AP Unit 4 Solutions
Worked solutions to all thirty Unit 4 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
W(t) is water in liters, t is minutes, and W′(6) = −2.5. Interpret this derivative. Can you determine W(6)?
Identify output liters and input minutes.
Why: Derivative units are liters per minute.
Interpret −2.5 as decrease at the specific instant t = 6.
Why: The sign gives direction of change, not a negative amount.
Do not assign a water amount.
Why: Many tank amounts can have the same instantaneous rate.
Final answer
At minute 6, water is decreasing at 2.5 L/min. W(6) cannot be determined from this rate alone.
Review the teaching: Topic 4.1
Problem 2
Use the table to estimate T′(3) with a centered secant. Interpret the result. T is °C and t is minutes.
| t | 2.8 | 3.0 | 3.2 |
|---|---|---|---|
| T(t) | 19.1 | 19.8 | 20.5 |
Compute (20.5 − 19.1)/(3.2 − 2.8) = 3.5.
Why: The secant uses values equally spaced around the target.
Label the value an estimate and use °C/min.
Why: Finite table spacing does not supply the exact tangent slope.
Say the temperature is rising at about 3.5 °C/min at minute 3.
Why: The positive sign describes the change in temperature.
Final answer
T′(3) ≈ 3.5 °C/min; at minute 3, the temperature is increasing at about 3.5 °C per minute.
Review the teaching: Topic 4.1
Problem 3
B(T) is a battery’s operating time in hours at temperature T in °C. B′(20) = −0.08. Explain its units and estimate the change when temperature rises from 20°C to 20.5°C.
Derivative units are hours divided by °C.
Why: The independent variable is temperature, not time.
Estimate ΔB ≈ B′(20)ΔT = (−0.08)(0.5) = −0.04 hour.
Why: A derivative gives the local change per unit input.
Convert 0.04 hour to 2.4 minutes and state decrease.
Why: Multiply by 60 minutes/hour and preserve the sign’s meaning.
Final answer
Hours per °C; the operating time decreases by approximately 0.04 hour (2.4 minutes).
Review the teaching: Topic 4.1
Problem 4
The graph of a differentiable quantity P has a horizontal tangent at t = 4 and lies above the horizontal axis there. What do P(4) > 0 and P′(4) = 0 each say? Must P have a local maximum?
Read the positive graph height as P(4) > 0.
Why: A graph height represents the output, not the slope.
Read the horizontal tangent as P′(4) = 0.
Why: A derivative represents tangent slope.
Use P(t) = 1 + (t − 4)³ as a counterexample to a forced maximum.
Why: It has the stated height and slope but increases through t = 4.
Final answer
The amount is positive and its instantaneous rate is zero. A local maximum is not guaranteed.
Review the teaching: Topic 4.1
Problem 5
For s(t) = t³ − 6t² + 9t meters, find velocity, acceleration, direction, and speed at t = 2 seconds.
Differentiate: v(t) = 3t² − 12t + 9 and a(t) = 6t − 12.
Why: Velocity and acceleration are the first and second position derivatives.
Substitute t = 2 to obtain v = −3 and a = 0.
Why: Evaluate the derived rates at the requested time.
State leftward motion and speed |−3| = 3.
Why: Velocity has direction; speed is its nonnegative magnitude.
Final answer
v(2) = −3 m/s, a(2) = 0 m/s²; moving left with speed 3 m/s.
Review the teaching: Topic 4.2
Problem 6
A particle has v(t) = (t − 1)(t − 3) m/s for 0 ≤ t ≤ 4. On which open intervals does it move right, and when does it reverse direction?
Velocity zeros are 1 and 3.
Why: Only zeros can separate the sign intervals for this continuous polynomial.
The velocity signs are positive, negative, positive on (0,1), (1,3), (3,4).
Why: Check the signs of both factors.
Positive velocity means rightward motion; both roots reverse direction.
Why: The sign changes at each root.
