Lecture 34 Notes: Vector Functions

Vector Functions: Velocity, Acceleration, Particle Paths, and Simple Harmonic Motion

This lecture primarily focuses on vector functions with a smaller portion about differential equations.

Vector Functions

• Constant Vectors: Previously studied in Lectures 13 and 16.
• Vector Quantities: Can depend on parameters like time (t) or position.
• General Form: A vector function in 3D space: \vec{V}(t) = x(t) \hat{i} + y(t) \hat{j} + z(t) \hat{k}
  • x(t), y(t), and z(t) are Cartesian components.
  • For every input value of t, the function returns a constant vector.

Example

\vec{V}(t) = t \hat{i} + (t+1) \hat{j} + 1 \hat{k}

• When t = 0, \vec{V}(0) = 0 \hat{i} + 1 \hat{j} + 1 \hat{k} , which is a constant vector.

Differentiation and Integration of Vector Functions

• Performed component-wise.

Differentiation

If \vec{V}(t) = u(t) \hat{i} + v(t) \hat{j} + w(t) \hat{k} , then

\frac{d \vec{V}(t)}{dt} = \vec{V}'(t) = u'(t) \hat{i} + v'(t) \hat{j} + w'(t) \hat{k}

Integration

\int \vec{V}(t) dt = \int u(t) dt \hat{i} + \int v(t) dt \hat{j} + \int w(t) dt \hat{k}

Example

Given \vec{V}(t) = t^2 \hat{i} + 2t \hat{j} + 1 \hat{k} , then

• Derivative: \vec{V}'(t) = 2t \hat{i} + 2 \hat{j} + 0 \hat{k}

• Integral: \int \vec{V}(t) dt = \frac{t^3}{3} \hat{i} + t^2 \hat{j} + t \hat{k} + \vec{C}

  • Where \vec{C} = c1 \hat{i} + c2 \hat{j} + c_3 \hat{k} is a constant vector of integration.

Particle Movement and Vector Functions

• A vector function can describe the position of a particle at any time t.
• The derivative of the position vector gives the velocity vector.
• The second derivative of the position vector (or the derivative of the velocity vector) gives the acceleration vector.

Notation

• If \vec{r}(t) is the position vector, then:
  • Velocity: \vec{\dot{r}}(t) = \frac{d \vec{r}}{dt}
  • Acceleration: \vec{\ddot{r}}(t) = \frac{d^2 \vec{r}}{dt^2}

Particles Moving in a Straight Line (3D)

• A linear path is described by: \vec{r}(t) = \vec{s} + \vec{u}t , where \vec{s} and \vec{u} are constant vectors.
• This line passes through the point \vec{s} and the point \vec{u} + \vec{s}
• The equation of a line that passes through points A and B is: \vec{r}(t) = \vec{A} + (\vec{B} - \vec{A})t

Velocity and Acceleration

• Given \vec{r}(t) = \vec{s} + \vec{u}t , the velocity is \vec{v}(t) = \vec{u}
• The acceleration is \vec{a}(t) = 0

Example

• If \vec{s} =
• Velocity: \vec{v}(t) =
• Acceleration: \vec{a}(t) =

Parabolic Paths

• Described by: \vec{r}(t) = \vec{s} + \vec{u}t + \frac{1}{2} \vec{a}t^2 , where \vec{s} , \vec{u} , and \vec{a} are constant vectors.

Velocity and Acceleration

• Velocity: \vec{v}(t) = \vec{u} + \vec{a}t
• Acceleration: \vec{a}(t) = \vec{a}

Example

• Given a particle moving along a parabolic path, where \vec{s} =
• The acceleration is: \vec{a}(t) =

Projectile Motion under Gravity

• If \vec{s} = 0 , \vec{u} = u1 \hat{i} + u2 \hat{j} , and \vec{a} = -g \hat{j} (where \hat{j} points vertically upwards), then the position vector is:
  \vec{r}(t) = u1t \hat{i} + (u2t - \frac{1}{2}gt^2) \hat{j}
• The velocity vector is: \vec{v}(t) = u1 \hat{i} + (u2 - gt) \hat{j}
• The acceleration vector is: \vec{a}(t) = -g \hat{j}

Parabola

• From x(t) = u1t and y(t) = u2t - \frac{1}{2}gt^2 , we can derive the parabolic equation:
  y = (\frac{u2}{u1})x - (\frac{g}{2u_1^2})x^2

Circular Paths

• Described using cosine and sine functions.
• Position vector: \vec{r}(t) = r \cos(\omega t) \hat{i} + r \sin(\omega t) \hat{j}
  • r is the radius of the circle.
  • \omega is the angular frequency.
  • \omega t is the angular velocity.

Velocity

• Velocity vector: \vec{v}(t) = -r \omega \sin(\omega t) \hat{i} + r \omega \cos(\omega t) \hat{j}
  • This vector is tangential to the circular path.

Acceleration

• Acceleration vector: \vec{a}(t) = -r \omega^2 \cos(\omega t) \hat{i} - r \omega^2 \sin(\omega t) \hat{j} = -\omega^2 \vec{r}(t)
  • The acceleration points towards the center of the circle (centripetal acceleration).
  • Relates to centripetal force in physics.

Cartesian Components

• \ddot{x}(t) = -\omega^2 x(t)
• \ddot{y}(t) = -\omega^2 y(t)
• These are differential equations with solutions x(t) = r \cos(\omega t) and y(t) = r \sin(\omega t)

Simple Harmonic Motion

• Described by the differential equation: \frac{d^2y}{dx^2} + y = 0
• General solution: y(t) = A \sin(x) + B \cos(x) , where A and B are constants.
• Two initial conditions are required to find a particular solution.

Example

• Initial conditions: y(0) = 0 and y'(0) = 1
• Derivative: y'(x) = A \cos(x) - B \sin(x)
• Applying y(0) = 0 : 0 = A \sin(0) + B \cos(0) = 0 + B(1)
  • Therefore, B = 0
• The solution becomes: y(x) = A \sin(x)
• Derivative: y'(x) = A \cos(x)
• Applying y'(0) = 1 : 1 = A \cos(0) = A(1)
  • Therefore, A = 1
• Final solution: y(x) = \sin(x)