📝 Unit 3: Derivative Applications — Unit Test

Motion · Related Rates · Optimization · Marginal Analysis
✅ Graded
⏱️ 75 min  |  Total: /60 marks
K/U
/15
Thinking
/15
Comm.
/15
Applic.
/15
Part A: Knowledge & Understanding [15]
1 [3]
Position \(s(t)=2t^3-15t^2+24t\) (m, s). Find \(v(t)\) and \(a(t)\) and the values of \(t\) where the particle is at rest.
2 [3]
A spherical balloon's volume increases at 12 cm³/s. Find \(\frac{dr}{dt}\) when \(r=3\) cm.
3 [3]
Find the absolute maximum of \(f(x)=x^3-3x^2+1\) on \([-1,3]\).
4 [3]
A rectangle has perimeter 40 m. Find dimensions for maximum area.
5 [3]
Cost \(C(x)=0.005x^3-0.6x^2+30x+200\). Find the marginal cost at \(x=20\).
Part B: Thinking [15]
6 [5]
A 10 m ladder leans against a wall. The bottom slides away at 0.5 m/s. How fast is the top sliding down when the bottom is 6 m from the wall?
7 [5]
A right circular cone with height = 3·radius is filling with water at 8 m³/min. Find the rate at which the water level rises when the depth is 6 m.
8 [5]
Find the dimensions of the rectangle of maximum area that can be inscribed in a semicircle of radius 5, with one side along the diameter.
Part C: Communication [15]
9 [4]
Describe the standard 5-step procedure for solving an optimization problem. Apply the labels generally (no numbers).
10 [4]
Explain the difference between "speeding up" and "moving in the positive direction." Give an example where a particle is speeding up while moving in the negative direction.
11 [4]
Describe step-by-step how to solve a related-rates problem. Use the language "what we know," "what we want," and "relate via equation."
12 [3]
Justify why the second-derivative test confirms a maximum vs minimum. Apply to \(f(x)=x^3-3x\).
Part D: Application [15]
13 [5]
A farmer has 200 m of fencing for three sides of a rectangular pasture (river forms the fourth side). Find dimensions maximizing area.
14 [5]
A cylindrical can must hold 1 L (1000 cm³). Find the dimensions (radius and height) that minimize the surface area (closed top and bottom).
15 [5]
Demand: \(p(x)=200-0.5x\). Cost: \(C(x)=20x+1000\). Find production level \(x\) that maximizes profit. State maximum profit.

Evaluation Rubric

LevelDescription%
4Thorough, insightful80–100
3Considerable70–79
2Some60–69
1Limited50–59
RInsufficient<50
📕 Answer Key

1. \(v=6t^2-30t+24=6(t-1)(t-4)\); \(a=12t-30\). At rest: \(t=1,4\) s.

2. \(V=\frac{4}{3}\pi r^3\); \(\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\). At \(r=3\): \(12=36\pi\,r'\) → \(r'=1/(3\pi)\approx 0.106\) cm/s.

3. \(f'=3x^2-6x=3x(x-2)=0\) → \(x=0,2\). Values: \(f(-1)=-3,\,f(0)=1,\,f(2)=-3,\,f(3)=1\). Max = 1 at \(x=0\) or 3.

4. \(2x+2y=40\) → \(y=20-x\). \(A=x(20-x)\); \(A'=20-2x=0\) → \(x=10\), \(y=10\) (square, A = 100 m²).

5. \(C'=0.015x^2-1.2x+30\); at 20: \(6-24+30=12\) → $12/unit.

6. \(x^2+y^2=100\); at \(x=6\), \(y=8\). \(2x x'+2y y'=0\) → \(y'=-(6\cdot 0.5)/8=-0.375\) m/s.

7. \(r=h/3\), \(V=\frac{\pi h^3}{27}\). \(\frac{dV}{dt}=\frac{\pi h^2}{9}\frac{dh}{dt}\). \(8=\frac{36\pi}{9}h'\) → \(h'=8/(4\pi)=2/\pi\approx 0.637\) m/min.

8. Width = \(2x\), height = \(\sqrt{25-x^2}\). \(A=2x\sqrt{25-x^2}\); maximizing gives \(x=5/\sqrt{2}\), so width \(=5\sqrt 2\), height \(=5/\sqrt 2\). \(A_{max}=25\).

9. Read; identify constraint and quantity to optimize; express objective in one variable using the constraint; differentiate and find critical numbers; check endpoints/second derivative; answer with units.

10. Speeding up = \(|v|\) increasing = \(v,a\) same sign. A particle with \(v=-2, a=-3\) is moving negative AND speeding up.

11. List variables/rates known; identify rate wanted; write equation relating them; differentiate w.r.t. time; substitute given instant; solve. Include units.

12. If \(f''(c)<0\), concave down → max at \(c\); if \(>0\), min. \(f(x)=x^3-3x\): \(f'=3x^2-3=0\) at \(x=\pm 1\); \(f''=6x\); \(f''(1)=6>0\) → min; \(f''(-1)=-6<0\) → max.

13. \(2x+y=200\) (river replaces one side). \(A=x(200-2x)=200x-2x^2\); \(A'=200-4x=0\) → \(x=50\), \(y=100\). \(A_{max}=5000\) m².

14. \(\pi r^2 h=1000\); \(S=2\pi r^2+2\pi r h\). Sub \(h=1000/(\pi r^2)\): \(S=2\pi r^2+2000/r\). \(S'=4\pi r-2000/r^2=0\) → \(r^3=500/\pi\) → \(r\approx 5.42\) cm; \(h\approx 10.84\) cm (h = 2r).

15. \(R=xp=200x-0.5x^2\); \(P=R-C=200x-0.5x^2-20x-1000=180x-0.5x^2-1000\); \(P'=180-x=0\) → \(x=180\). \(P_{max}=180(180)-0.5(180)^2-1000=32400-16200-1000=$15{,}200\).