Final answer
Right on (0,1) and (3,4); direction reverses at t = 1 and t = 3.
Review the teaching: Topic 4.2
Problem 7
The entries are exact instantaneous velocity and acceleration values. Classify direction and whether speed increases or decreases at each time.
| t | v(t) | a(t) |
|---|---|---|
| 1 | −3 | 2 |
| 2 | 4 | −1 |
| 3 | −2 | −5 |
Use velocity signs: negative, positive, negative.
Why: Velocity controls direction.
Compare signs of v and a: opposite, opposite, same.
Why: For v ≠ 0, opposite signs decrease speed and same signs increase it.
Assign the three classifications with no interpolation between rows.
Why: The table gives the requested instants, not complete interval behavior.
Final answer
At 1: left, slowing. At 2: right, slowing. At 3: left, speeding up.
Review the teaching: Topic 4.2
Problem 8
A velocity graph is the straight line through (0,−4), (2,0), and (4,4), with seconds and m/s. Determine acceleration and describe motion on (0,2) and (2,4).
Calculate slope (4 − (−4))/(4 − 0) = 2.
Why: The slope of a velocity graph is acceleration.
Before t = 2, velocity is negative and acceleration positive.
Why: Opposite signs imply decreasing speed.
After t = 2, both are positive.
Why: Same signs imply increasing speed; velocity crosses zero at 2.
Final answer
a = 2 m/s². On (0,2), moving left and slowing; on (2,4), moving right and speeding up. Direction reverses at 2 s.
Review the teaching: Topic 4.2
Problem 9
For v(t) = (t − 2)², t ≥ 0, find the rest time. Does the particle reverse direction? Where is it slowing down?
Solve (t − 2)² = 0: t = 2.
Why: Rest means zero velocity.
For t ≠ 2, velocity is positive on both sides.
Why: A repeated zero touches the axis without a sign change.
a(t) = 2(t − 2) is negative on (0,2), so speed decreases there.
Why: For positive v, negative a reduces speed.
Final answer
Rest at t = 2; no direction reversal; slowing on (0,2).
Review the teaching: Topic 4.2
Problem 10
A cost model is C(q) = 100 + 8q + 0.02q² dollars. Find marginal cost at q = 50 and the exact model cost of increasing from 50 to 51 items. Compare them.
C′(q) = 8 + 0.04q, so C′(50) = 10.
Why: Marginal cost is the derivative with respect to output count.
C(51) − C(50) = 8 + 0.02(51² − 50²) = 10.02.
Why: An exact finite increase uses a difference of function values.
The derivative estimate is C′(50)·1 = 10 dollars.
Why: A local rate times a small input change estimates output change.
Final answer
$10/item marginal cost; exact increase $10.02. The one-item derivative estimate is $10.
Review the teaching: Topic 4.3
Problem 11
A concentration model is A(t) = 5e^(−0.3t) + 0.2t mg/L, t in hours. Use a numerical derivative or the derivative formula to find and interpret A′(2). Round to three decimals.
A′(t) = −1.5e^(−0.3t) + 0.2.
Why: Apply the chain rule to the exponential and differentiate the linear term.
Evaluate at t = 2: −1.5e^(−0.6) + 0.2 ≈ −0.623217.
Why: Use the calculator to evaluate the expression; retain guard digits.
Round to −0.623 and interpret as a concentration decrease at hour 2.
Why: The sign and output/input units answer the contextual question.
Final answer
A′(2) ≈ −0.623 mg/L per hour: concentration is decreasing at about 0.623 mg/L each hour at t = 2.
Review the teaching: Topic 4.3
Problem 12
A population N is measured in people and t in years. At t = 2, N′(2) = 40 and N″(2) = −3. Interpret both values and explain how the population can grow while its growth rate falls.
N′(2) = 40 is a positive instantaneous population rate.
Why: First derivative units are people per year.
N″(2) = −3 describes a declining growth rate at that instant.
Why: A second derivative measures the change in the first derivative and has people per year squared units.
