AP Unit 7 Solutions
Worked solutions to all thirty Unit 7 problems. Each solution gives the mathematical steps and the reason for using them. Compare your setup and justification as well as the final answer.
Problem 1
A tank’s inflow is 5 L/min. Outflow is proportional to its current volume V with coefficient 0.1 min⁻¹. Write a differential equation and find the rate when V=20 L.
- Net rate is inflow minus outflow. Reason: Conservation of volume gives the derivative model.
- Substitute V=20: 5−0.1(20)=3. Reason: Evaluate the current rate from the current state.
Answer: V′=5−0.1V; rate = 3 L/min.
Final answer
V′=5−0.1V; rate = 3 L/min.
Review the teaching: Topic 7.1
Problem 2
A body’s temperature T moves toward a room temperature of 20°C at a rate proportional to the temperature difference, with positive coefficient k. Write the model and explain the sign when T=30.
- The difference from the room is T−20. Reason: The model acts on temperature difference.
- Choose −k(T−20). Reason: A body hotter than the room must cool, so the derivative is negative.
Answer: T′=−k(T−20); at T=30, T′=−10k<0.
Final answer
T′=−k(T−20); at T=30, T′=−10k<0.
Review the teaching: Topic 7.1
Problem 3
A particle’s acceleration is 2t m/s² at time t seconds, and its initial velocity is 3 m/s. Write an initial-value problem for velocity.
- Acceleration is the derivative of velocity. Reason: The motion definitions connect the two quantities.
- Record the initial velocity separately. Reason: An initial condition is a value, not an added rate.
Answer: v′(t)=2t, v(0)=3.
Final answer
v′(t)=2t, v(0)=3.
Review the teaching: Topic 7.1
Problem 4
Verify that y=3e^(2x) solves y′=2y and y(0)=3.
- Differentiate y to get y′=6e^(2x). Reason: Use the chain rule.
- Compute 2y=6e^(2x), and y(0)=3. Reason: Both sides agree throughout the domain and the initial value agrees.
Answer: It satisfies both the differential equation and initial condition for all real x.
Final answer
It satisfies both the differential equation and initial condition for all real x.
Review the teaching: Topic 7.2
Problem 5
Does y=x²+1 solve y′=2x and y(0)=0?
- Its derivative is 2x. Reason: Differentiation verifies the differential equation.
- At zero, y=1 rather than 0. Reason: Check the initial condition independently.
Answer: It solves the equation but fails the initial condition.
Final answer
It solves the equation but fails the initial condition.
Review the teaching: Topic 7.2
Problem 6
Verify that y=Ceˣ is a family of solutions to y′=y. Which member satisfies y(0)=−2?
- Differentiate Ceˣ to obtain Ceˣ=y. Reason: The equality holds for any real constant.
- Set C=y(0)=−2. Reason: The initial condition chooses this family member.
Answer: Every constant C works; the selected solution is y=−2eˣ.
Final answer
Every constant C works; the selected solution is y=−2eˣ.
Review the teaching: Topic 7.2
Problem 7
Does y=eˣ+x solve y′=y? Explain without using a graph.
- Differentiate: y′=eˣ+1, while y=eˣ+x. Reason: Verification compares the derivative to the equation’s right side.
- These agree only at x=1, not on an interval. Reason: A solution must satisfy the equation throughout its domain.
Answer: No.
Final answer
No.
Review the teaching: Topic 7.2
Problem 8
For y′=x−y, give slopes at (0,0), (1,2), and (2,−1), and describe the corresponding segments.
- Substitute each point: 0−0=0, 1−2=−1, 2−(−1)=3. Reason: A slope field evaluates the derivative at both coordinates.
- Orient short segments with those slopes. Reason: The derivative is the local tangent direction.
Answer: Slopes 0, −1, 3: horizontal, falling, and rising steeply.
Final answer
Slopes 0, −1, 3: horizontal, falling, and rising steeply.
