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 (tt) or position.
  • General Form: A vector function in 3D space: V⃗(t)=x(t)i^+y(t)j^+z(t)k^\vec{V}(t) = x(t) \hat{i} + y(t) \hat{j} + z(t) \hat{k}
    • x(t)x(t), y(t)y(t), and z(t)z(t) are Cartesian components.
    • For every input value of tt, the function returns a constant vector.
Example

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

  • When t=0t = 0, V⃗(0)=0i^+1j^+1k^\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 V⃗(t)=u(t)i^+v(t)j^+w(t)k^\vec{V}(t) = u(t) \hat{i} + v(t) \hat{j} + w(t) \hat{k}, then

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

Integration

∫V⃗(t)dt=∫u(t)dti^+∫v(t)dtj^+∫w(t)dtk^\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 V⃗(t)=t2i^+2tj^+1k^\vec{V}(t) = t^2 \hat{i} + 2t \hat{j} + 1 \hat{k}, then

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

  • Integral: ∫V⃗(t)dt=t33i^+t2j^+tk^+C⃗\int \vec{V}(t) dt = \frac{t^3}{3} \hat{i} + t^2 \hat{j} + t \hat{k} + \vec{C}

    • Where C⃗=c<em>1i^+c</em>2j^+c3k^\vec{C} = c<em>1 \hat{i} + c</em>2 \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 tt.
  • 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 r⃗(t)\vec{r}(t) is the position vector, then:
    • Velocity: r˙⃗(t)=dr⃗dt\vec{\dot{r}}(t) = \frac{d \vec{r}}{dt}
    • Acceleration: r¨⃗(t)=d2r⃗dt2\vec{\ddot{r}}(t) = \frac{d^2 \vec{r}}{dt^2}

Particles Moving in a Straight Line (3D)

  • A linear path is described by: r⃗(t)=s⃗+u⃗t\vec{r}(t) = \vec{s} + \vec{u}t, where s⃗\vec{s} and u⃗\vec{u} are constant vectors.
  • This line passes through the point s⃗\vec{s} and the point u⃗+s⃗\vec{u} + \vec{s}
  • The equation of a line that passes through points A and B is: r⃗(t)=A⃗+(B⃗−A⃗)t\vec{r}(t) = \vec{A} + (\vec{B} - \vec{A})t
Velocity and Acceleration
  • Given r⃗(t)=s⃗+u⃗t\vec{r}(t) = \vec{s} + \vec{u}t, the velocity is v⃗(t)=u⃗\vec{v}(t) = \vec{u}
  • The acceleration is a⃗(t)=0\vec{a}(t) = 0
Example
  • If s⃗=<1,1,1>\vec{s} = <1, 1, 1> and u⃗=<1,0,1>\vec{u} = <1, 0, 1>, the straight line through s⃗\vec{s} and u⃗\vec{u} is:
    r⃗(t)=<1,1,1>+t(<1,0,1>−<1,1,1>)=<1,1,1>+t<0,−1,0>\vec{r}(t) = <1, 1, 1> + t(<1, 0, 1> - <1, 1, 1>) = <1, 1, 1> + t<0, -1, 0>
  • Velocity: v⃗(t)=<0,−1,0>\vec{v}(t) = <0, -1, 0>
  • Acceleration: a⃗(t)=<0,0,0>\vec{a}(t) = <0, 0, 0>

