MA 261 - Lesson 7: Motion in Space (14.3)

Review Example

Review: Derivative of Vector-Valued Function

Given \(\vec{r}(t) = \langle t\sin t, t\cos t, t \rangle\), what does \(\vec{r}'\left(\frac{\pi}{2}\right)\) represent?

Solution:

Recall that for a scalar function \(f(x)\), \(f'(a)\) represents the slope of the tangent line at \((a, f(a))\).

For a vector-valued function, \(\vec{r}'(a)\) represents the direction vector of the tangent line.

Computing the derivative:

\[ \vec{r}'(t) = \left\langle \frac{d}{dt}(t\sin t), \frac{d}{dt}(t\cos t), \frac{d}{dt}(t) \right\rangle \] \[ = \langle \sin t + t\cos t, \cos t - t\sin t, 1 \rangle \]

Evaluating at \(t = \frac{\pi}{2}\):

\[ \vec{r}'\left(\frac{\pi}{2}\right) = \left\langle \sin\frac{\pi}{2} + \frac{\pi}{2}\cos\frac{\pi}{2}, \cos\frac{\pi}{2} - \frac{\pi}{2}\sin\frac{\pi}{2}, 1 \right\rangle \] \[ = \left\langle 1 + \frac{\pi}{2}(0), 0 - \frac{\pi}{2}(1), 1 \right\rangle = \left\langle 1, -\frac{\pi}{2}, 1 \right\rangle \]

Graph Description: A three-dimensional curve with a tangent line shown at a specific point. The curve spirals around as it ascends, and the purple tangent line indicates the direction of motion at that instant.

Finding the Equation of a Tangent Line

Example: Tangent Line at a Specific Point

Find the equation of the tangent line to \(\vec{r}(t) = \langle t\sin t, t\cos t, t \rangle\) at \(\left(\frac{\pi}{2}, 0, \frac{\pi}{2}\right)\).

Solution:

We need two components for the equation of a line:

A point on the line: \(\left(\frac{\pi}{2}, 0, \frac{\pi}{2}\right)\)
A direction vector: \(\vec{r}'\left(\frac{\pi}{2}\right)\)

First, verify that the point corresponds to \(t = \frac{\pi}{2}\):

\[ \vec{r}\left(\frac{\pi}{2}\right) = \left\langle \frac{\pi}{2}\sin\frac{\pi}{2}, \frac{\pi}{2}\cos\frac{\pi}{2}, \frac{\pi}{2} \right\rangle = \left\langle \frac{\pi}{2}, 0, \frac{\pi}{2} \right\rangle \checkmark \]

From the previous problem, we found:

\[ \vec{r}'\left(\frac{\pi}{2}\right) = \left\langle 1, -\frac{\pi}{2}, 1 \right\rangle \]

The equation of the tangent line is:

\[ \langle x, y, z \rangle = t\left\langle 1, -\frac{\pi}{2}, 1 \right\rangle + \left\langle \frac{\pi}{2}, 0, \frac{\pi}{2} \right\rangle \]

Particle Moving in 3D/2D

Definition: Position, Velocity, and Acceleration

For a particle moving in space:

\(\vec{r}(t)\) = position vector
\(\vec{v}(t) = \vec{r}'(t)\) = velocity vector
\(\vec{a}(t) = \vec{v}'(t) = \vec{r}''(t)\) = acceleration vector

Graph Description: A three-dimensional coordinate system showing a curved trajectory. The position vector extends from the origin to a point on the curve, while arrows along the curve indicate the direction of motion (velocity vectors).

Computing Velocity and Acceleration Vectors

Example: Finding Velocity and Acceleration

Given \(\vec{r}(t) = \langle e^{2t}, t^2, \sin(\pi t) \rangle\) is the position vector.

