NUM8ERS Tutoring Center Dubai logo – you can count on us
Book Now
Sign In

AP Calculus BC — Unit 9.5 (Integrating Vector-Valued Functions)

Key Concepts Packet + Worked Examples + Interactive MCQ Practice (36 Questions)

Unit 9.5 Key Concepts Packet

1) Vector Antiderivatives (Component-Wise Integration)

If \( \mathbf{r}(t)=\langle f(t),g(t)\rangle \), then \[ \int \mathbf{r}(t)\,dt=\left\langle \int f(t)\,dt,\; \int g(t)\,dt \right\rangle =\langle F(t),G(t)\rangle+\mathbf{C}, \] where \( \mathbf{C}=\langle C_1,C_2\rangle \) is a constant vector.
  • In 3D: \( \int \langle f,g,h\rangle dt=\langle \int fdt,\int gdt,\int hdt\rangle+\langle C_1,C_2,C_3\rangle \).
  • A “constant of integration” for vectors is a vector, not a single scalar.

2) Position from Velocity (Initial Conditions)

If \( \mathbf{v}(t)=\mathbf{r}'(t) \), then \[ \mathbf{r}(t)=\int \mathbf{v}(t)\,dt+\mathbf{C}. \] If you know \( \mathbf{r}(t_0) \), substitute \(t=t_0\) to solve for \( \mathbf{C} \).
  • Net displacement on \( [a,b] \): \( \Delta \mathbf{r}=\mathbf{r}(b)-\mathbf{r}(a)=\int_a^b \mathbf{v}(t)\,dt \).
  • Average velocity: \( \mathbf{v}_{avg}=\dfrac{\mathbf{r}(b)-\mathbf{r}(a)}{b-a} \).

3) Velocity & Position from Acceleration

If \( \mathbf{a}(t)=\mathbf{v}'(t) \), then \[ \mathbf{v}(t)=\int \mathbf{a}(t)\,dt+\mathbf{C}_v, \quad \mathbf{r}(t)=\int \mathbf{v}(t)\,dt+\mathbf{C}_r. \] Use initial conditions like \( \mathbf{v}(t_0) \) and \( \mathbf{r}(t_0) \).

4) Displacement vs. Distance Traveled

  • Displacement vector: \( \int_a^b \mathbf{v}(t)\,dt=\mathbf{r}(b)-\mathbf{r}(a) \).
  • Distance traveled (a scalar): \( \int_a^b \|\mathbf{v}(t)\|\,dt \), where \( \|\mathbf{v}(t)\|=\sqrt{(v_x)^2+(v_y)^2} \) (or include \(v_z\) in 3D).
  • Displacement can be \( \langle 0,0\rangle \) even if the particle moved (it may return to its start).

Worked Examples + Notes

Example 1 — Position from Velocity

A particle has velocity \( \mathbf{v}(t)=\langle 2t,\,3\rangle \) and position \( \mathbf{r}(0)=\langle 1,\,-2\rangle \). Find \( \mathbf{r}(t) \).

  • Integrate component-wise: \( \mathbf{r}(t)=\int\langle 2t,3\rangle dt=\langle t^2,3t\rangle+\mathbf{C} \).
  • Use \( \mathbf{r}(0)=\langle 1,-2\rangle \): \( \langle 0,0\rangle+\mathbf{C}=\langle 1,-2\rangle \Rightarrow \mathbf{C}=\langle 1,-2\rangle \).
  • \( \mathbf{r}(t)=\langle t^2+1,\;3t-2\rangle \).

Example 2 — From Acceleration to Velocity and Position

A particle has acceleration \( \mathbf{a}(t)=\langle 6t,\;0\rangle \), velocity \( \mathbf{v}(0)=\langle 2,\,-1\rangle \), and position \( \mathbf{r}(0)=\langle 0,\,3\rangle \). Find \( \mathbf{r}(t) \).

  • \( \mathbf{v}(t)=\int\mathbf{a}(t)\,dt=\langle 3t^2,0\rangle+\mathbf{C}_v \).
  • Use \( \mathbf{v}(0)=\langle 2,-1\rangle \Rightarrow \mathbf{C}_v=\langle 2,-1\rangle \), so \( \mathbf{v}(t)=\langle 3t^2+2,\,-1\rangle \).
  • \( \mathbf{r}(t)=\int\mathbf{v}(t)\,dt=\langle t^3+2t,\,-t\rangle+\mathbf{C}_r \).
  • Use \( \mathbf{r}(0)=\langle 0,3\rangle \Rightarrow \mathbf{C}_r=\langle 0,3\rangle \).
  • \( \mathbf{r}(t)=\langle t^3+2t,\;3-t\rangle \).

