01
Given:
Rate in: \(R(t) = 20\sin\!\left(\dfrac{t^2}{35}\right)\) ft³/hr |
Rate out: \(D(t) = -0.04t^3 + 0.4t^2 + 0.96t\) ft³/hr |
Initial water: \(W(0) = 30\) ft³, on \(0 \le t \le 8\)
🔑 Core Concept
Net Change Theorem
Amount at time \(t\) = Initial amount + \(\displaystyle\int_0^t [\text{Rate in} - \text{Rate out}]\,dt\)
$$W(t) = 30 + \int_0^t [R(x) - D(x)]\,dx$$
a
Total rainwater flowing IN over 8 hours
2 pts
▾
Concept
Total volume = definite integral of the inflow rate R(t) over [0, 8]. This is simply the area under R(t), regardless of outflow.
TRAP: Do NOT subtract D(t) here. The question asks only for rainwater flowing IN, not net change.
TI-Nspire CX II — Definite Integral
1
Open Calc scratchpad or Calculator app
2
Press menu → 4: Calculus → 2: Numerical Integral ∫
3
Type exactly: ∫(20sin(t²/35),t,0,8)
4
OR use the template: ∫ symbol from math palette, enter lower=0, upper=8
→ Result ≈ 76.570 ft³
✎ Model Answer
\[\int_0^8 R(t)\,dt = \int_0^8 20\sin\!\left(\frac{t^2}{35}\right)dt \approx \boxed{76.570 \text{ ft}^3}\]
b
Increasing or decreasing at t = 3?
3 pts
▾
Key Idea
Sign of W′(t) = R(t) − D(t)
If R(3) − D(3) > 0 → increasing. If < 0 → decreasing. You must state numerical values.
Compute R(3) = \(20\sin(9/35) \approx 20\sin(0.2571) \approx 20(0.2543) \approx 5.087\)
Compute D(3) = \(-0.04(27)+0.4(9)+0.96(3) = -1.08+3.6+2.88 = 5.4\)
W′(3) = R(3) − D(3) ≈ 5.087 − 5.4 = −0.313 < 0
Conclude: amount is decreasing at t = 3
TI-Nspire CX II — Evaluating Functions
1
In Calculator: 20sin(3²/35) → enter
2
-0.04(3)³+0.4(3)²+0.96(3) → enter
3
Subtract: ans1 − ans2
R(3) ≈ 5.0870 | D(3) = 5.4 | Difference ≈ −0.313
MUST include: Write the actual numerical values — just saying "R < D" without numbers gets 0 pts on the justification.
✎ Model Answer
\[W'(3) = R(3) - D(3) \approx 5.087 - 5.4 = -0.313 < 0\]
Since \(W'(3) < 0\), the amount of water in the pipe is decreasing at \(t = 3\) hours.
c
Time of minimum water in pipe (0 ≤ t ≤ 8)
3 pts
▾
Strategy
Find where W′(t) = R(t) − D(t) = 0, then use Candidate Test
Check critical points AND endpoints. Compare W(t) values at each.
Solve R(t) − D(t) = 0 using calculator. Find all zeros on (0,8).
Zeros occur at: t ≈ 0 (trivial) and analyze sign changes
On (0, ~3.27): R(t) < D(t) → W decreasing. On (~3.27, 8): R(t) > D(t) → W increasing.
Minimum candidates: t ≈ 3.271 (critical point) and endpoints t = 0, t = 8
W(0) = 30. Compute W(3.271) and W(8).
TI-Nspire CX II — Finding Zeros & Integrals
1
Find zero: Graph 20sin(x²/35)-(-0.04x³+0.4x²+0.96x)
2
Press menu → 6: Analyze Graph → 1: Zero, set bounds around the crossing near x≈3.3
3
Critical point: t ≈ 3.271
4
Compute W(3.271): 30 + ∫(20sin(t²/35)-(-0.04t³+0.4t²+0.96t), t, 0, 3.271)
W(3.271) ≈ 27.964 ft³ < W(0) = 30 < W(8) ≈ 48.543 ft³
Justification required: Must show sign change of W′ from negative to positive, OR compare all candidate values.
✎ Model Answer
W′(t) = R(t) − D(t) = 0 at t ≈ 3.271.