A positive rate can decline while remaining positive.
Why: The two specified signs describe different quantities; no converse inference from monotonicity is needed.
Final answer
At year 2, population grows at 40 people/year, while that growth rate decreases at 3 people/year². The amount increases even though its positive rate is becoming smaller.
Review the teaching: Topic 4.3
Problem 13
x(t) and y(t) satisfy x² + xy + y² = 12. Differentiate with respect to t, keeping both variables symbolic.
Differentiate x² and y² as 2xx′ and 2yy′.
Why: Use the chain rule because each variable depends on time.
Differentiate xy as x′y + xy′.
Why: The product rule is needed for two changing factors.
Collect x′ and y′ terms: (2x + y)x′ + (x + 2y)y′ = 0.
Why: The constant right side has derivative zero; grouping exposes the rate relation.
Final answer
(2x + y)x′ + (x + 2y)y′ = 0.
Review the teaching: Topic 4.4
Problem 14
Let R(t) = x(t)/y(t). At an instant x = 6, y = 3, x′ = 2, y′ = −1, find R′.
R′ = (x′y − xy′)/y².
Why: Both numerator and denominator change, so use the quotient rule.
Substitute: [2(3) − 6(−1)]/9 = 12/9.
Why: Instantaneous values enter the differentiated equation.
Simplify to 4/3; R is rising at that instant.
Why: The positive rate fits an increasing numerator and decreasing positive denominator.
Final answer
R′ = 4/3 per unit time.
Review the teaching: Topic 4.4
Problem 15
A student substitutes x = 3 and y = 4 into x² + y² = 25, then differentiates 25 = 25 to get 0 = 0. Explain the mistake and write the useful rate equation.
Keep x(t), y(t) in the equation before differentiation.
Why: The distances change even though their present values are given.
Differentiate x² + y² = 25 to obtain 2xx′ + 2yy′ = 0.
Why: The chain rule retains the unknown rates.
Only now substitute x = 3 and y = 4.
Why: The instantaneous relation becomes 6x′ + 8y′ = 0.
Final answer
The values are a snapshot, so substituting early erases time dependence. Use 2xx′ + 2yy′ = 0, then 6x′ + 8y′ = 0 at the instant.
Review the teaching: Topic 4.4
Problem 16
A rectangle has x = 8 cm, y = 5 cm, x′ = −0.5 cm/s, y′ = 0.8 cm/s. Find and interpret the area rate.
A = xy gives A′ = x′y + xy′.
Why: Area depends on both changing sides.
A′ = (−0.5)(5) + (8)(0.8) = −2.5 + 6.4 = 3.9.
Why: Substitute the instant’s lengths and signed rates.
State that the area increases at 3.9 cm²/s.
Why: The positive net rate outweighs the shrinking side’s contribution.
Final answer
A′ = 3.9 cm²/s; the area is increasing at 3.9 square centimeters per second.
Problem 17
A 10-ft ladder leans against a vertical wall. Its foot moves away at 2 ft/s. When its foot is 6 ft from the wall, how fast is the top moving?
Let x be the foot distance and y the height; x² + y² = 100. At x = 6, y = 8.
Why: The fixed ladder is the hypotenuse, and height is positive.
Differentiate: 2xx′ + 2yy′ = 0.
Why: Both legs change with time.
y′ = −xx′/y = −6(2)/8 = −1.5.
Why: Use the rate relation after inserting the snapshot.
The top descends at 1.5 ft/s.
Why: Negative height rate means downward motion.
Final answer
The top’s height rate is −1.5 ft/s, so it moves downward at 1.5 ft/s.
Problem 18
Air expands a spherical balloon at dV/dt = 24π cm³/s. Find dr/dt when r = 2 cm, using V = (4/3)πr³.
Differentiate: V′ = 4πr²r′.
Why: The radius changes with time, so use the chain rule.
24π = 4π(2²)r′, giving r′ = 24/16 = 1.5.