Review the teaching: Topic 7.3
Problem 9
| y | −2 | −1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| slope | −2 | −1 | 0 | 1 | 2 |
- Slopes stay the same as x changes and match the sign and value of y. Reason: The graph and its equivalent table show slope determined by height, ruling out dependence only on x or the opposite sign.
- At y=1 the slope is positive, and positive solutions keep positive slopes. Reason: The matching equation y′=y or its verified solution y=eˣ describes the rising curve.
Answer: y′=y; the solution through (0,1) rises for increasing x.
Final answer
y′=y; the solution through (0,1) rises for increasing x.
Problem 10
For y′=x+y, where are the slope-field segments horizontal?
- Horizontal means slope zero. Reason: A horizontal tangent has derivative 0.
- Solve x+y=0 for y. Reason: The equation identifies all locations with that slope.
Answer: Along the line y=−x.
Final answer
Along the line y=−x.
Review the teaching: Topic 7.3
Problem 11
For y′=1/(x−y), can a slope-field segment be drawn at (1,1)? What is the slope at (1,0)?
- At (1,1), the denominator is zero. Reason: The differential equation is undefined there.
- At (1,0), the slope is 1/(1−0)=1. Reason: Evaluate the rule only at a defined point.
Answer: No finite segment at (1,1); slope 1 at (1,0).
Final answer
No finite segment at (1,1); slope 1 at (1,0).
Review the teaching: Topic 7.3
Problem 12
For y′=2−y, identify the equilibrium and describe a solution starting at y(0)=1 for x≥0.
- The derivative vanishes for the constant y=2; below 2 it is positive. Reason: This gives equilibrium and increasing behavior.
- Differentiate the equation: y″=−y′<0 below 2. Reason: The chain rule gives concavity.
- The solution y=2−e^(−x) confirms the forward limit 2. Reason: Solving or direct verification makes the asymptotic statement precise.
Answer: Equilibrium y=2; the solution rises toward 2 and is concave down.
Final answer
Equilibrium y=2; the solution rises toward 2 and is concave down.
Review the teaching: Topic 7.4
Problem 13
For y′=2−y, describe a solution starting at y(0)=3 for x≥0.
- Above 2, y′<0, while y″=−y′>0. Reason: The equation gives both derivative signs.
- The verified solution y=2+e^(−x) approaches 2. Reason: Its exponential term tends to zero.
Answer: It decreases toward 2 and is concave up.
Final answer
It decreases toward 2 and is concave up.
Review the teaching: Topic 7.4
Problem 14
For y′=xy, find y′ and y″ at a solution point (1,2).
- Evaluate y′=xy=2. Reason: Use the derivative rule at the specified point.
- Differentiate: y″=y+xy′=2+1·2=4. Reason: Product rule applies because y is a function of x.
Answer: y′=2 and y″=4.
Final answer
y′=2 and y″=4.
Review the teaching: Topic 7.4
Problem 15
Find the general solution of y′=2xy, including any equilibrium lost by division.
- For y≠0, separate dy/y=2x dx and integrate ln|y|=x²+C. Reason: Separation uses reverse chain-rule integration.
- Exponentiate with a signed constant A. Reason: The absolute value allows either sign.
- Check y=0 directly and include it as A=0. Reason: Division by y excluded that equilibrium.
Answer: y=Ae^(x²) for any real A, including A=0.
Final answer
y=Ae^(x²) for any real A, including A=0.
Review the teaching: Topic 7.6
Problem 16
Find the general solution of y′=x(1+y²).
- Separate dy/(1+y²)=x dx. Reason: The derivative factors into a y-dependent and an x-dependent part.
- Integrate arctan y=x²/2+C and solve for y. Reason: The arctangent derivative reverses the left side.
- Use an interval avoiding tangent poles. Reason: A solution must stay finite and differentiable.
Answer: y=tan(x²/2+C), on intervals where the tangent is defined.
Final answer
y=tan(x²/2+C), on intervals where the tangent is defined.
Review the teaching: Topic 7.6
Problem 17
Solve y′=3(1−y) generally, including its equilibrium.
- Set z=y−1; then z′=−3z. Reason: Shifting exposes a separable proportional equation.