W′(t) < 0 for 0 < t < 3.271 (decreasing)
W′(t) > 0 for 3.271 < t < 8 (increasing)
Therefore W has a minimum at t ≈ 3.271 hours.
W(3.271) ≈ 27.964 ft³
W′(t) < 0 for 0 < t < 3.271 (decreasing)
W′(t) > 0 for 3.271 < t < 8 (increasing)
Therefore W has a minimum at t ≈ 3.271 hours.
W(3.271) ≈ 27.964 ft³
d
Equation for overflow time w (t > 8)
3 pts
▾
Key Idea
Set up W(w) = 50 using the net change from t = 0, including what you found at t = 8.
DO NOT SOLVE — the problem says "write but do not solve." You lose points if you compute a numerical answer.
Start the integral at 0 (using W(0)=30) OR start at 8 (using W(8) ≈ 48.543). Both are acceptable.
✎ Model Answer (starting from t = 0)
\[30 + \int_0^w \bigl[R(t) - D(t)\bigr]\,dt = 50\]
where \(R(t) = 20\sin\!\left(\dfrac{t^2}{35}\right)\) and \(D(t) = -0.04t^3+0.4t^2+0.96t\).
where \(R(t) = 20\sin\!\left(\dfrac{t^2}{35}\right)\) and \(D(t) = -0.04t^3+0.4t^2+0.96t\).
02
Given: Velocity vector \(\mathbf{v}(t) = \langle t^2\cos t,\; e^{0.5t} \rangle\). At \(t=1\), particle is at \((3,5)\).
🔑 Parametric Essentials
$$x(t) = x(1) + \int_1^t v_x(\tau)\,d\tau \qquad y(t) = y(1) + \int_1^t v_y(\tau)\,d\tau$$
$$\text{Slope} = \frac{dy/dt}{dx/dt} = \frac{v_y(t)}{v_x(t)} \qquad \text{Speed} = \sqrt{v_x^2 + v_y^2}$$
$$\text{Distance} = \int_a^b \sqrt{[v_x(t)]^2 + [v_y(t)]^2}\,dt$$
a
x-coordinate at t = 2
2 pts
▾
x(2) = x(1) + \(\int_1^2 v_x(t)\,dt\) = \(3 + \int_1^2 t^2\cos t\,dt\)
Use calculator to evaluate the integral.
TI-Nspire CX II
1
3 + ∫(t²·cos(t), t, 1, 2)
→ x(2) ≈ 3 + 0.9056 ≈ 3.906
✎ Model Answer
\[x(2) = 3 + \int_1^2 t^2\cos t\,dt \approx 3 + 0.906 \approx \boxed{3.906}\]
b
Time when tangent slope = 2 (on 0 ≤ t ≤ 1)
3 pts
▾
Formula
Slope = dy/dx = v_y / v_x
$$\frac{e^{0.5t}}{t^2\cos t} = 2$$
Solve numerically for t ∈ (0, 1).
TI-Nspire CX II — Solve equation
1
Graph e^(0.5x)/(x²cos(x)) - 2 on [0,1]
2
Find zero: menu → 6 → 1: Zero
3
OR use menu → 3: Algebra → 1: Solve: solve(e^(0.5t)/(t²cos(t))=2, t) | 0<t<1
→ t ≈ 0.503
✎ Model Answer
\[\frac{dy/dt}{dx/dt} = \frac{e^{0.5t}}{t^2\cos t} = 2 \implies t \approx \boxed{0.503}\]
c
Time when speed = 3
3 pts
▾
Formula
$$\text{Speed} = \sqrt{(t^2\cos t)^2 + (e^{0.5t})^2} = 3$$
TI-Nspire CX II
1
Graph √((x²cos(x))²+(e^(0.5x))²) - 3
2
Find positive zero using menu → 6 → 1: Zero
→ t ≈ 1.128
Speed is a scalar (always ≥ 0). You're solving Speed = 3, not velocity = 3.
