📌 Core Concepts
A rate function f(t) [gal/sec]. Table of values. You need to: interpret a definite integral, approximate using Riemann sums, find average value, and evaluate/interpret a derivative.
Given Data Table
| t (sec) | 0 | 60 | 90 | 120 | 135 | 150 |
|---|
| f(t) (gal/sec) | 0 | 0.1 | 0.15 | 0.1 | 0.05 | 0 |
Model g(t) is also given for parts (c) and (d):
$$g(t) = \left(\cos\!\left(\frac{t^2}{500}\right)\right)\!\left(\frac{t}{120}\right), \quad 0 \le t \le 150$$
Part (a) — Interpret the integral · Right Riemann Sum
Interpretation:
∫₆₀¹³⁵ f(t) dt = the total gallons of gasoline pumped from t = 60 to t = 135 seconds.
Units: (gal/sec)(sec) = gallons. Always state units!
Right Riemann Sum: Use the RIGHT endpoint of each subinterval.
$$\int_{60}^{135} f(t)\,dt \approx f(90)\cdot(90-60) + f(120)\cdot(120-90) + f(135)\cdot(135-120)$$
- 1Subintervals: [60,90] → width 30; [90,120] → width 30; [120,135] → width 15
- 2Right endpoints: t = 90, 120, 135 → f-values: 0.15, 0.1, 0.05
- 3Sum = 0.15(30) + 0.1(30) + 0.05(15) = 4.5 + 3 + 0.75
✦ Model Answer
The integral ∫₆₀¹³⁵ f(t) dt represents the total amount of gasoline (in gallons) pumped from t = 60 to t = 135 seconds.
Right Riemann Sum:
= f(90)·30 + f(120)·30 + f(135)·15
= (0.15)(30) + (0.1)(30) + (0.05)(15)
= 4.5 + 3.0 + 0.75
≈ 8.25 gallons
⚠️ Common Trap: Using LEFT endpoints instead of RIGHT endpoints. The problem says "right Riemann sum." Left endpoints would use f(60), f(90), f(120).
Part (b) — MVT / Rolle's Theorem Existence
Must there exist c in (60, 120) with f′(c) = 0? This is a Rolle's Theorem question.
Rolle's Theorem: If f is continuous on [a,b], differentiable on (a,b), and f(a) = f(b), then ∃c ∈ (a,b) such that f′(c) = 0.
- 1f is differentiable (given), so it is continuous.
- 2Check: f(60) = 0.1 and f(120) = 0.1 → f(60) = f(120)
- 3By Rolle's Theorem, there must exist c in (60,120) with f′(c) = 0.
✦ Model Answer
Since f is differentiable on (60, 120) (and hence continuous on [60, 120]), and f(60) = f(120) = 0.1, by Rolle's Theorem there must exist a value c, 60 < c < 120, such that f′(c) = 0. Yes — by Rolle's Theorem
✅ Key: You MUST name the theorem and state both conditions: differentiability AND equal endpoint values.
Part (c) — Average Rate of Flow using g(t) Calculator
$$\text{Average value} = \frac{1}{150-0}\int_0^{150} g(t)\,dt$$
🔢 TI-Nspire CX II Steps:
- Press [home] → [Calculator] (scratchpad or new doc)
- Type: integral(g(t), t, 0, 150) / 150
- Or use template: [MENU] → 4: Calculus → 2: Integral
- Enter: lower=0, upper=150, expression=(cos(t²/500))·(t/120), variable=t
- Divide result by 150
✦ Model Answer
$$\text{Average rate} = \frac{1}{150}\int_0^{150} g(t)\,dt = \frac{1}{150}\int_0^{150}\cos\!\left(\tfrac{t^2}{500}\right)\!\cdot\!\tfrac{t}{120}\,dt$$
≈ 0.0754 gallons per second
✅ Scoring Note: You MUST write the setup (the integral formula with 1/150 out front). Just writing the numerical answer earns no setup point.
