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9.4 - Defining and Differentiating Vector-Valued Functions

Introduction to Vectors

A vector is a quantity that has both magnitude, meaning size or length, and direction.
Vectors are drawn as directed line segments, with an arrow showing the direction.
9.4 - Defining and Differentiating Vector-Valued Functions
Key properties: - Initial point: where the vector starts - Terminal point: where the vector ends, at the arrow tip - Direction: shown by the arrow

When Are Vectors Equal?

Two vectors are equal if and only if they have:
  1. The same magnitude, meaning length
  1. The same direction
Important: position doesn’t matter. Vectors sitting at different locations can still be equal.

Vector Components

Every vector can be broken into horizontal and vertical components.
Component form:
⟨x,y⟩\langle x, y \rangle
where: - xx = the horizontal component, the change in the xx-direction - yy = the vertical component, the change in the yy-direction
Note: the angle brackets ⟨ ⟩\langle \, \rangle are what distinguish vector notation from coordinates (x,y)(x, y).

Finding Component Form

Given an initial point (x1,y1)(x_1, y_1) and a terminal point (x2,y2)(x_2, y_2):
Component form=⟨x2−x1, y2−y1⟩\boxed{\text{Component form} = \langle x_2 - x_1, \, y_2 - y_1 \rangle}

Example 1: Component Form

Given: a vector with initial point (1,2)(1, 2) and terminal point (5,4)(5, 4).
⟨5−1, 4−2⟩=⟨4,2⟩\langle 5 - 1, \, 4 - 2 \rangle = \langle 4, 2 \rangle
Interpretation: the vector moves 4 units right and 2 units up.

Magnitude of a Vector

The magnitude, or length, of a vector is written with double bars: ∥r∥\|r\|
Formula, straight from the Pythagorean theorem:
∥r∥=x2+y2\boxed{\|r\| = \sqrt{x^2 + y^2}}
where xx and yy are the components of the vector.

Example 1, Continued: Magnitude

Given: the component form ⟨4,2⟩\langle 4, 2 \rangle.
∥r∥=42+22=16+4=20\|r\| = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20}
Simplified:
∥r∥=4⋅5=25\|r\| = \sqrt{4 \cdot 5} = 2\sqrt{5}
Interpretation: the vector has a length of 252\sqrt{5} units.

Vector-Valued Functions

A vector-valued function is a function whose output is a vector that depends on a parameter, usually tt:
r(t)=⟨f(t),g(t)⟩\boxed{r(t) = \langle f(t), g(t) \rangle}
where: - f(t)f(t) = the horizontal component, a function of tt - g(t)g(t) = the vertical component, a function of tt - tt = the parameter, which often represents time
Connection to parametric equations: these are very similar. Instead of writing x=f(t)x = f(t) and y=g(t)y = g(t) separately, we write r(t)=⟨f(t),g(t)⟩r(t) = \langle f(t), g(t) \rangle.

Derivative of a Vector-Valued Function

To find the derivative of a vector-valued function, differentiate each component separately:
r′(t)=⟨ddt[f(t)], ddt[g(t)]⟩\boxed{r'(t) = \left\langle \frac{d}{dt}[f(t)], \, \frac{d}{dt}[g(t)] \right\rangle}
That’s it. Take the derivative of the xx-component and the derivative of the yy-component.

Example 2: Basic Derivative

Given: r(t)=⟨2t2+4t, 3t3−4t⟩r(t) = \langle 2t^2 + 4t, \, 3t^3 - 4t \rangle. Find r′(t)r'(t).
Differentiate each component:
ddt[2t2+4t]=4t+4\frac{d}{dt}[2t^2 + 4t] = 4t + 4ddt[3t3−4t]=9t2−4\frac{d}{dt}[3t^3 - 4t] = 9t^2 - 4
Answer:
r′(t)=⟨4t+4, 9t2−4⟩r'(t) = \langle 4t + 4, \, 9t^2 - 4 \rangle