Why: Substitute the current radius and known volume rate.
Use cm/s and say increasing.
Why: A radius rate is length per time; its positive sign indicates expansion.
Final answer
dr/dt = 1.5 cm/s; the radius is increasing.
Review the teaching: Topic 4.5
Problem 19
A conical tank is 12 ft high with top radius 4 ft. Water enters at 2π ft³/min. Find the rate of depth increase when h = 3 ft.
Similar triangles give r = h/3. Thus V = (1/3)πr²h = πh³/27.
Why: The radius of the water changes with depth, and the water cone is similar to the full tank.
V′ = (πh²/9)h′.
Why: Differentiate the one-variable volume relation.
2π = (π·9/9)h′, so h′ = 2 ft/min.
Why: Insert h = 3 only after differentiation and interpret a depth rate.
Final answer
dh/dt = 2 ft/min.
Review the teaching: Topic 4.5
Problem 20
A 6-ft person walks away from a 15-ft lamp at 4 ft/s. Let x be the lamp-to-person distance and s the shadow length. Find s′ and the speed of the shadow tip.
Similar triangles give 15/(x + s) = 6/s, so 9s = 6x.
Why: The large triangle extends to the shadow tip, a distance x + s from the lamp.
Differentiate: 9s′ = 6x′, hence s′ = (2/3)4 = 8/3.
Why: The proportional relationship holds throughout the motion.
The tip distance is x + s, so its rate is 4 + 8/3 = 20/3.
Why: Shadow length and tip distance have different definitions.
Final answer
Shadow grows at 8/3 ft/s; tip moves away from lamp at 20/3 ft/s.
Review the teaching: Topic 4.5
Problem 21
At an instant, one car is 3 miles east of an intersection and travels east at 40 mph. Another is 4 miles north and travels north at 30 mph. How fast is their separation changing?
Let x = 3, y = 4 and d² = x² + y², so d = 5.
Why: The roads are perpendicular and separation is the hypotenuse.
Differentiate: dd′ = xx′ + yy′.
Why: Differentiate squares with respect to the shared time.
d′ = [3(40) + 4(30)]/5 = 48.
Why: Use outward-positive coordinate rates and divide by the current separation.
The positive answer means they move apart at 48 mph.
Why: The rate concerns the distance between cars, not either car’s speed.
Final answer
Separation increases at 48 mph.
Review the teaching: Topic 4.5
Problem 22
Use linearization at a = 16 to estimate √16.32 and justify overestimate or underestimate.
f(16) = 4 and f′(16) = 1/(2√16) = 1/8.
Why: These values define the tangent line.
L(16.32) = 4 + (1/8)(0.32) = 4.04.
Why: Use the input change from the anchor.
f″(x) = −1/(4x^(3/2)) < 0 for positive x.
Why: Concavity is down between anchor and target, so the tangent lies above the curve.
Final answer
√16.32 ≈ 4.04, an overestimate.
Review the teaching: Topic 4.6
Problem 23
A differentiable function has f(2) = 5 and f′(2) = −3. It is concave up throughout [2,2.1]. Approximate f(2.1) and state the error direction.
L(x) = 5 − 3(x − 2).
Why: The tangent model uses the value and derivative at 2.
L(2.1) = 5 − 0.3 = 4.7.
Why: The input change is 0.1.
The estimate is below the actual value.
Why: Concavity up keeps the curve above its tangent on this interval even with a negative slope.
Final answer
f(2.1) ≈ 4.7, an underestimate.
Review the teaching: Topic 4.6
Problem 24
A circle has measured radius 10 cm with a small increase of 0.02 cm. Use a differential to estimate the area increase, then find the exact increase.
A′(r) = 2πr, so dA = 2π(10)(0.02) = 0.4π.
Why: A differential estimates output change from the local rate.
ΔA = π[(10.02)² − 10²] = 0.4004π.
Why: The finite change uses the actual function values.
A″(r) = 2π > 0, so the tangent underestimate is consistent.