Part (d) — g′(140) and Interpretation Calculator
🔢 TI-Nspire CX II Steps:
- [MENU] → 4: Calculus → 1: Derivative
- Or type: d/dt(cos(t²/500)·(t/120))|t=140
- Shortcut: derivative(g(t), t, 1)|t=140
✦ Model Answer
g′(140) ≈ −0.0240 gallons per second per second
Interpretation: At time t = 140 seconds, the rate of flow of gasoline is decreasing at approximately 0.0240 gallons per second per second (the flow rate is slowing down).
⚠️ Interpretation Trap: Must state: (1) the direction (decreasing), (2) the units (gal/sec/sec or gal/sec²), (3) the context (rate of flow at t = 140). Missing any of these costs points.
📌 Setup
x(0) = 1, y(t) = 2 sin t, dx/dt = e^(cos t). For 0 ≤ t ≤ π.
dy/dt = 2 cos t (derivative of y(t) = 2 sin t).
Speed = √[(dx/dt)² + (dy/dt)²]
Part (a) — Acceleration Vector at t = 1 Calculator
$$\vec{a}(t) = \left\langle \frac{d^2x}{dt^2},\; \frac{d^2y}{dt^2} \right\rangle$$
- 1dx/dt = e^(cos t) → d²x/dt² = e^(cos t) · (−sin t)
- 2dy/dt = 2cos t → d²y/dt² = −2sin t
- 3At t = 1: d²x/dt² = e^(cos 1)·(−sin 1); d²y/dt² = −2 sin 1
🔢 TI-Nspire CX II:
- d²x/dt²: derivative(e^(cos(t))·(-sin(t)), t, 1) ... or just evaluate e^(cos(1))·(-sin(1))
- Actually compute: e^(cos(1))·(-sin(1)) at t=1 directly
- e^(cos(1)) ≈ e^(0.5403) ≈ 1.7166; ×(-sin(1)) ≈ ×(-0.8415) ≈ -1.4439
- d²y/dt² = -2sin(1) ≈ -1.6829
✦ Model Answer
$$\frac{d^2x}{dt^2} = e^{\cos t}\cdot(-\sin t), \quad \frac{d^2y}{dt^2} = -2\sin t$$
At t = 1:
$$\vec{a}(1) = \langle e^{\cos 1}\cdot(-\sin 1),\;-2\sin 1\rangle$$
≈ ⟨−1.444, −1.683⟩
✅ Key: Show the derivatives symbolically FIRST, then evaluate. The setup earns points even if the calculator computation is off.
Part (b) — First time speed = 1.5 Calculator
$$\text{Speed} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = \sqrt{e^{2\cos t} + 4\cos^2 t} = 1.5$$
🔢 TI-Nspire CX II — Solve equation:
- [MENU] → 3: Algebra → 1: Solve
- solve(√(e^(2cos(t)) + 4cos²(t)) = 1.5, t) | 0 ≤ t ≤ π
- Or graph y = √(e^(2cos(x)) + 4cos²(x)) and y = 1.5 on [0, π], find intersection
- [MENU] → 6: Analyze Graph → 4: Intersection
✦ Model Answer
Set speed = 1.5:
$$\sqrt{e^{2\cos t} + 4\cos^2 t} = 1.5$$
Solving numerically on [0, π]:
t ≈ 1.8377
⚠️ Trap: Make sure you restrict to [0, π]. Also, write the speed equation setup explicitly — do not just write the answer.
Part (c) — Slope of tangent at t=1 · x-coordinate at t=1 Calculator
$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2\cos t}{e^{\cos t}}$$
$$x(1) = x(0) + \int_0^1 \frac{dx}{dt}\,dt = 1 + \int_0^1 e^{\cos t}\,dt$$
🔢 TI-Nspire CX II:
- Slope: (2cos(1)) / e^(cos(1))
- x(1): 1 + integral(e^(cos(t)), t, 0, 1)
✦ Model Answer
$$\text{slope} = \frac{dy}{dx}\bigg|_{t=1} = \frac{2\cos 1}{e^{\cos 1}} \approx \mathbf{0.635}$$
$$x(1) = 1 + \int_0^1 e^{\cos t}\,dt \approx 1 + 1.2638$$
slope ≈ 0.635 | x(1) ≈ 2.264
✅ Key: For x(1), use the Fundamental Theorem: x(1) = x(0) + ∫₀¹ (dx/dt) dt. You MUST use the initial condition x(0) = 1.