Solution:

Step 1: Find the velocity vector \(\vec{v}(t) = \vec{r}'(t)\):

\[ \vec{v}(t) = \vec{r}'(t) = \left\langle \frac{d}{dt}(e^{2t}), \frac{d}{dt}(t^2), \frac{d}{dt}(\sin(\pi t)) \right\rangle \] \[ = \langle 2e^{2t}, 2t, \pi\cos(\pi t) \rangle \]

At \(t = 0\):

\[ \vec{v}(0) = \langle 2e^0, 2(0), \pi\cos(0) \rangle = \langle 2, 0, \pi \rangle \]

At \(t = \frac{1}{2}\):

\[ \vec{v}\left(\frac{1}{2}\right) = \left\langle 2e^1, 2\left(\frac{1}{2}\right), \pi\cos\left(\frac{\pi}{2}\right) \right\rangle = \langle 2e, 1, 0 \rangle \]

Step 2: Find the acceleration vector \(\vec{a}(t) = \vec{v}'(t) = \vec{r}''(t)\):

\[ \vec{a}(t) = \vec{v}'(t) = \left\langle \frac{d}{dt}(2e^{2t}), \frac{d}{dt}(2t), \frac{d}{dt}(\pi\cos(\pi t)) \right\rangle \] \[ = \langle 4e^{2t}, 2, -\pi^2\sin(\pi t) \rangle \]

At \(t = 0\):

\[ \vec{a}(0) = \langle 4e^0, 2, -\pi^2\sin(0) \rangle = \langle 4, 2, 0 \rangle \]

At \(t = \frac{1}{2}\):

\[ \vec{a}\left(\frac{1}{2}\right) = \left\langle 4e^1, 2, -\pi^2\sin\left(\frac{\pi}{2}\right) \right\rangle = \langle 4e, 2, -\pi^2 \rangle \]

Circular Motion

Example: Uniform Circular Motion

Consider \(\vec{r}(t) = \langle \cos t, \sin t \rangle\) for \(0 \leq t \leq 2\pi\) (2D motion).

Solution:

The velocity vector is:

\[ \vec{v}(t) = \vec{r}'(t) = \langle -\sin t, \cos t \rangle \]

At specific times:

\(\vec{r}(0) = \langle 1, 0 \rangle\), \(\vec{v}(0) = \langle 0, 1 \rangle\)
\(\vec{r}\left(\frac{\pi}{4}\right) = \left\langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle\), \(\vec{v}\left(\frac{\pi}{4}\right) = \left\langle -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle\)
\(\vec{r}\left(\frac{\pi}{2}\right) = \langle 0, 1 \rangle\), \(\vec{v}\left(\frac{\pi}{2}\right) = \langle -1, 0 \rangle\)
\(\vec{r}(\pi) = \langle -1, 0 \rangle\), \(\vec{v}(\pi) = \langle 0, -1 \rangle\)

Graph Description: A unit circle in the xy-plane with position vectors drawn from the origin to points on the circle, and velocity vectors shown tangent to the circle at those points. The velocity vectors are perpendicular to the position vectors, indicating circular motion.

Important Property: This represents circular motion because \(x^2(t) + y^2(t) = \cos^2 t + \sin^2 t = 1\) for all \(t\).

Key observation: \(\vec{v}(t) \perp \vec{r}(t)\) for all \(t\), meaning \(\vec{v}(t) \cdot \vec{r}(t) = 0\) for all \(t\).

General Circular Motion

Theorem: Characterization of Circular Motion

If \(\vec{r}(t) = \langle x(t), y(t) \rangle\) is in circular motion, then:

Eliminating \(t\) yields the equation \(x^2 + y^2 = R^2\) (a circle of radius \(R\))
\(\vec{r}(t) \perp \vec{v}(t)\) for all values of \(t\)

Example 1: Standard Circular Motion

For \(\vec{r}(t) = \langle \cos(t^2), \sin(t^2) \rangle\):

The velocity is:

\[ \vec{v}(t) = \langle -2t\sin(t^2), 2t\cos(t^2) \rangle \]

We can verify: \(x^2 + y^2 = \cos^2(t^2) + \sin^2(t^2) = 1\), and \(\vec{r}(t) \cdot \vec{v}(t) = 0\).