- Integrate to z=Ae^(−3x), and restore y. Reason: Separate variables for z≠0 and include z=0 afterward.
Answer: y=1+Ae^(−3x), including A=0.
Final answer
y=1+Ae^(−3x), including A=0.
Review the teaching: Topic 7.6
Problem 18
Solve y′=xy with y(0)=2.
- Separate and integrate: y=Ae^(x²/2). Reason: The equation is separable, with the zero equilibrium included in the family.
- Apply y(0)=2 to get A=2. Reason: The initial condition selects the constant.
- Differentiate to verify y′=xy. Reason: The expression is defined and satisfies the equation for all x.
Answer: y=2e^(x²/2), all real x.
Final answer
y=2e^(x²/2), all real x.
Review the teaching: Topic 7.7
Problem 19
Solve y′=y² with y(0)=1 and give the largest interval containing 0 on which the solution is defined.
- Separate dy/y²=dx and integrate −1/y=x+C. Reason: The initial value excludes the zero equilibrium.
- Use x=0,y=1 to find C=−1; solve for y. Reason: The initial condition sets the integration constant.
- The denominator vanishes at x=1; choose the connected interval containing 0. Reason: One differentiable solution cannot pass through that singularity.
Answer: y=1/(1−x), interval (−∞,1).
Final answer
y=1/(1−x), interval (−∞,1).
Review the teaching: Topic 7.7
Problem 20
Solve y′=y/x with y(1)=3 and give a connected domain containing the initial point.
- Separate dy/y=dx/x to obtain ln|y|=ln|x|+C and y=Ax on x>0. Reason: The equation is defined only when x≠0; work on the initial point’s component.
- Use y(1)=3 to get A=3. Reason: The initial condition chooses the member.
- Restrict to (0,∞). Reason: The original equation is undefined at zero even though the formula 3x extends there.
Answer: y=3x for x>0.
Final answer
y=3x for x>0.
Review the teaching: Topic 7.7
Problem 21
Give a solution to y′=e^(−x²), y(0)=2, using an accumulation function, and verify it.
- Add the initial amount 2 to change accumulated from 0. Reason: An integral of the derivative gives net change.
- Its derivative is e^(−x²), and at x=0 the integral is zero. Reason: Continuity permits the Fundamental Theorem and verifies both requirements.
Answer: y(x)=2+∫₀ˣ e^(−t²)dt.
Final answer
y(x)=2+∫₀ˣ e^(−t²)dt.
Review the teaching: Topic 7.7
Problem 22
Solve y′=3x²−2 with y(1)=4.
- Integrate y=x³−2x+C. Reason: Reverse the derivative rules.
- Substitute 4=1−2+C, so C=5. Reason: The initial condition selects the particular solution.
Answer: y=x³−2x+5.
Final answer
y=x³−2x+5.
Review the teaching: Topic 7.7
Problem 23
Solve y′=2xy with y(0)=0.
- Test the constant zero function: its derivative and 2xy are both zero. Reason: An equilibrium can satisfy the initial condition directly.
- In the general family y=Ae^(x²), the condition gives A=0. Reason: This confirms the equilibrium is included without illegal division by zero.
Answer: y=0.
Final answer
y=0.
Review the teaching: Topic 7.7
Problem 24
Solve y′=−x/y with y(0)=2 and give the interval of this branch.
- Separate y dy=−x dx to get y²=−x²+C. Reason: The derivative is defined only for y≠0.
- Use y(0)=2 to obtain C=4 and choose the positive root. Reason: The initial condition selects both constant and branch.
- Restrict to −2<x<2. Reason: At the endpoints y=0 and the original differential equation is undefined.
Answer: y=√(4−x²), interval (−2,2).
Final answer
y=√(4−x²), interval (−2,2).
Review the teaching: Topic 7.7
Problem 25
A population follows P′=kP, P(0)=100, P(3)=200, with t in hours. Find k and P(6).
- Write P=100e^(kt); set 200=100e^(3k). Reason: Proportional growth gives the exponential family.