Properties of Vector Derivatives

These rules mirror the ones for regular derivatives.
Constant multiple:
ddt[c⋅r(t)]=c⋅r′(t)\frac{d}{dt}[c \cdot r(t)] = c \cdot r'(t)
Sum and difference:
ddt[r(t)±s(t)]=r′(t)±s′(t)\frac{d}{dt}[r(t) \pm s(t)] = r'(t) \pm s'(t)
Product rule:
ddt[u(t)⋅r(t)]=u′(t)⋅r(t)+u(t)⋅r′(t)\frac{d}{dt}[u(t) \cdot r(t)] = u'(t) \cdot r(t) + u(t) \cdot r'(t)
Chain rule:
ddt[r(u(t))]=r′(u(t))⋅u′(t)\frac{d}{dt}[r(u(t))] = r'(u(t)) \cdot u'(t)

Example 3: Chain Rule with Vectors

Given: r(t)=⟨t3,2⟩r(t) = \langle t^3, 2 \rangle. Find the derivative of r(2t)r(2t).
Step 1: Find r′(t)r'(t)
r′(t)=⟨3t2,0⟩r'(t) = \langle 3t^2, 0 \rangle
Step 2: Find r′(2t)r'(2t)
Substitute 2t2t for tt:
r′(2t)=⟨3(2t)2,0⟩=⟨12t2,0⟩r'(2t) = \langle 3(2t)^2, 0 \rangle = \langle 12t^2, 0 \rangle
Step 3: Apply the chain rule
Multiply by the derivative of the inside function, since ddt[2t]=2\frac{d}{dt}[2t] = 2:
ddt[r(2t)]=r′(2t)⋅2\frac{d}{dt}[r(2t)] = r'(2t) \cdot 2
For vectors, multiply each component by 2:
=⟨12t2,0⟩⋅2=⟨24t2,0⟩= \langle 12t^2, 0 \rangle \cdot 2 = \langle 24t^2, 0 \rangle
Answer: ddt[r(2t)]=⟨24t2,0⟩\frac{d}{dt}[r(2t)] = \langle 24t^2, 0 \rangle

Scalar Multiplication

When you multiply a vector by a scalar, meaning a regular number, multiply each component by that scalar:
c⋅⟨a,b⟩=⟨ca,cb⟩c \cdot \langle a, b \rangle = \langle ca, cb \rangle
Example:
2⋅⟨3,−5⟩=⟨6,−10⟩2 \cdot \langle 3, -5 \rangle = \langle 6, -10 \rangle

Finding the Slope of a Parametric Path

When a vector-valued function describes a particle’s position, we can find the slope of its path at any time.
Key insight: the components are just parametric equations.
If r(t)=⟨f(t),g(t)⟩r(t) = \langle f(t), g(t) \rangle, then: - x=f(t)x = f(t), the horizontal position - y=g(t)y = g(t), the vertical position
Slope formula:
dydx=dy/dtdx/dt=g′(t)f′(t)\boxed{\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}}
This is exactly the same formula from parametric equations.

Example 4: Slope of a Particle’s Path

Given: r(t)=⟨t2,sin⁡t⟩r(t) = \langle t^2, \sin t \rangle. Find the slope of the particle’s path at t=π2t = \frac{\pi}{2}.
Step 1: Identify the components
  • x=t2x = t^2, the horizontal component
  • y=sin⁡ty = \sin t, the vertical component
Step 2: Find the derivatives
dxdt=2tdydt=cos⁡t\frac{dx}{dt} = 2t \qquad \frac{dy}{dt} = \cos t
Step 3: Apply the slope formula
dydx=dy/dtdx/dt=cos⁡t2t\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\cos t}{2t}
Step 4: Evaluate at t=π2t = \frac{\pi}{2}
dydx∣t=π/2=cos⁡(π/2)2(π/2)=0π=0\frac{dy}{dx}\bigg|_{t=\pi/2} = \frac{\cos(\pi/2)}{2(\pi/2)} = \frac{0}{\pi} = 0
Answer: the slope at t=π2t = \frac{\pi}{2} is 0, meaning a horizontal tangent line.