✎ Model Answer
\[\sqrt{(t^2\cos t)^2 + e^t} = 3 \implies t \approx \boxed{1.128}\]
d
Total distance traveled from t = 0 to t = 1
3 pts
▾
Key Formula
$$\text{Distance} = \int_0^1 \sqrt{[v_x(t)]^2 + [v_y(t)]^2}\,dt = \int_0^1 \sqrt{t^4\cos^2 t + e^t}\,dt$$
TI-Nspire CX II
1
∫(√(t⁴·(cos(t))²+e^t), t, 0, 1)
→ ≈ 1.949
Distance ≠ displacement. Distance uses the speed integral; displacement uses the velocity components separately.
✎ Model Answer
\[\int_0^1 \sqrt{t^4\cos^2 t + e^t}\,dt \approx \boxed{1.949}\]
03
| t (min) | 0 | 12 | 20 | 24 | 40 |
|---|---|---|---|---|---|
| v(t) (m/min) | 0 | 200 | 240 | −220 | 150 |
Bob's velocity: \(B(t) = t^3 - 6t^2 + 300\), \(0 \le t \le 10\)
a
Estimate v′(16)
2 pts
▾
Key Concept
Use the difference quotient (symmetric if possible)
Since 16 is between data points 12 and 20, use those two neighbors:
$$v'(16) \approx \frac{v(20) - v(12)}{20 - 12} = \frac{240 - 200}{8} = 5 \text{ m/min}^2$$
Always use the closest data points surrounding the target value. Include correct units (m/min²).
✎ Model Answer
\[v'(16) \approx \frac{v(20)-v(12)}{20-12} = \frac{240-200}{8} = \boxed{5 \text{ m/min}^2}\]
b
Meaning of ∫|v(t)|dt and Right Riemann Sum
3 pts
▾
Interpretation
\(\displaystyle\int_0^{40}|v(t)|\,dt\) = total distance Johanna travels (in meters) over 40 minutes. (Not displacement — absolute value removes the sign.)
Right Riemann Sum Setup
4 subintervals: [0,12], [12,20], [20,24], [24,40]. Use right endpoint of each.
$$\int_0^{40}|v(t)|\,dt \approx |v(12)|\cdot12 + |v(20)|\cdot8 + |v(24)|\cdot4 + |v(40)|\cdot16$$
$$= 200(12) + 240(8) + 220(4) + 150(16)$$
$$= 2400 + 1920 + 880 + 2400 = \mathbf{7600 \text{ m}}$$
TRAP: The integral has |v(t)|. v(24) = −220, so |v(24)| = 220. Many students forget the absolute value!
✎ Model Answer
Meaning: total distance (in meters) Johanna travels from t=0 to t=40 min.
Right Riemann Sum:
= 200(12) + 240(8) + |−220|(4) + 150(16)
= 2400 + 1920 + 880 + 2400 = 7600 meters
Right Riemann Sum:
= 200(12) + 240(8) + |−220|(4) + 150(16)
= 2400 + 1920 + 880 + 2400 = 7600 meters
c
Bob's acceleration at t = 5
2 pts
▾
Concept
Acceleration = derivative of velocity: B′(t)
$$B(t) = t^3 - 6t^2 + 300 \implies B'(t) = 3t^2 - 12t$$
$$B'(5) = 3(25) - 12(5) = 75 - 60 = 15 \text{ m/min}^2$$
✎ Model Answer
B′(t) = 3t² − 12t
B′(5) = 3(25) − 12(5) = 75 − 60 = 15 m/min²
B′(5) = 3(25) − 12(5) = 75 − 60 = 15 m/min²
d
Bob's average velocity on [0, 10]
3 pts
▾
Formula
$$\bar{v} = \frac{1}{10-0}\int_0^{10} B(t)\,dt = \frac{1}{10}\int_0^{10}(t^3-6t^2+300)\,dt$$
\(\int_0^{10}(t^3-6t^2+300)\,dt = \left[\frac{t^4}{4} - 2t^3 + 300t\right]_0^{10}\)
= (10000/4 − 2000 + 3000) − 0 = 2500 − 2000 + 3000 = 3500
Average = 3500 / 10 = 350 m/min
✎ Model Answer
\[\frac{1}{10}\int_0^{10}(t^3-6t^2+300)\,dt = \frac{1}{10}\left[\frac{t^4}{4}-2t^3+300t\right]_0^{10} = \frac{3500}{10} = \boxed{350 \text{ m/min}}\]
04
\(\dfrac{dy}{dx} = 2x - y\) | Initial condition: \(f(2) = 3\)
a
Sketch slope field at 6 indicated points
2 pts
▾
Method
At each point (x, y), compute the slope = 2x − y and draw a short segment with that slope.