Part (d) — Total Distance Traveled Calculator
$$\text{Distance} = \int_0^{\pi} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt = \int_0^{\pi}\sqrt{e^{2\cos t} + 4\cos^2 t}\,dt$$
🔢 TI-Nspire CX II:
- integral(√(e^(2cos(t)) + 4cos²(t)), t, 0, π)
- [MENU] → 4: Calculus → 2: Integral, then input expression
✦ Model Answer
$$\text{Total Distance} = \int_0^{\pi}\sqrt{e^{2\cos t} + 4\cos^2 t}\,dt$$
≈ 4.916
⚠️ Trap: Distance ≠ displacement. For total distance in parametric motion, always use the arc-length integral — never just subtract positions.
📌 Setup
dM/dt = (1/4)(40 − M), M(0) = 5, M(t) < 40 for all t. This is a Newton's Law of Cooling/Heating type DE. Equilibrium at M = 40.
Part (a) — Sketch Solution Curve on Slope Field
Key behaviors to draw correctly:
- 1Start at point (0, 5) — below equilibrium M = 40
- 2Since M < 40: dM/dt = (1/4)(40−M) > 0 → always increasing
- 3Curve is concave down (rate of increase slows as M → 40)
- 4Asymptote: M → 40 as t → ∞ — never reaches 40
✦ Model Answer (verbal description of sketch)
Draw a smooth increasing curve starting at (0, 5), following the slope field marks, concave down, approaching but never reaching M = 40 as a horizontal asymptote.
⚠️ Trap: Curve must NOT cross M = 40. If it does, you lose the point.
Part (b) — Tangent Line Approximation for M(2)
$$M(0) = 5, \quad \frac{dM}{dt}\bigg|_{t=0} = \frac{1}{4}(40-5) = \frac{35}{4} = 8.75$$
$$L(t) = M(0) + M'(0)\cdot t = 5 + 8.75t$$
✦ Model Answer
At t = 0: M′(0) = (1/4)(40 − 5) = 35/4
Tangent line: L(t) = 5 + (35/4)t
$$L(2) = 5 + \frac{35}{4}\cdot 2 = 5 + \frac{35}{2} = 5 + 17.5$$
M(2) ≈ 22.5°C
Part (c) — Second Derivative · Over/Underestimate
$$\frac{d^2M}{dt^2} = \frac{d}{dt}\!\left[\frac{1}{4}(40-M)\right] = \frac{1}{4}\!\cdot\!\left(-\frac{dM}{dt}\right) = \frac{1}{4}\!\cdot\!\left(-\frac{1}{4}(40-M)\right) = -\frac{1}{16}(40-M)$$
Since M < 40 for all t: (40 − M) > 0 → d²M/dt² < 0 → concave DOWN
✦ Model Answer
$$\frac{d^2M}{dt^2} = -\frac{1}{16}(40 - M)$$
Since M(t) < 40 for all t, we have 40 − M > 0, so d²M/dt² < 0 for all t. The graph of M is concave down. Therefore, the tangent line lies above the curve, and the approximation L(2) = 22.5 is an overestimate.
Concave down → tangent line is ABOVE → approximation OVERESTIMATES.
✅ Key: You must write the expression for d²M/dt² in terms of M, then explain the sign, then conclude over/under.
Part (d) — Separation of Variables · Particular Solution
- 1Separate: dM/(40−M) = (1/4) dt
- 2Integrate both sides: −ln|40−M| = t/4 + C
- 3Solve: |40−M| = e^(−t/4 − C) = Ae^(−t/4)
- 4Since M < 40: 40−M = Ae^(−t/4)
- 5Apply M(0) = 5: 40−5 = A → A = 35
- 6Solve for M: M = 40 − 35e^(−t/4)
✦ Model Answer
$$\frac{dM}{40-M} = \frac{1}{4}\,dt$$
$$-\ln|40-M| = \frac{t}{4} + C$$
$$40 - M = Ae^{-t/4}$$
M(0) = 5 → A = 35
$$M(t) = 40 - 35e^{-t/4}$$
⚠️ Big Traps: (1) Don't forget the +C. (2) Don't drop the absolute value until you've justified the sign. (3) Write ln correctly — missing the negative sign on the left is a common error.
