Let's solve **Question 31** and **Question 33** step-by-step, and provide graphs for each. --- ## **Question 31** **Given:** \[ x = t, \quad y = e^{t-1}, \quad z = 2t - t^2 \] Point: \((0, 1, 0)\) ### **Step 1: Find \( t \) at the point** At the point \((0, 1, 0)\): - \(x = t = 0\) - \(y = e^{0-1} = e^{-1} = 1\) (**Correct so t = 1 not 0!**) - \(z = 2t - t^2 = 2 \cdot 0 - 0^2 = 0\) (**But this gives contradiction**) Let's check \(x\): - \(x = t\), so for the point \((0, 1, 0)\), \(t = 0\). - \(y = e^{t-1} = e^{-1}\), but that's \(1/e\), not 1. Let's check \(t = 1\): - \(x = 1\), \(y = e^{1-1} = 1\), \(z = 2 \cdot 1 - 1 = 1\) So, none of the \(t\) values match all three. But let's use \(t = 1\) for the point \((1, 1, 1)\), since the others don't fit. (There could be a typo in the question.) ### **Step 2: Find the derivative (tangent vector)** \[ \begin{align*} \frac{dx}{dt} &= 1 \\ \frac{dy}{dt} &= e^{t-1} \\ \frac{dz}{dt} &= 2 - 2t \\ \end{align*} \] At \(t = 1\): \[ \left(\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}\right) = (1, e^{0}, 2 - 2 \cdot 1) = (1, 1, 0) \] ### **Step 3: Parametric Equations for the Tangent Line** At \(t = 1\), point is \((1, 1, 1)\), and direction vector is \((1, 1, 0)\): \[ \begin{align*} x &= 1 + s \\ y &= 1 + s \\ z &= 1 \\ \end{align*} \] where \(s\) is the parameter along the tangent line. --- ## **Question 33** **Given:** \[ x = t \cos t,\quad y = t,\quad z = t \sin t \] Point: \((-\pi, \pi, 0)\) ### **Step 1: Find \( t \) at the point** - \(y = t = \pi\) - \(x = t \cos t = \pi \cos \pi = \pi \cdot (-1) = -\pi\) - \(z = t \sin t = \pi \cdot 0 = 0\) So, \(t = \pi\). ### **Step 2: Find the derivative (tangent vector)** \[ \begin{align*} \frac{dx}{dt} &= \cos t - t \sin t \\ \frac{dy}{dt} &= 1 \\ \frac{dz}{dt} &= \sin t + t \cos t \\ \end{align*} \] At \(t = \pi\): - \(\cos \pi = -1\), \(\sin \pi = 0\) \[ \frac{dx}{dt} = -1 - \pi \cdot 0 = -1 \\ \frac{dy}{dt} = 1 \\ \frac{dz}{dt} = 0 + \pi \cdot (-1) = -\pi \\ \] So, tangent vector: \((-1, 1, -\pi)\) ### **Step 3: Parametric Equations for the Tangent Line** At point \((-\pi, \pi, 0)\), direction \((-1, 1, -\pi)\): \[ \begin{align*} x &= -\pi - s \\ y &= \pi + s \\ z &= -\pi s \\ \end{align*} \] --- ## **Graphs** Below are graphs illustrating both the curve and tangent line for each question. --- ### **Python Code for Plotting (You Can Run in Jupyter Notebook or Google Colab)** ```python import numpy as np import matplotlib.pyplot as plt from mpl_toolkits.mplot3d import Axes3D # Question 31 fig = plt.figure(figsize=(12, 5)) # Curve 31 ax1 = fig.add_subplot(121, projection='3d') t = np.linspace(-2, 3, 200) x = t y = np.exp(t-1) z = 2*t - t**2 ax1.plot(x, y, z, label='Curve 31') # Tangent line at t=1 s = np.linspace(-2, 2, 20) x_tan = 1 + s y_tan = 1 + s z_tan = 1 + 0*s ax1.plot(x_tan, y_tan, z_tan, 'r--', label='Tangent at (1,1,1)') ax1.scatter([1], [1], [1], color='k') ax1.set_title('Q31: Curve & Tangent') ax1.legend() # Question 33 ax2 = fig.add_subplot(122, projection='3d') t = np.linspace(-2*np.pi, 2*np.pi, 400) x = t * np.cos(t) y = t z = t * np.sin(t) ax2.plot(x, y, z, label='Curve 33') # Tangent line at t=pi s = np.linspace(-2, 2, 20) x_tan = -np.pi - s y_tan = np.pi + s z_tan = 0 - np.pi * s ax2.plot(x_tan, y_tan, z_tan, 'r--', label='Tangent at (-π,π,0)') ax2.scatter([-np.pi], [np.pi], [0], color='k') ax2.set_title('Q33: Curve & Tangent') ax2.legend() plt.show() ``` --- ## **Final Answers** ### **Question 31:** Tangent line at \((1,1,1)\): \[ \boxed{ x = 1 + s,\quad y = 1 + s,\quad z = 1 } \] ### **Question 33:** Tangent line at \((-\pi, \pi, 0)\): \[ \boxed{ x = -\pi - s,\quad y = \pi + s,\quad z = -\pi s } \] --- If you want the graph as an image, let me know!