| (x, y) | Slope = 2x − y | Direction |
|---|---|---|
| (0, 1) | 0 − 1 = −1 | ↘ downward |
| (0, 2) | 0 − 2 = −2 | ↘ steep down |
| (0, −1) | 0 − (−1) = 1 | ↗ upward |
| (1, 1) | 2 − 1 = 1 | ↗ upward |
| (1, 2) | 2 − 2 = 0 | → horizontal |
| (1, −1) | 2 − (−1) = 3 | ↗ steep up |
Draw neat short segments — not arrows, not curves. Length shouldn't matter; slope direction is what's graded.
✎ Grader Looks For
Correct sign of slope at each point. Relative steepness should be consistent (e.g., slope −2 steeper than slope −1). Segments, not curves.
b
Find d²y/dx² and concavity in Quadrant II
3 pts
▾
Differentiate dy/dx = 2x − y with respect to x:
\[\frac{d^2y}{dx^2} = \frac{d}{dx}(2x - y) = 2 - \frac{dy}{dx}\]
Substitute dy/dx = 2x − y:
\[\frac{d^2y}{dx^2} = 2 - (2x - y) = 2 - 2x + y\]
In Quadrant II: x < 0, y > 0. So 2 − 2x > 2 and y > 0 → d²y/dx² > 0 → concave up
Must state: "because −2x > 0 when x < 0, and y > 0, so d²y/dx² = 2−2x+y > 0." Full reasoning required for full credit.
✎ Model Answer
\[\frac{d^2y}{dx^2} = 2 - 2x + y\]
In Quadrant II: x < 0 ⟹ −2x > 0, and y > 0. Therefore d²y/dx² = 2 − 2x + y > 0.
All solution curves are concave up in Quadrant II.
All solution curves are concave up in Quadrant II.
c
Relative min, max, or neither at x = 2? (f(2) = 3)
3 pts
▾
Find f′(2): plug into DE: dy/dx = 2(2) − 3 = 4 − 3 = 1 ≠ 0
Since f′(2) = 1 ≠ 0, there is no critical point at x = 2.
Many students assume they need to check the second derivative. But since f′(2) ≠ 0, x=2 is NOT a critical point — so f has NEITHER a min nor a max there!
✎ Model Answer
f′(2) = 2(2) − f(2) = 4 − 3 = 1 ≠ 0.
Since f′(2) ≠ 0, f has neither a relative minimum nor a relative maximum at x = 2.
Since f′(2) ≠ 0, f has neither a relative minimum nor a relative maximum at x = 2.
d
Find m and b so that y = mx + b is a solution
3 pts
▾
Method
If y = mx + b, then dy/dx = m. Substitute into the DE:
$$m = 2x - (mx + b) = (2-m)x - b$$
For this to hold for ALL x: coefficients of x and constants must match.
Coefficient of x: 0 = 2 − m → m = 2
Constant: m = −b → 2 = −b → b = −2
✎ Model Answer
Substituting y = mx + b and dy/dx = m into the DE:
m = 2x − mx − b → 0 = (2−m)x + (−b−m)
Matching coefficients: m = 2, b = −2
Solution: y = 2x − 2
m = 2x − mx − b → 0 = (2−m)x + (−b−m)
Matching coefficients: m = 2, b = −2
Solution: y = 2x − 2
05
\(f(x) = \dfrac{1}{x^2 - kx}\), \(f'(x) = \dfrac{2-kx}{(x^2-kx)^2} \cdot (-1)... = \dfrac{-(2x-k)}{(x^2-kx)^2}\) wait — given: \(f'(x) = \dfrac{-(2x-k)}{(x^2-kx)^2}\)
Given Derivative
$$f'(x) = \frac{-(2x-k)}{(x^2-kx)^2}$$
Critical points where f′(x) = 0: numerator = 0 → 2x − k = 0 → x = k/2
a
k = 3: Tangent line at x = 4
2 pts
▾
\(f(x) = \dfrac{1}{x^2 - 3x}\). Find \(f(4) = \dfrac{1}{16-12} = \dfrac{1}{4}\)
\(f'(4) = \dfrac{-(8-3)}{(16-12)^2} = \dfrac{-5}{16}\)
Tangent line: \(y - \frac{1}{4} = -\frac{5}{16}(x-4)\)
✎ Model Answer
f(4) = 1/4, f′(4) = −5/16
\[y - \frac{1}{4} = -\frac{5}{16}(x-4)\] \[y = -\frac{5}{16}x + \frac{5}{4} + \frac{1}{4} = \boxed{-\frac{5}{16}x + \frac{3}{2}}\]
\[y - \frac{1}{4} = -\frac{5}{16}(x-4)\] \[y = -\frac{5}{16}x + \frac{5}{4} + \frac{1}{4} = \boxed{-\frac{5}{16}x + \frac{3}{2}}\]
b
k = 4: Relative min, max, or neither at x = 2?