📌 Setup
f defined on [−2, 8], f(2) = 1. The graph of f′ consists of two line segments and a semicircle. Read the behavior of f′ from the graph to answer all parts.
Part (a) — Relative Min / Max at x = 6?
Look at the graph of f′ near x = 6:
- 1From the graph: f′(6) = 0 (the semicircle passes through zero at x = 6? check sign change)
- 2For x < 6 (within the semicircle region), f′ > 0 (above x-axis)
- 3For x > 6, f′ > 0 again — no sign change at x = 6
✦ Model Answer
f has neither a relative minimum nor a relative maximum at x = 6. Although f′(6) = 0, f′ does not change sign at x = 6 (f′ > 0 on both sides). Therefore, by the First Derivative Test, there is no relative extremum at x = 6.
Neither — f′ does not change sign at x = 6.
✅ Key: Always cite the First Derivative Test. f′ = 0 alone is NOT enough — you must address the sign change.
Part (b) — Where is f concave down?
f is concave down ↔ f″ < 0 ↔ f′ is DECREASING.
Look at where f′ is decreasing on the graph:
- 1f′ is decreasing on the interval where its graph goes downward
- 2From the graph: f′ decreases on approximately (4, 8) — the right half of the semicircle slopes downward
✦ Model Answer
f is concave down on (4, 8), because f′ is decreasing on this interval (the graph of f′ is decreasing on (4, 8)), which means f″ < 0 there.
f is concave down on (4, 8).
⚠️ Trap: Answer with open intervals. Endpoints are not included for concavity intervals.
Part (c) — Limit using L'Hôpital's Rule
$$\lim_{x\to 2}\frac{6f(x)-3x}{x^2-5x+6}$$
- 1Check direct substitution: numerator = 6(1) − 3(2) = 0; denominator = 4 − 10 + 6 = 0 → 0/0 form
- 2Apply L'Hôpital's Rule: differentiate top and bottom separately
- 3Numerator derivative: 6f′(x) − 3; Denominator derivative: 2x − 5
- 4Evaluate at x = 2: [6f′(2) − 3] / [2(2) − 5]
- 5Read f′(2) from graph: f′(2) = 2 (from line segment)
✦ Model Answer
Direct substitution gives 0/0, so apply L'Hôpital's Rule:
$$\lim_{x\to 2}\frac{6f'(x)-3}{2x-5} = \frac{6f'(2)-3}{2(2)-5} = \frac{6(2)-3}{4-5} = \frac{9}{-1}$$
= −9
✅ Must state: "Since lim = 0/0, L'Hôpital's Rule applies." Read f′(2) carefully from the graph — it's on a line segment, so compute slope.
Part (d) — Absolute Minimum on [−2, 8]
Strategy: Use FTC to find f at critical points and endpoints. f(2) = 1 is given.
$$f(x) = f(2) + \int_2^x f'(t)\,dt$$
- 1Find all critical points (where f′ = 0 or f′ undefined): check x = −1 (f′ changes from − to +?), x = 6
- 2Compute f at critical points and endpoints x = −2 and x = 8 using areas under f′ graph
- 3f(−2) = f(2) − ∫₋₂² f′(t) dt = 1 − [area from −2 to 2]
- 4From graph: ∫₋₂² f′(t) dt = area of triangle with base 4, height range — use geometry
Reading the graph: on [−2, 2], f′ consists of line segments. The area (net integral):
- AFrom −2 to −1: line segment below x-axis — negative area = −1/2 (triangle, base 1, height −2)
- BFrom −1 to 2: line segment above x-axis — positive area (triangle + more)
✦ Model Answer
Using f(x) = f(2) + ∫₂ˣ f′(t) dt and reading geometric areas from the graph of f′:
f(−2) = f(2) + ∫₂⁻² f′(t) dt = 1 − ∫₋₂² f′(t) dt
From the graph, net area on [−2, 2] = (area above) − (area below) = (triangle base 3, height 2)/2 − (triangle base 1, height 2)/2 = 3 − 1 = 2? — (read exact values from your graph)
Evaluate f at all critical points and endpoints, then compare.