- Take logs for k, then evaluate at t=6. Reason: The second time interval doubles the amount again.
Answer: k=(ln2)/3 h⁻¹; P(6)=400.
Final answer
k=(ln2)/3 h⁻¹; P(6)=400.
Review the teaching: Topic 7.8
Problem 26
A substance has half-life 4 days and starts at 80 g. Give its differential equation and its amount after 12 days.
- A half-life requires e^(4k)=1/2, so k=−ln2/4 day⁻¹. Reason: Fit the exponential decay constant.
- After three half-lives, M=80(1/2)³=10 g. Reason: The model multiplies by one-half every four days.
Answer: M′=−(ln2/4)M; M(12)=10 g.
Final answer
M′=−(ln2/4)M; M(12)=10 g.
Review the teaching: Topic 7.8
Problem 27
For y′=0.05y with y(0)=100, use a calculator to find y(10) and the doubling time, to three decimals.
- Use y=100e^(0.05t), giving 100e^0.5≈164.8721271. Reason: The initial-value exponential model gives the quantity.
- Solve 2=e^(0.05T): T=ln2/0.05≈13.8629436. Reason: Doubling time follows from the ratio 2; round at the end.
Answer: y(10)≈164.872; doubling time ≈13.863 time units.
Final answer
y(10)≈164.872; doubling time ≈13.863 time units.
Review the teaching: Topic 7.8
Problem 28
A cooling model is T′=−0.2(T−20), T(0)=80, with T in °C and time in minutes. Find T(t) and its limiting temperature.
- Set D=T−20; then D′=−0.2D and D(0)=60. Reason: The temperature difference is the decaying quantity.
- Solve D=60e^(−0.2t) and add 20. Reason: Restore the original variable.
- As t→∞ the exponential tends to zero. Reason: The solution approaches the room temperature.
Answer: T(t)=20+60e^(−0.2t); limit 20°C.
Final answer
T(t)=20+60e^(−0.2t); limit 20°C.
Review the teaching: Topic 7.8
Problem 29
A particle has acceleration a(t)=2t m/s² and velocity v(0)=3 m/s. Find v(t) and its velocity at 2 s.
- Integrate v′=2t to obtain v=t²+C. Reason: Acceleration is the derivative of velocity.
- The condition gives C=3; substitute t=2. Reason: An initial velocity fixes the particular motion model.
Answer: v(t)=t²+3 m/s; v(2)=7 m/s.
Final answer
v(t)=t²+3 m/s; v(2)=7 m/s.
Review the teaching: Topic 7.8
Problem 30
A positive quantity has a continuous relative rate k=0.05 per year. Explain why its one-year percent increase is not exactly 5%, and calculate it to three decimals.
- The one-year multiplier is e^0.05. Reason: The quantity changes continuously according to y′=0.05y.
- Subtract 1 and multiply by 100 to convert relative change to percent. Reason: Interval percent change compares the final amount with the initial amount.
Answer: 100(e^0.05−1)% ≈ 5.127%.
Final answer
100(e^0.05−1)% ≈ 5.127%.
Review the teaching: Topic 7.8
Explain and apply: worked solutions
Compare the setup and the reasoning, as well as the final answer. Use the rubric to find your next step; equivalent correct reasoning earns the same credit.
Estimate two members of a solution family
The supplied field represents y′ = 2 − y on 0 ≤ x ≤ 3 and 0 ≤ y ≤ 4. At every plotted x, slopes at y = 0, 1, 2, 3 and 4 are respectively 2, 1, 0, −1 and −2; intermediate half-unit rows follow the same rule. Two marked initial points are (0,0.5) and (0,3.5). Use qualitative graphical estimation, not a fixed-step numerical method.
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Follow a turning curve through a two-coordinate field
The supplied slope field represents y′ = x − y on 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3. For example, slopes at (0,1), (1,1) and (2,1) are −1, 0 and +1. The selected solution passes through (0,1). Use qualitative estimation; an exact formula or numerical stepping algorithm is not required.
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Return to the unit problem →Try this problem in its lesson →