the sign of the rate. Total change discards it.

Rate behaviour	Net change	Total change
rate always positive	$\int_a^b r\,dt$	equal to net change
rate always negative	$\int_a^b r\,dt$ , negative	$\left\lvert \int_a^b r\,dt \right\rvert$
rate changes sign once	signed sum	signed sum plus twice the backward part
two competing rates $B$ and $D$	$\int (B - D)\,dt$	$\int B\,dt + \int D\,dt$

Units Are the Check

A rate always carries “quantity per time.” Multiplying by time returns the quantity, so the answer to a net change question carries the units of the quantity itself, not the rate.

Rate and units	Interval	Net change carries
gallons per minute	minutes	gallons
people per year	years	people
cubic feet per second	seconds	cubic feet
degrees per hour	hours	degrees

If your answer comes out in gallons per minute, you forgot to integrate. If it comes out in gallons per minute squared, you integrated something twice.

Recovering the Quantity Itself

Net change gives you a difference, not a level. To find $Q(b)$ you still need one known value:

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

Two Rates, One Quantity

$\text{net change} = \int_a^b \big(\text{in} - \text{out}\big)\,dt \qquad\qquad \text{total moved} = \int_a^b \text{in}\,dt + \int_a^b \text{out}\,dt$

The distinction is physical, not cosmetic. A problem about the tank’s level wants the first. A problem about wear on the pump wants the second.

Common Misconception Check

Myth 1: “Net change and total change are the same whenever the rate is positive.” Truth: They are equal in that case, which is exactly why the distinction gets forgotten until a rate turns negative.
Myth 2: “Net change gives the final amount.” Truth: It gives the difference. Adding the starting amount is a separate step that the stem usually supplies.
Myth 3: “A net change of zero means nothing happened.” Truth: It means the gains and losses balanced. The total change can be large while the net change is zero.

Topic-Specific Study Method

Read the Rate’s Sign Structure First

Exam disguise	What you see on the page	What your brain must translate	Required execution
“net change in the amount of water”	inflow and outflow rates	difference of rates	integrate $(\text{in} - \text{out})$
“total amount of water that flowed in”	one rate only	plain accumulation	integrate the inflow alone
“by how much did the population change”	birth and death rates	net change in disguise	integrate the difference, state people
“total change in the population”	the same two rates	both integrals, magnitudes added	integrate each, then add
“how many people are in the town at $t = 10$ ”	an initial count plus a rate	level, not a change	add the initial count to the integral

① Underline the word “net” or its absence before writing anything. That single word selects the formula.

② Write the rate with its units on the integral sign, so the answer’s units are forced rather than remembered.

③ For level questions, copy the initial value onto the page as a separate line. It is easy to lose inside a long computation.

Training & Error Diagnosis

Standard Execution:

Net rate. $R - D = 3 + 0.5t$ , which is positive throughout, so no splitting is needed.

Integrate. $\int_0^6 (3 + 0.5t)\,dt = \left[3t + 0.25t^{2}\right]_0^6 = 18 + 9 = 27$ .

Units. Gallons per minute times minutes gives gallons.

Check. Both rates are positive and the inflow dominates, so a positive net change of $27$ gallons is plausible.

Solution: The rate is zero at $t = 5$ . Net change is $\int_0^{10}(40 - 8t)\,dt = \left[40t - 4t^{2}\right]_0^{10} = 400 - 400 = 0$ . Total change requires the split: $\int_0^5 (40 - 8t)\,dt = 200 - 100 = 100$ and $\int_5^{10}(40 - 8t)\,dt = (400 - 400) - 100 = -100$ . Adding absolute values gives $200$ people. The population ends where it began after gaining and losing $100$ people.

Common Traps & Failure Signals:

Absolute value outside (Signal: $\left|\int r\right|$ written for a total change): the absolute value belongs on each piece.
Level for a change (Signal: an answer of “ $500$ people” when the stem gave no starting count): a change needs no initial value; a level does.
Units dropped (Signal: “ $27$ ” with no gallons): context points are awarded for units.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Net Only: You used the signed integral for a total change.
- Level Confused: You reported the accumulated change as the final amount and never added the starting value the stem provided, so the arithmetic was right and the answer described the wrong quantity entirely.
- Sign Ignored: You assumed the rate stayed positive because the first term looked large, and the single reversal at $t = 5$ silently erased half of the total change.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; The net change theorem appears often in multiple choice, as a rate integrated over a span, but no released FRQ part since 2017 asks for it under that name
10-Year Total Count: 0 of 9 Exams · 0 Parts; Every released FRQ that touches accumulation phrases the task as an amount or an average instead, and those are scored under topics 8.1 and 8.6
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

Net Change From a Linear Rate

A quantity $Q$ changes at a rate of $Q'(t) = 6 - 2t$ units per hour. The net change in $Q$ on $0 \le t \le 4$ is

MCQ 2

Total Change With a Reversal

Using the rate $Q'(t) = 6 - 2t$ again, the total change in $Q$ on $0 \le t \le 4$ is

MCQ 3

Two Rates, One Quantity

A reservoir receives water at $5$ cubic meters per hour and drains at $2$ cubic meters per hour. Over $6$ hours, the net change in stored water and the total water moved are, respectively,

FRQ 4

A Tank, a Net Change, and a Level

Water flows into a tank at $R(t) = 4 + t$ gallons per minute and drains out at $D(t) = 1 + 0.5t$ gallons per minute, for $0 \le t \le 6$ . At $t = 0$ the tank holds $30$ gallons.

(a) Find the net change in the volume of water in the tank on $0 \le t \le 6$ and state its units.

(b) Find the volume of water in the tank at $t = 6$ .

(c) Explain why the answer to (a) is not the answer to (b), and identify the quantity the stem supplied to connect them.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	locate the word “net” or its absence	the wrong of two formulas gets used
2	write the rate with units	the answer’s units become a guess
3	solve $r(t) = 0$ if a total change is wanted	the reversal is missed and the pieces cancel
4	integrate the net rate for net change	the answer reports total movement
5	add absolute pieces for total change	the answer can come out smaller than the net change
6	add the initial value for a level question	a difference is reported where a level was asked

Sentences That Score

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

Ready to check your understanding?

Strengthen this topic with AP-style practice, or continue to next topic.

Start Topic Practice Next Topic: 8.6 The integral of a rate of change (e.g., rate of flow) gives the total net change (e.g., total quantity).