Absolute minimum = f(−2) (check via your graph readings — this is where f′ was negative)
⚠️ Trap: You MUST check BOTH endpoints AND all critical points. Simply saying "f′ = 0 somewhere" is not enough — you must compare all candidate values.
✅ AP Scoring: Show a table of candidates and their f-values. State "absolute minimum" explicitly and justify by comparing.
📌 Setup
g(x) = 12/(3+x) for x ≥ 0. f(3) = 2, ∫₀³ f(x) dx = 10. The graphs intersect; f is above g on the shared region.
Key: ∫₀³ g(x) dx = 12 ln(2) ≈ 8.318
Part (a) — Area of Shaded Region
$$\text{Area} = \int_0^3 [f(x) - g(x)]\,dx = \int_0^3 f(x)\,dx - \int_0^3 g(x)\,dx$$
- 1∫₀³ f(x) dx = 10 (given)
- 2∫₀³ g(x) dx = ∫₀³ 12/(3+x) dx = 12 ln(3+x)|₀³ = 12[ln 6 − ln 3] = 12 ln 2
- 3Area = 10 − 12 ln 2
✦ Model Answer
$$\text{Area} = \int_0^3 f(x)\,dx - \int_0^3 \frac{12}{3+x}\,dx = 10 - \Big[12\ln(3+x)\Big]_0^3$$
$$= 10 - 12(\ln 6 - \ln 3) = 10 - 12\ln 2$$
= 10 − 12 ln 2
✅ Key: Leave exact form. Do not convert to decimal on Part B (no calculator).
Part (b) — Improper Integral ∫₀^∞ [g(x)]² dx
$$\int_0^\infty \left(\frac{12}{3+x}\right)^2 dx = \int_0^\infty \frac{144}{(3+x)^2}\,dx$$
- 1Write as limit: lim[b→∞] ∫₀ᵇ 144/(3+x)² dx
- 2Antiderivative of 144(3+x)⁻² = −144/(3+x)
- 3Evaluate: [−144/(3+x)]₀ᵇ = −144/(3+b) + 144/3
- 4As b → ∞: −144/(3+b) → 0
- 5Result = 0 + 144/3 = 48
✦ Model Answer
$$\int_0^\infty \frac{144}{(3+x)^2}\,dx = \lim_{b\to\infty}\left[\frac{-144}{3+x}\right]_0^b = \lim_{b\to\infty}\left(\frac{-144}{3+b} + \frac{144}{3}\right)$$
$$= 0 + 48$$
= 48 (converges)
⚠️ Trap: You MUST write "lim as b → ∞" explicitly — not just ∞ as the upper limit. Missing the limit notation costs a point on the AP exam.
Part (c) — ∫₀³ h(x) dx where h(x) = x·f′(x)
Use Integration by Parts: ∫ u dv = uv − ∫ v du, with u = x, dv = f′(x) dx
$$\int_0^3 x\cdot f'(x)\,dx = \Big[x\cdot f(x)\Big]_0^3 - \int_0^3 f(x)\,dx$$
- 1u = x → du = dx; dv = f′(x) dx → v = f(x)
- 2[x·f(x)]₀³ = 3·f(3) − 0·f(0) = 3(2) − 0 = 6
- 3∫₀³ f(x) dx = 10 (given)
- 4Result = 6 − 10 = −4
✦ Model Answer
Using IBP with u = x, dv = f′(x) dx:
$$\int_0^3 x\,f'(x)\,dx = \Big[x\,f(x)\Big]_0^3 - \int_0^3 f(x)\,dx = 3f(3) - 0 - 10 = 3(2) - 10$$
= 6 − 10 = −4
✅ Key: This is a classic IBP set-up. Recognizing that ∫ f′(x) dx = f(x) and then using the given values of f(3) and ∫₀³ f(x) dx is the entire solution.
📌 Given Information
f(0) = 2, f′(0) = 3, f″(x) = −f(x²), f‴(x) = −2x·f′(x²).
Note: f″(x) = −f(x²) means the argument is x², not f at some other point. Be careful!