Example 3 — Displacement and Average Velocity

If \( \mathbf{v}(t)=\langle t,\;t^2\rangle \) on \(1\le t\le 3\), find the displacement and average velocity.

  • Displacement: \( \int_1^3\langle t,t^2\rangle dt=\left\langle \int_1^3 tdt,\int_1^3 t^2dt\right\rangle=\left\langle 4,\tfrac{26}{3}\right\rangle \).
  • Average velocity: \( \mathbf{v}_{avg}=\dfrac{\Delta\mathbf{r}}{3-1}=\left\langle 2,\tfrac{13}{3}\right\rangle \).

Example 4 — Displacement vs Distance Traveled

If \( \mathbf{v}(t)=\langle 3,4\rangle \) for \(0\le t\le 2\), find displacement and distance traveled.

  • Displacement: \( \int_0^2\langle 3,4\rangle dt=\langle 6,8\rangle \).
  • Speed: \( \|\mathbf{v}\|=\sqrt{3^2+4^2}=5 \).
  • Distance traveled: \( \int_0^2 5\,dt=10 \).

Unit 9.5 Multiple-Choice Practice (36 Questions)

1.If \( \mathbf{r}(t)=\langle f(t),g(t)\rangle \), then \( \displaystyle \int \mathbf{r}(t)\,dt \) equals:
  1. Vector integration is component-wise.
  2. \( \int\langle f,g\rangle dt=\langle \int fdt,\int gdt\rangle+\mathbf{C} \), where \( \mathbf{C} \) is a constant vector.
2.An antiderivative of \( \mathbf{r}(t)=\langle t^2,\sin t\rangle \) is:
  1. Integrate each component separately.
  2. \( \int t^2dt=\tfrac{t^3}{3}\) and \( \int \sin t\,dt=-\cos t\).
  3. So an antiderivative is \( \left\langle \tfrac{t^3}{3},-\cos t\right\rangle+\mathbf{C} \).
3.A particle has velocity \( \mathbf{v}(t)=\langle 2t,3\rangle \) and position \( \mathbf{r}(0)=\langle 1,-2\rangle \). Then \( \mathbf{r}(t) \) is:
  1. \( \mathbf{r}(t)=\int\mathbf{v}(t)\,dt=\langle t^2,3t\rangle+\mathbf{C} \).
  2. Use \( \mathbf{r}(0)=\langle 1,-2\rangle \Rightarrow \mathbf{C}=\langle 1,-2\rangle \).
  3. \( \mathbf{r}(t)=\langle t^2+1,3t-2\rangle \).
4.If \( \mathbf{a}(t)=\langle 6t,0\rangle \) and \( \mathbf{v}(0)=\langle 2,-1\rangle \), then \( \mathbf{v}(t) \) is:
  1. \( \mathbf{v}(t)=\int\mathbf{a}(t)\,dt=\langle 3t^2,0\rangle+\mathbf{C} \).
  2. Use \( \mathbf{v}(0)=\langle 2,-1\rangle \Rightarrow \mathbf{C}=\langle 2,-1\rangle \).
  3. \( \mathbf{v}(t)=\langle 3t^2+2,-1\rangle \).
5.If \( \mathbf{v}(t)=\langle \cos t,\sin t\rangle \) and \( \mathbf{r}(0)=\langle 0,0\rangle \), then \( \mathbf{r}(\pi) \) is:
  1. \( \mathbf{r}(t)=\int\langle \cos t,\sin t\rangle dt=\langle \sin t,-\cos t\rangle+\mathbf{C} \).
  2. Use \( \mathbf{r}(0)=\langle 0,0\rangle \): at \(t=0\), \( \langle 0,-1\rangle+\mathbf{C}=\langle 0,0\rangle \Rightarrow \mathbf{C}=\langle 0,1\rangle \).
  3. \( \mathbf{r}(t)=\langle \sin t,1-\cos t\rangle \Rightarrow \mathbf{r}(\pi)=\langle 0,2\rangle \).
6.The displacement from \(t=1\) to \(t=3\) for \( \mathbf{v}(t)=\langle t,t^2\rangle \) is:
  1. Displacement \(=\int_1^3\mathbf{v}(t)\,dt=\left\langle \int_1^3 t\,dt,\int_1^3 t^2\,dt\right\rangle \).
  2. \( \int_1^3 t\,dt=\left[\tfrac{t^2}{2}\right]_1^3=\tfrac{9-1}{2}=4 \).
  3. \( \int_1^3 t^2\,dt=\left[\tfrac{t^3}{3}\right]_1^3=\tfrac{27-1}{3}=\tfrac{26}{3} \).
  4. \( \Delta\mathbf{r}=\left\langle 4,\tfrac{26}{3}\right\rangle \).
7.If \( \mathbf{r}(t)=\langle t^3,4t\rangle \), then the average velocity on \(0\le t\le 2\) is:
  1. \( \mathbf{v}_{avg}=\dfrac{\mathbf{r}(2)-\mathbf{r}(0)}{2-0} \).
  2. \( \mathbf{r}(2)=\langle 8,8\rangle \) and \( \mathbf{r}(0)=\langle 0,0\rangle \).
  3. \( \mathbf{v}_{avg}=\dfrac{\langle 8,8\rangle}{2}=\langle 4,4\rangle \).
8.If \( \mathbf{v}(t)=\langle e^t,\cos t\rangle \), then the net change in the \(x\)-coordinate on \(0\le t\le 1\) is:
  1. Net change in \(x\) equals \( \int_0^1 v_x(t)\,dt=\int_0^1 e^t dt \).
  2. \( \int_0^1 e^t dt=[e^t]_0^1=e-1 \).
9.Which statement is always true for velocity \( \mathbf{v}(t) \) and position \( \mathbf{r}(t) \)?
  1. Since \( \mathbf{v}(t)=\mathbf{r}'(t) \), apply the Fundamental Theorem of Calculus component-wise.
  2. \( \int_a^b \mathbf{v}(t)\,dt=\mathbf{r}(b)-\mathbf{r}(a) \).
10.If \( \mathbf{v}(t)=\langle \tfrac{1}{t},2t\rangle \) and \( \mathbf{r}(1)=\langle 0,3\rangle \), then \( \mathbf{r}(t) \) is:
  1. Integrate: \( \mathbf{r}(t)=\langle \ln t,\;t^2\rangle+\mathbf{C} \).
  2. Use \( \mathbf{r}(1)=\langle 0,3\rangle \): \( \langle 0,1\rangle+\mathbf{C}=\langle 0,3\rangle \Rightarrow \mathbf{C}=\langle 0,2\rangle \).
  3. \( \mathbf{r}(t)=\langle \ln t,\;t^2+2\rangle \).
11.If \( \mathbf{v}(t)=\langle 3t^2,-4t\rangle \) and \( \mathbf{r}(0)=\langle 1,5\rangle \), then \( \mathbf{r}(2) \) is:
  1. \( \mathbf{r}(t)=\int\langle 3t^2,-4t\rangle dt=\langle t^3,-2t^2\rangle+\mathbf{C} \).
  2. Use \( \mathbf{r}(0)=\langle 1,5\rangle \Rightarrow \mathbf{C}=\langle 1,5\rangle \).
  3. \( \mathbf{r}(2)=\langle 8+1,\;-8+5\rangle=\langle 9,-3\rangle \).
12.If \( \mathbf{a}(t)=\langle \sin t,2\rangle \) and \( \mathbf{v}(0)=\langle 1,-1\rangle \), then \( \mathbf{v}(\pi) \) is:
  1. \( \mathbf{v}(t)=\int\langle \sin t,2\rangle dt=\langle -\cos t,2t\rangle+\mathbf{C} \).
  2. Use \( \mathbf{v}(0)=\langle 1,-1\rangle \): \( \langle -1,0\rangle+\mathbf{C}=\langle 1,-1\rangle \Rightarrow \mathbf{C}=\langle 2,-1\rangle \).
  3. \( \mathbf{v}(t)=\langle -\cos t+2,2t-1\rangle \Rightarrow \mathbf{v}(\pi)=\langle 3,2\pi-1\rangle \).
13.If \( \mathbf{a}(t)=\langle 4,-6t\rangle \), \( \mathbf{v}(0)=\langle 0,2\rangle \), and \( \mathbf{r}(0)=\langle -1,1\rangle \), then \( \mathbf{r}(t) \) is:
  1. \( \mathbf{v}(t)=\int\mathbf{a}(t)dt=\langle 4t,-3t^2\rangle+\mathbf{C}_v \).
  2. Use \( \mathbf{v}(0)=\langle 0,2\rangle \Rightarrow \mathbf{C}_v=\langle 0,2\rangle \), so \( \mathbf{v}(t)=\langle 4t,-3t^2+2\rangle \).
  3. \( \mathbf{r}(t)=\int\mathbf{v}(t)dt=\langle 2t^2,-t^3+2t\rangle+\mathbf{C}_r \).
  4. Use \( \mathbf{r}(0)=\langle -1,1\rangle \Rightarrow \mathbf{C}_r=\langle -1,1\rangle \).
  5. \( \mathbf{r}(t)=\langle 2t^2-1,-t^3+2t+1\rangle \).
14.If \( \mathbf{v}(t)=\langle 3,4\rangle \) for \(0\le t\le2\), then the distance traveled is:
  1. Speed is \( \|\mathbf{v}\|=\sqrt{3^2+4^2}=5 \).
  2. Distance traveled is \( \int_0^2 5\,dt=10 \).
15.If \( \int_a^b \mathbf{v}(t)\,dt=\langle 0,0\rangle \), which statement must be true?
  1. \( \int_a^b \mathbf{v}(t)\,dt=\mathbf{r}(b)-\mathbf{r}(a) \).
  2. If the integral is \( \langle 0,0\rangle \), then \( \mathbf{r}(b)=\mathbf{r}(a) \).
  3. This does not force the distance traveled to be zero; the particle could loop and return.
16.Compute \( \displaystyle \int_0^{\pi}\langle \sin t,\cos t\rangle\,dt \).
  1. Integrate component-wise.
  2. \( \int_0^{\pi}\sin t\,dt=[-\cos t]_0^{\pi}=2 \).
  3. \( \int_0^{\pi}\cos t\,dt=[\sin t]_0^{\pi}=0 \).
  4. \( \int_0^{\pi}\langle \sin t,\cos t\rangle dt=\langle 2,0\rangle \).
17.If \( \mathbf{v}(t)=\langle \sec^2 t,1\rangle \) and \( \mathbf{r}(0)=\langle 2,-1\rangle \), then \( \mathbf{r}(t) \) is:
  1. \( \mathbf{r}(t)=\int\langle \sec^2 t,1\rangle dt=\langle \tan t,t\rangle+\mathbf{C} \).
  2. Use \( \mathbf{r}(0)=\langle 2,-1\rangle \): at \(t=0\), \( \langle 0,0\rangle+\mathbf{C}=\langle 2,-1\rangle \Rightarrow \mathbf{C}=\langle 2,-1\rangle \).
  3. \( \mathbf{r}(t)=\langle \tan t+2,t-1\rangle \).
18.If \( \mathbf{r}(t)=\langle \ln t,t^2\rangle \), then \( \mathbf{v}(t)=\mathbf{r}'(t) \) is:
  1. Differentiate component-wise: \( (\ln t)'=\tfrac{1}{t} \) and \( (t^2)'=2t \).
  2. \( \mathbf{v}(t)=\left\langle \tfrac{1}{t},2t\right\rangle \).
19.Compute the displacement \( \mathbf{r}(2)-\mathbf{r}(0) \) if \( \mathbf{v}(t)=\langle 2t,1-t\rangle \).
  1. Displacement is \( \int_0^2\mathbf{v}(t)\,dt=\left\langle \int_0^2 2t\,dt,\int_0^2 (1-t)\,dt\right\rangle \).
  2. \( \int_0^2 2t\,dt=[t^2]_0^2=4 \).
  3. \( \int_0^2 (1-t)\,dt=[t-\tfrac{t^2}{2}]_0^2=2-2=0 \).
  4. \( \Delta\mathbf{r}=\langle 4,0\rangle \).
20.If \( \mathbf{r}(1)=\langle 2,0\rangle \) and \( \mathbf{v}(t)=\langle 2,3t^2\rangle \), then \( \mathbf{r}(3) \) is:
  1. Use net change: \( \mathbf{r}(3)=\mathbf{r}(1)+\int_1^3 \mathbf{v}(t)\,dt \).
  2. \( \int_1^3 \langle 2,3t^2\rangle dt=\left\langle \int_1^3 2dt,\int_1^3 3t^2dt\right\rangle=\langle 4,26\rangle \).