Example 5: Slope with Trigonometric Functions

Given: r(t)=⟨t2,sin⁡t⟩r(t) = \langle t^2, \sin t \rangle. Find the slope at t=3π4t = \frac{3\pi}{4}.
Step 1: Set up the slope formula
dydx=cos⁡t2t\frac{dy}{dx} = \frac{\cos t}{2t}
Step 2: Evaluate at t=3π4t = \frac{3\pi}{4}
Numerator:
cos⁡(3π4)=−22\cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}
Denominator:
2⋅3π4=3π22 \cdot \frac{3\pi}{4} = \frac{3\pi}{2}
Step 3: Divide
dydx∣t=3π/4=−2/23π/2=−22⋅23π=−23π\frac{dy}{dx}\bigg|_{t=3\pi/4} = \frac{-\sqrt{2}/2}{3\pi/2} = -\frac{\sqrt{2}}{2} \cdot \frac{2}{3\pi} = -\frac{\sqrt{2}}{3\pi}
Answer: the slope is −23π-\frac{\sqrt{2}}{3\pi}, which is about −0.15-0.15.

Velocity and Position

In physics applications: - r(t)r(t) represents position at time tt - r′(t)r'(t) represents velocity at time tt
Interpretation: the derivative vector r′(t)r'(t) points in the direction of motion, and its magnitude is the speed.

Summary: Finding Slopes vs. Finding Derivatives

To find the derivative vector r′(t)r'(t):

r(t)=⟨f(t),g(t)⟩⇒r′(t)=⟨f′(t),g′(t)⟩r(t) = \langle f(t), g(t) \rangle \quad \Rightarrow \quad r'(t) = \langle f'(t), g'(t) \rangle
This gives you a vector.

To find the slope dydx\frac{dy}{dx}:

dydx=dy/dtdx/dt=g′(t)f′(t)\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}
This gives you a number, the slope of the tangent line.

Common Mistakes to Avoid

❌ Confusing component form with coordinates
Vectors use angle brackets: ⟨3,4⟩\langle 3, 4 \rangle. Points use parentheses: (3,4)(3, 4).
❌ Forgetting the chain rule
When finding ddt[r(2t)]\frac{d}{dt}[r(2t)], don’t forget to multiply by 2.
❌ Flipping the slope formula
Wrong: dydx=dx/dtdy/dt\frac{dy}{dx} = \frac{dx/dt}{dy/dt}
Correct: dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}
❌ Mixing up position and velocity
r(t)r(t) is position, and r′(t)r'(t) is velocity.

Problem-Solving Strategy

To find component form: subtract the initial point from the terminal point: ⟨x2−x1,y2−y1⟩\langle x_2 - x_1, y_2 - y_1 \rangle
To find magnitude: use ∥r∥=x2+y2\|r\| = \sqrt{x^2 + y^2}
To find the derivative: differentiate each component separately, and keep the vector bracket notation
To find the slope at a specific time: 1. Identify x=f(t)x = f(t) and y=g(t)y = g(t) 2. Find dxdt\frac{dx}{dt} and dydt\frac{dy}{dt} 3. Compute dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt} 4. Evaluate at the given tt-value

Key Takeaways

  • Vector: has both magnitude (length) and direction
  • Component form: ⟨x,y⟩\langle x, y \rangle represents the horizontal and vertical components
  • Magnitude formula: ∥r∥=x2+y2\|r\| = \sqrt{x^2 + y^2}
  • Vector-valued function: r(t)=⟨f(t),g(t)⟩r(t) = \langle f(t), g(t) \rangle
  • Derivative: differentiate each component separately, so r′(t)=⟨f′(t),g′(t)⟩r'(t) = \langle f'(t), g'(t) \rangle
  • The chain rule still applies. Don’t forget to multiply by the derivative of the inside function
  • Slope formula: dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, the same as parametric equations
  • Scalar multiplication: multiply each component by the scalar
  • Applications: vector-valued functions model particle motion and paths
  • Notation matters. Use angle brackets ⟨ ⟩\langle \, \rangle for vectors, not parentheses

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