#Vector-Valued Functions: Your Guide to Planar Motion 🚀

Hey there! Let's dive into vector-valued functions, where we combine parametric functions, planar motion, and vectors to describe how things move in 2D. Think of it as unlocking the secrets of motion! 🔑

#Position Vector

#What is it?

The position vector, denoted as p(t), tells you exactly where a particle is at any given time, t. It's like a GPS for a moving object! We can express it in two ways:

• p(t) = x(t)i + y(t)j: Here, x(t) and y(t) are the coordinates of the particle at time t, and i and j are the unit vectors in the x and y directions. Think of it as breaking down the position into its horizontal and vertical components.
• p(t) = < x(t), y(t) >: This is just another way to write the same thing, using vector notation. It's often more compact and easier to work with.

Position vector

#Velocity Vector

#What is it?

The velocity vector, denoted as v(t), describes how fast and in what direction a particle is moving at any time t. It's like the speedometer and compass of the particle! 🧭

• v(t) = < x'(t), y'(t) >: The x and y components of the vector represent the horizontal and vertical velocities of the particle, respectively. Notice that we are taking the derivative of the position vector to get the velocity vector.

Velocity vector

#Direction Matters!

• Horizontal Velocity (x'(t)):
  • If x'(t) is positive, the particle is moving to the right ▶️.
  • If x'(t) is negative, the particle is moving to the left ◀️.
• Vertical Velocity (y'(t)):
  • If y'(t) is positive, the particle is moving upwards 🔼.
  • If y'(t) is negative, the particle is moving downwards 🔽.

#Connecting the Concepts

#Position, Velocity, and the Big Picture

• Position tells you where the particle is.
• Velocity tells you how the position is changing (speed and direction).

Understanding the relationship between position and velocity is crucial. Remember, velocity is the derivative of position. This concept pops up everywhere in calculus problems! 💡

#Final Exam Focus

#High-Priority Topics

• Position and velocity vectors: Make sure you can find them, interpret their components, and calculate their magnitudes.
• Direction of motion: Know how the signs of x'(t) and y'(t) indicate the direction of the particle.
• Speed vs. Velocity: Remember that speed is the magnitude of the velocity vector.

#Common Question Types

• Finding position and velocity at a given time. Plug in the time value into the vector-valued functions.
• Determining when a particle is moving up, down, left, or right. Look at the signs of the velocity components.
• Calculating the speed of the particle. Use the magnitude of the velocity vector.

#Last-Minute Tips

• Time Management: Don't get bogged down on one problem. If you're stuck, move on and come back to it later.
• Common Pitfalls: Be careful with signs and make sure you are using the correct formula for speed.
• Challenging Questions: Break down complex problems into smaller, more manageable steps. Visualize the motion if possible.

#Practice Questions

Practice Question

#Multiple Choice Questions

1. A particle's position is given by p(t) = < t², 3t >. What is the speed of the particle at t = 2? (A) √13 (B) 5 (C) √20 (D) 4

2. The velocity of a particle is given by v(t) = < cos(t), sin(t) >. At t = π/2, in which direction is the particle moving? (A) Up and to the right (B) Up and to the left (C) Down and to the right (D) Down and to the left

#Free Response Question

A particle moves in the xy-plane so that its position at any time t, 0 ≤ t ≤ 5, is given by x(t) = t³ - 6t² + 9t + 1 and y(t) = t² - 4t + 3. (a) Find the velocity vector v(t) of the particle at any time t. (b) Find the speed of the particle at t = 2. (c) Find the total distance traveled by the particle along the x-axis from t = 0 to t = 3. ### Scoring Rubric: (a) Velocity Vector (2 points)

• 1 point: x'(t) = 3t² - 12t + 9
• 1 point: y'(t) = 2t - 4
• v(t) = < 3t² - 12t + 9, 2t - 4 >

(b) Speed at t=2 (2 points)

• 1 point: v(2) = < 3(2)² - 12(2) + 9, 2(2) - 4 > = < -3, 0 >
• 1 point: speed = √((-3)² + 0²) = 3

(c) Total Distance (3 points)

• 1 point: Find when x'(t) = 0: 3t² - 12t + 9 = 0 => t = 1, 3
• 1 point: Calculate the distance from t=0 to t=1: |x(1) - x(0)| = |5 - 1| = 4
• 1 point: Calculate the distance from t=1 to t=3: |x(3) - x(1)| = |1 - 5| = 4
• Total distance: 4 + 4 = 8

Alright, you've got this! Remember, vector-valued functions are all about understanding motion in a clear, organized way. Keep practicing, and you'll be ready to ace that exam! 🎉