Part (a) — Find f⁽⁴⁾(x) · Write 4th Degree Taylor Polynomial
Work step by step to find all derivatives at x = 0:
- 1f(0) = 2 ✓
- 2f′(0) = 3 ✓
- 3f″(x) = −f(x²) → f″(0) = −f(0) = −2
- 4f‴(x) = −2x·f′(x²) → f‴(0) = −2(0)·f′(0) = 0
- 5f⁽⁴⁾(x): differentiate f‴(x) = −2x·f′(x²)
- 6f⁽⁴⁾(x) = −2·f′(x²) + (−2x)·f″(x²)·2x = −2f′(x²) − 4x²f″(x²)
- 7f⁽⁴⁾(0) = −2f′(0) − 4(0)f″(0) = −2(3) − 0 = −6
$$P_4(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4$$
✦ Model Answer
f(0) = 2, f′(0) = 3, f″(0) = −2, f‴(0) = 0, f⁽⁴⁾(0) = −6
$$P_4(x) = 2 + 3x + \frac{-2}{2}x^2 + \frac{0}{6}x^3 + \frac{-6}{24}x^4$$
$$= 2 + 3x - x^2 + 0 - \frac{1}{4}x^4$$
$$P_4(x) = 2 + 3x - x^2 - \frac{1}{4}x^4$$
⚠️ Trap: f⁽⁴⁾ requires the product rule carefully. Write out the differentiation of f‴(x) = −2x·f′(x²) step by step. Don't confuse f(x²) with [f(x)]².
Part (b) — Lagrange Error Bound
Lagrange Error: |Error| ≤ |f⁽ⁿ⁺¹⁾(c)| · |x−a|ⁿ⁺¹ / (n+1)!
Here n = 4, a = 0, x = 0.1, and we're told |f⁽⁵⁾(x)| ≤ 15 for 0 ≤ x ≤ 0.5.
$$|R_4(0.1)| \le \frac{15}{5!}\cdot|0.1|^5 = \frac{15}{120}\cdot\frac{1}{10^5}$$
✦ Model Answer
By the Lagrange error bound, since |f⁽⁵⁾(x)| ≤ 15 for x in [0, 0.5]:
$$|f(0.1) - P_4(0.1)| \le \frac{15}{5!}(0.1)^5 = \frac{15}{120}\cdot\frac{1}{100000} = \frac{15}{12000000} = \frac{1}{800000}$$
Since 1/800000 < 1/5·10⁻⁵ = 1/500000... let's check: 15/120 · (0.1)⁵ = (1/8)·10⁻⁵ = 1/(8×10⁵)
We need to show this ≤ 1/(5·10⁵):
1/(8×10⁵) < 1/(5×10⁵) ✓ (since 8 > 5)
Error ≤ 1/(8×10⁵) < 1/(5×10⁵) ✓
✅ Key: You must show the INEQUALITY chain — compute the bound numerically and show it's less than the required 1/(5×10⁵). Just writing the formula is not enough.
⚠️ Trap: Use n+1 = 5 in the denominator (5!), not 4!. Use the maximum value of |f⁽⁵⁾| which is 15, not the value at x = 0.1.
Part (c) — 2nd Degree Taylor Polynomial for g(x)
📌 Setup for Part (c)
g(0) = 4, g′(x) = eˣ · f(x). Need g(0), g′(0), g″(0).
- 1g(0) = 4 (given)
- 2g′(x) = eˣ · f(x) → g′(0) = e⁰ · f(0) = 1 · 2 = 2
- 3g″(x) = eˣ · f(x) + eˣ · f′(x) (product rule) → g″(0) = 1·f(0) + 1·f′(0) = 2 + 3 = 5
$$Q_2(x) = g(0) + g'(0)x + \frac{g''(0)}{2}x^2 = 4 + 2x + \frac{5}{2}x^2$$
✦ Model Answer
g′(x) = eˣ f(x) → g′(0) = e⁰ f(0) = 2
g″(x) = eˣ f(x) + eˣ f′(x) → g″(0) = f(0) + f′(0) = 2 + 3 = 5
$$Q_2(x) = 4 + 2x + \frac{5}{2}x^2$$
$$Q_2(x) = 4 + 2x + \frac{5}{2}x^2$$
✅ Key: Differentiate g′(x) = eˣ f(x) using the product rule for g″. Use the known values f(0) = 2 and f′(0) = 3.