Why: Concavity up explains the missing positive quadratic contribution.
Final answer
Estimated increase 0.4π cm²; exact increase 0.4004π cm². The differential underestimates.
Review the teaching: Topic 4.6
Problem 25
Use the tangent to f(x) = eˣ at 0 to estimate e^0.08. Use a calculator to find the actual value and absolute error to six decimals. Explain the direction.
f(0) = f′(0) = 1 gives L(x) = 1 + x.
Why: The tangent matches the value and slope at the anchor.
L(0.08) = 1.08; calculator evaluation gives e^0.08 ≈ 1.083287068.
Why: Compute the comparison without rounding intermediate values.
Absolute error is |1.083287068 − 1.08| ≈ 0.003287.
Why: Absolute error measures the size of the discrepancy.
f″(x) = eˣ > 0, so the tangent estimate is low.
Why: Concavity up over the interval supplies the analytic error direction.
Final answer
Estimate 1.08; actual ≈ 1.083287; absolute error ≈ 0.003287. It is an underestimate.
Review the teaching: Topic 4.6
Problem 26
Find lim as x → 0 of (e^(2x) − 1)/x and justify the method.
Substitution gives 0/0.
Why: This is a qualifying indeterminate quotient.
The derivative quotient is 2e^(2x)/1.
Why: The chain rule applies to the numerator; the denominator derivative is nonzero.
Its limit is 2, so the original limit is 2.
Why: L’Hospital’s hypotheses hold near zero and the derivative quotient has a limit.
Final answer
2.
Review the teaching: Topic 4.7
Problem 27
Find lim as x → ∞ of ln x/x², stating its initial form.
Both ln x and x² tend to +∞.
Why: Identify the form before using the rule.
L’Hospital gives (1/x)/(2x) = 1/(2x²).
Why: The functions are differentiable for x > 0 and denominator derivative 2x is nonzero.
1/(2x²) tends to 0.
Why: The derivative quotient’s limit determines the original quotient’s limit.
Final answer
Infinite-over-infinite form; limit 0.
Review the teaching: Topic 4.7
Problem 28
Find lim as x → 0 of (1 − cos(3x))/x² using justified applications of L’Hospital’s Rule.
The original quotient is 0/0; first derivative quotient is 3sin(3x)/(2x).
Why: The numerator uses the chain rule and denominator derivative is nonzero off zero.
The new quotient is still 0/0; second derivative quotient is 9cos(3x)/2.
Why: Recheck the form and hypotheses before another application.
The final limit is 9/2.
Why: Direct substitution now succeeds and both applications were valid.
Final answer
9/2.
Review the teaching: Topic 4.7
Problem 29
A student applies L’Hospital to lim as x → 0⁺ of (1 + x)/x and answers 1. Explain the invalid step and find the limit.
The numerator tends to 1 while the denominator tends to 0 from above.
Why: This fails the rule’s required indeterminate-form hypothesis.
Rewrite (1 + x)/x = 1/x + 1.
Why: Algebra directly shows the behavior.
As x → 0⁺, 1/x → +∞, so the quotient tends to +∞.
Why: The side of approach determines the sign of the unbounded behavior.
Final answer
Invalid: the form is 1/0, not indeterminate. The right-hand limit is +∞.
Review the teaching: Topic 4.7
Problem 30
Evaluate lim as x → 1 of (x² − 1)/(x − 1) two ways. Explain why L’Hospital is valid but unnecessary.
For x ≠ 1, factor to get x + 1, whose limit is 2.
Why: Cancelling a common factor on a punctured neighborhood preserves the limit.
The original substitution is 0/0; derivative quotient is 2x/1 → 2.
Why: The functions are differentiable and denominator derivative is nonzero, so the rule also applies.
Prefer factoring here when it is quicker.
Why: A qualifying form allows the theorem; it does not require it.
Final answer
2, by factoring or one valid L’Hospital application.
Review the teaching: Topic 4.7