Parabolic Paths

  • Described by: r⃗(t)=s⃗+u⃗t+12a⃗t2\vec{r}(t) = \vec{s} + \vec{u}t + \frac{1}{2} \vec{a}t^2, where s⃗\vec{s}, u⃗\vec{u}, and a⃗\vec{a} are constant vectors.
Velocity and Acceleration
  • Velocity: v⃗(t)=u⃗+a⃗t\vec{v}(t) = \vec{u} + \vec{a}t
  • Acceleration: a⃗(t)=a⃗\vec{a}(t) = \vec{a}
Example
  • Given a particle moving along a parabolic path, where s⃗=<0,0,0>\vec{s} = <0,0,0>, u⃗=<0,1,0>\vec{u} = <0, 1, 0>, and a⃗=<1,1,1>\vec{a} = <1, 1, 1>, the velocity is:
    v⃗(t)=<0,1,0>+<1,1,1>t\vec{v}(t) = <0, 1, 0> + <1, 1, 1>t
  • The acceleration is: a⃗(t)=<1,1,1>\vec{a}(t) = <1, 1, 1>
Projectile Motion under Gravity
  • If s⃗=0\vec{s} = 0, u⃗=u<em>1i^+u</em>2j^\vec{u} = u<em>1 \hat{i} + u</em>2 \hat{j}, and a⃗=−gj^\vec{a} = -g \hat{j} (where j^\hat{j} points vertically upwards), then the position vector is:
    r⃗(t)=u<em>1ti^+(u</em>2t−12gt2)j^\vec{r}(t) = u<em>1t \hat{i} + (u</em>2t - \frac{1}{2}gt^2) \hat{j}
  • The velocity vector is: v⃗(t)=u<em>1i^+(u</em>2−gt)j^\vec{v}(t) = u<em>1 \hat{i} + (u</em>2 - gt) \hat{j}
  • The acceleration vector is: a⃗(t)=−gj^\vec{a}(t) = -g \hat{j}
Parabola
  • From x(t)=u<em>1tx(t) = u<em>1t and y(t)=u</em>2t−12gt2y(t) = u</em>2t - \frac{1}{2}gt^2, we can derive the parabolic equation:
    y=(u<em>2u</em>1)x−(g2u12)x2y = (\frac{u<em>2}{u</em>1})x - (\frac{g}{2u_1^2})x^2

Circular Paths

  • Described using cosine and sine functions.
  • Position vector: r⃗(t)=rcos⁡(ωt)i^+rsin⁡(ωt)j^\vec{r}(t) = r \cos(\omega t) \hat{i} + r \sin(\omega t) \hat{j}
    • rr is the radius of the circle.
    • ω\omega is the angular frequency.
    • ωt\omega t is the angular velocity.
Velocity
  • Velocity vector: v⃗(t)=−rωsin⁡(ωt)i^+rωcos⁡(ωt)j^\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: a⃗(t)=−rω2cos⁡(ωt)i^−rω2sin⁡(ωt)j^=−ω2r⃗(t)\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
  • x¨(t)=−ω2x(t)\ddot{x}(t) = -\omega^2 x(t)
  • y¨(t)=−ω2y(t)\ddot{y}(t) = -\omega^2 y(t)
  • These are differential equations with solutions x(t)=rcos⁡(ωt)x(t) = r \cos(\omega t) and y(t)=rsin⁡(ωt)y(t) = r \sin(\omega t)

Simple Harmonic Motion

  • Described by the differential equation: d2ydx2+y=0\frac{d^2y}{dx^2} + y = 0
  • General solution: y(t)=Asin⁡(x)+Bcos⁡(x)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)=0y(0) = 0 and y′(0)=1y'(0) = 1
  • Derivative: y′(x)=Acos⁡(x)−Bsin⁡(x)y'(x) = A \cos(x) - B \sin(x)
  • Applying y(0)=0y(0) = 0: 0=Asin⁡(0)+Bcos⁡(0)=0+B(1)0 = A \sin(0) + B \cos(0) = 0 + B(1)
    • Therefore, B=0B = 0
  • The solution becomes: y(x)=Asin⁡(x)y(x) = A \sin(x)
  • Derivative: y′(x)=Acos⁡(x)y'(x) = A \cos(x)
  • Applying y′(0)=1y'(0) = 1: 1=Acos⁡(0)=A(1)1 = A \cos(0) = A(1)
    • Therefore, A=1A = 1
  • Final solution: y(x)=sin⁡(x)y(x) = \sin(x)