3 pts
▾
Critical point: 2x − 4 = 0 → x = 2. Domain excludes x = 0 and x = 4.
Sign of f′(x) = -(2x-4)/(x²-4x)²: denominator always positive (when ≠ 0).
For x < 2: (2x−4) < 0 → numerator −(2x−4) > 0 → f′ > 0 (increasing)
For x > 2 (near 2): (2x−4) > 0 → f′ < 0 (decreasing)
Sign change + → − at x = 2 → relative maximum
✎ Model Answer
f′ changes sign from positive to negative at x = 2.
Therefore f has a relative maximum at x = 2.
Therefore f has a relative maximum at x = 2.
c
Find k for critical point at x = 5
3 pts
▾
Critical point at x = k/2 = 5 → k = 10
Verify x = 5 is in domain: denominator = x² − 10x = 25 − 50 = −25 ≠ 0 ✓
✎ Model Answer
f′(x) = 0 when 2x − k = 0 → x = k/2 = 5 → k = 10
d
k = 6: Partial fractions and ∫f(x)dx
4 pts
▾
Partial Fractions Method
$$f(x) = \frac{1}{x^2-6x} = \frac{1}{x(x-6)} = \frac{A}{x} + \frac{B}{x-6}$$
Multiply both sides by x(x−6): 1 = A(x−6) + Bx
x = 0: 1 = A(−6) → A = −1/6
x = 6: 1 = B(6) → B = 1/6
\(f(x) = \dfrac{-1/6}{x} + \dfrac{1/6}{x-6}\)
\(\int f(x)\,dx = -\dfrac{1}{6}\ln|x| + \dfrac{1}{6}\ln|x-6| + C = \dfrac{1}{6}\ln\!\left|\dfrac{x-6}{x}\right| + C\)
Don't forget the absolute values in ln|x| and ln|x−6|. Missing them costs a point.
✎ Model Answer
\[\frac{1}{x(x-6)} = \frac{-1/6}{x} + \frac{1/6}{x-6}\]
\[\int f(x)\,dx = -\frac{1}{6}\ln|x| + \frac{1}{6}\ln|x-6| + C = \boxed{\frac{1}{6}\ln\left|\frac{x-6}{x}\right| + C}\]
06
\(f(x) = \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{3^n \cdot n} = x/3 - x^2/18 + x^3/81 - \cdots\)
🔑 Key Series Facts to Memorize
$$e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$
$$\ln(1+x) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n} \text{ for } |x| \le 1, x \ne -1$$
Notice: our series is ln(1 + x/3) by substitution u = x/3.
a
Find radius of convergence R using the Ratio Test
3 pts
▾
Ratio Test Formula
$$L = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| < 1 \text{ for convergence}$$
\(a_n = \dfrac{(-1)^{n+1}x^n}{3^n \cdot n}\), so \(a_{n+1} = \dfrac{(-1)^{n+2}x^{n+1}}{3^{n+1}(n+1)}\)
\(\left|\dfrac{a_{n+1}}{a_n}\right| = \left|\dfrac{x^{n+1}}{3^{n+1}(n+1)} \cdot \dfrac{3^n \cdot n}{x^n}\right| = \dfrac{|x| \cdot n}{3(n+1)}\)
As n→∞: limit = |x|/3. Converges when |x|/3 < 1 → |x| < 3
R = 3
Show ALL steps of the ratio. Write out the limit explicitly. Don't skip to R=3.