  3. \( \mathbf{r}(3)=\langle 2,0\rangle+\langle 4,26\rangle=\langle 6,26\rangle \).
21.If \( \mathbf{v}(t)=\langle t-1,2\rangle \), then an antiderivative is:
  1. Integrate: \( \int(t-1)dt=\tfrac{t^2}{2}-t \) and \( \int 2dt=2t \).
  2. \( \int\langle t-1,2\rangle dt=\left\langle \tfrac{t^2}{2}-t,2t\right\rangle+\mathbf{C} \).
22.If \( \mathbf{r}'(t)=\mathbf{v}(t) \), then \( \int_a^b \mathbf{v}(t)\,dt \) represents:
  1. By FTC: \( \int_a^b \mathbf{v}(t)\,dt=\mathbf{r}(b)-\mathbf{r}(a) \).
  2. That is exactly the displacement vector.
23.For \( \mathbf{v}(t)=\langle 3t,4t\rangle \), the distance traveled on \(0\le t\le 1\) is:
  1. Speed: \( \|\mathbf{v}(t)\|=\sqrt{(3t)^2+(4t)^2}=5t \) for \(t\ge 0\).
  2. Distance traveled: \( \int_0^1 5t\,dt=\left[\tfrac{5}{2}t^2\right]_0^1=\tfrac{5}{2} \).
24.If \( \mathbf{a}(t)=\langle 0,6\rangle \) and \( \mathbf{v}(2)=\langle 5,-1\rangle \), then \( \mathbf{v}(t) \) is:
  1. \( \mathbf{v}(t)=\int\langle 0,6\rangle dt=\langle C_1,6t+C_2\rangle \).
  2. Use \( \mathbf{v}(2)=\langle 5,-1\rangle \): \(C_1=5\) and \(12+C_2=-1\Rightarrow C_2=-13\).
  3. \( \mathbf{v}(t)=\langle 5,6t-13\rangle \).
25.If \( \mathbf{v}(t)=\langle t^2,2t\rangle \), then \( \int_0^2 \mathbf{v}(t)\,dt \) equals:
  1. \( \int_0^2 \langle t^2,2t\rangle dt=\left\langle \int_0^2 t^2dt,\int_0^2 2t\,dt\right\rangle \).
  2. \( \int_0^2 t^2dt=\left[\tfrac{t^3}{3}\right]_0^2=\tfrac{8}{3} \).
  3. \( \int_0^2 2t\,dt=[t^2]_0^2=4 \).
  4. \( \left\langle \tfrac{8}{3},4\right\rangle \).
26.If \( \mathbf{v}(t)=\langle -\sin t,\cos t\rangle \) and \( \mathbf{r}(0)=\langle 1,0\rangle \), then \( \mathbf{r}(t) \) is:
  1. Integrate: \( \mathbf{r}(t)=\int\langle -\sin t,\cos t\rangle dt=\langle \cos t,\sin t\rangle+\mathbf{C} \).
  2. Use \( \mathbf{r}(0)=\langle 1,0\rangle \): at \(t=0\), \( \langle 1,0\rangle+\mathbf{C}=\langle 1,0\rangle \Rightarrow \mathbf{C}=\langle 0,0\rangle \).
  3. \( \mathbf{r}(t)=\langle \cos t,\sin t\rangle \).
27.Which quantity equals the total distance traveled on \( [a,b] \)?
  1. Distance traveled is the integral of speed, not velocity.
  2. \( \text{Distance}=\int_a^b \|\mathbf{v}(t)\|\,dt \).
28.If \( \mathbf{r}(t)=\langle t, t^2, \sin t\rangle \), then \( \int \mathbf{r}(t)\,dt \) equals:
  1. Integrate each component: \( \int tdt=\tfrac{t^2}{2} \), \( \int t^2dt=\tfrac{t^3}{3} \), \( \int \sin tdt=-\cos t \).
  2. \( \int \langle t,t^2,\sin t\rangle dt=\left\langle \tfrac{t^2}{2},\tfrac{t^3}{3},-\cos t\right\rangle+\mathbf{C} \).
29.If \( \mathbf{v}(t)=\langle 1,2t\rangle \) and \( \mathbf{r}(0)=\langle -3,5\rangle \), then \( \mathbf{r}(4) \) is:
  1. Find \( \mathbf{r}(t)=\int\langle 1,2t\rangle dt=\langle t,t^2\rangle+\mathbf{C} \).
  2. Use \( \mathbf{r}(0)=\langle -3,5\rangle \Rightarrow \mathbf{C}=\langle -3,5\rangle \).
  3. \( \mathbf{r}(4)=\langle 4-3,16+5\rangle=\langle 1,21\rangle \).
30.A constant of integration for a vector antiderivative in 2D should be written as:
  1. Each component integrates to an antiderivative with its own constant.
  2. So the constant is a vector: \( \mathbf{C}=\langle C_1,C_2\rangle \).
31.If \( \mathbf{v}(t)=\langle 2t+1,\,e^t\rangle \), then \( \int_0^1 \mathbf{v}(t)\,dt \) equals:
  1. \( \int_0^1(2t+1)dt=[t^2+t]_0^1=2 \).
  2. \( \int_0^1 e^t dt=[e^t]_0^1=e-1 \).
  3. \( \int_0^1 \langle 2t+1,e^t\rangle dt=\langle 2,e-1\rangle \).
32.If \( \mathbf{a}(t)=\langle 2,0\rangle \), \( \mathbf{v}(0)=\langle -1,3\rangle \), and \( \mathbf{r}(0)=\langle 0,0\rangle \), then \( \mathbf{r}(t) \) is:
  1. \( \mathbf{v}(t)=\int\langle 2,0\rangle dt=\langle 2t+C_1,C_2\rangle \).
  2. Use \( \mathbf{v}(0)=\langle -1,3\rangle \Rightarrow C_1=-1,\;C_2=3 \), so \( \mathbf{v}(t)=\langle 2t-1,3\rangle \).
  3. \( \mathbf{r}(t)=\int\mathbf{v}(t)dt=\langle t^2-t,3t\rangle+\mathbf{C}_r \).
  4. Use \( \mathbf{r}(0)=\langle 0,0\rangle \Rightarrow \mathbf{C}_r=\langle 0,0\rangle \).
  5. \( \mathbf{r}(t)=\langle t^2-t,3t\rangle \).
33.If \( \mathbf{r}(t)=\langle t^2-3,\;2t+5\rangle \), then the displacement from \(t=1\) to \(t=4\) is:
  1. Displacement: \( \mathbf{r}(4)-\mathbf{r}(1) \).
  2. \( \mathbf{r}(4)=\langle 16-3,8+5\rangle=\langle 13,13\rangle \).
  3. \( \mathbf{r}(1)=\langle 1-3,2+5\rangle=\langle -2,7\rangle \).
  4. \( \mathbf{r}(4)-\mathbf{r}(1)=\langle 15,6\rangle \).
34.If a student computes \( \int_a^b \mathbf{v}(t)\,dt \) and gets a vector, which is the best interpretation?
  1. The velocity integral returns \( \mathbf{r}(b)-\mathbf{r}(a) \).
  2. That is net displacement (direction + magnitude), not distance traveled.
35.Which behavior is built into these MCQs?
  1. Spec: single-select with toggle-off behavior.
  2. Clicking the same selected choice again unselects it and clears red/green + feedback.
36.After a correct selection, the feedback text must display:
  1. Spec: Correct → show \( \)Correct, Answer: X\( \).
  2. Incorrect → show Incorrect only; toggled off → blank.

Answer Key

  • 1. B
  • 2. D
  • 3. A
  • 4. C
  • 5. E
  • 6. B
  • 7. D
  • 8. A
  • 9. C
  • 10. B
  • 11. E
  • 12. D
  • 13. A
  • 14. C
  • 15. E
  • 16. B
  • 17. C
  • 18. A
  • 19. D
  • 20. B
  • 21. E
  • 22. A
  • 23. C
  • 24. B
  • 25. D
  • 26. A
  • 27. E
  • 28. B
  • 29. C
  • 30. A
  • 31. D
  • 32. B
  • 33. E
  • 34. A
  • 35. E
  • 36. C