✎ Model Answer
\[\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}\frac{|x|}{3}\cdot\frac{n}{n+1} = \frac{|x|}{3}\]
Convergence requires |x|/3 < 1, so R = 3.
b
First 4 terms of f′(x), then express as rational function
4 pts
▾
Method: Term-by-Term Differentiation
Differentiate \(f(x) = \dfrac{x}{3} - \dfrac{x^2}{18} + \dfrac{x^3}{81} - \dfrac{x^4}{324} + \cdots\)
\(f'(x) = \dfrac{1}{3} - \dfrac{2x}{18} + \dfrac{3x^2}{81} - \dfrac{4x^3}{324} + \cdots\)
Simplify: \(f'(x) = \dfrac{1}{3} - \dfrac{x}{9} + \dfrac{x^2}{27} - \dfrac{x^3}{81} + \cdots\)
General term: \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{n-1}}{3^n} = \sum_{n=0}^{\infty}\frac{(-1)^n x^n}{3^{n+1}} = \frac{1}{3}\sum_{n=0}^{\infty}\left(\frac{-x}{3}\right)^n\)
Geometric series with ratio r = −x/3: \(f'(x) = \dfrac{1/3}{1-(-x/3)} = \dfrac{1/3}{1+x/3} = \dfrac{1}{3+x}\)
✎ Model Answer
First 4 terms: \(f'(x) = \dfrac{1}{3} - \dfrac{x}{9} + \dfrac{x^2}{27} - \dfrac{x^3}{81} + \cdots\)
As a rational function: \(\boxed{f'(x) = \dfrac{1}{x+3}}\) for |x| < 3
As a rational function: \(\boxed{f'(x) = \dfrac{1}{x+3}}\) for |x| < 3
c
First 4 terms of e^x f(x); 3rd-degree Taylor polynomial of g(x) = e^x f(x)
5 pts
▾
Method: Multiply Series Together
Use \(e^x = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \cdots\) and \(f(x) = \dfrac{x}{3} - \dfrac{x^2}{18} + \dfrac{x^3}{81} + \cdots\)
For the 3rd-degree Taylor polynomial of g = eˣ · f(x), we only need terms up to x³.
Multiply keeping only up to x³:
\(e^x \cdot f(x) = \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6}\right)\left(\frac{x}{3} - \frac{x^2}{18} + \frac{x^3}{81}\right)\)
\(e^x \cdot f(x) = \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6}\right)\left(\frac{x}{3} - \frac{x^2}{18} + \frac{x^3}{81}\right)\)
\(x^1\): \(1\cdot\frac{x}{3} = \frac{x}{3}\)
\(x^2\): \(1\cdot(-\frac{x^2}{18}) + x\cdot\frac{x}{3} = -\frac{x^2}{18} + \frac{x^2}{3} = \frac{5x^2}{18}\)
\(x^3\): \(1\cdot\frac{x^3}{81} + x\cdot(-\frac{x^2}{18}) + \frac{x^2}{2}\cdot\frac{x}{3} = \frac{1}{81} - \frac{1}{18} + \frac{1}{6} = \frac{2}{81} - \frac{9}{81} + \frac{13.5}{81}\)...
= \(\dfrac{1}{81} - \dfrac{1}{18} + \dfrac{1}{6} = \dfrac{2-9+27}{162} = \dfrac{20}{162} = \dfrac{10}{81}\)
= \(\dfrac{1}{81} - \dfrac{1}{18} + \dfrac{1}{6} = \dfrac{2-9+27}{162} = \dfrac{20}{162} = \dfrac{10}{81}\)
Be very careful collecting x³ terms — there are THREE contributions. Organize by powers systematically.
✎ Model Answer
3rd-degree Taylor polynomial for g(x) = e^x · f(x):
\[P_3(x) = \frac{x}{3} + \frac{5x^2}{18} + \frac{10x^3}{81}\]
\[P_3(x) = \frac{x}{3} + \frac{5x^2}{18} + \frac{10x^3}{81}\]