Example 2: Circular Motion with Different Radius

For \(\vec{r}(t) = \langle R\cos(f(t)), R\sin(f(t)) \rangle\) where \(f(t)\) is any function:

The velocity is:

\[ \vec{v}(t) = \langle -Rf'(t)\sin(f(t)), Rf'(t)\cos(f(t)) \rangle \]

We verify: \(x^2 + y^2 = R^2\cos^2(f(t)) + R^2\sin^2(f(t)) = R^2\), and \(\vec{r}(t) \cdot \vec{v}(t) = 0\).

Think-Pair-Share (iClicker)

Question: Does This Describe Circular Motion?

Given \(\vec{r}(t) = \langle 5\cos t, \sin t \rangle\) for \(0 \leq t \leq 2\pi\), does this describe circular motion?

Solution:

To check, we compute \(\vec{v}(t)\) and test if \(\vec{r}(t) \cdot \vec{v}(t) = 0\):

\[ \vec{v}(t) = \langle -5\sin t, \cos t \rangle \] \[ \vec{r}(t) \cdot \vec{v}(t) = (5\cos t)(-5\sin t) + (\sin t)(\cos t) \] \[ = -25\sin t\cos t + \sin t\cos t = -24\sin t\cos t \neq 0 \]

Answer: No, this is NOT circular motion.

To identify the curve, we eliminate the parameter \(t\):

From the parametric equations \(x = 5\cos t\) and \(y = \sin t\), we have:

\[ \frac{x}{5} = \cos t \quad \text{and} \quad y = \sin t \] \[ \left(\frac{x}{5}\right)^2 + y^2 = \cos^2 t + \sin^2 t = 1 \] \[ \frac{x^2}{25} + y^2 = 1 \]

This is the equation of an ellipse with semi-major axis 5 (along the x-axis) and semi-minor axis 1 (along the y-axis).

Graph Description: An ellipse centered at the origin with horizontal extent from (-5, 0) to (5, 0) and vertical extent from (0, -1) to (0, 1). The curve is labeled with key points at (5, 0), (0, 1), (-5, 0), and (0, -1).

Review: Integration of Vector-Valued Functions

Integration of vector-valued functions is performed component-wise, with each component producing a constant of integration.

Definition: If \(\vec{r}(t) = \langle f(t), g(t), h(t) \rangle\), then:

\[ \int \vec{r}(t) \, dt = \left\langle \int f(t) \, dt, \int g(t) \, dt, \int h(t) \, dt \right\rangle + \vec{C} \]

where \(\vec{C}\) is a constant vector of integration.

Example 1: Indefinite Integral

Find \(\int \vec{r}(t) \, dt\) where \(\vec{r}(t) = \langle t^2, e^t, \sqrt{t} \rangle\).

Solution:

\[ \int \langle t^2, e^t, \sqrt{t} \rangle \, dt = \left\langle \int t^2 \, dt, \int e^t \, dt, \int t^{1/2} \, dt \right\rangle + \vec{C} \] \[ = \left\langle \frac{t^3}{3}, e^t, \frac{2t^{3/2}}{3} \right\rangle + \vec{C} \]

Example 2: Definite Integral

Evaluate \(\int_0^1 \vec{r}(t) \, dt\) where \(\vec{r}(t) = \langle t^2, e^t, \sqrt{t} \rangle\).

Solution:

\[ \int_0^1 \langle t^2, e^t, \sqrt{t} \rangle \, dt = \left\langle \frac{t^3}{3}, e^t, \frac{2t^{3/2}}{3} \right\rangle \Big|_0^1 \] \[ = \left\langle \frac{1^3}{3}, e^1, \frac{2(1)^{3/2}}{3} \right\rangle - \left\langle \frac{0^3}{3}, e^0, \frac{2(0)^{3/2}}{3} \right\rangle \] \[ = \left\langle \frac{1}{3}, e, \frac{2}{3} \right\rangle - \langle 0, 1, 0 \rangle = \left\langle \frac{1}{3}, e-1, \frac{2}{3} \right\rangle \]

Application: Projectile Motion

Example: Ball Launched from Height

A ball 49 m above the ground is launched with initial velocity \(\vec{v}(0) = \langle 2, 4, 14.7 \rangle\) m/s. Find the position vector \(\vec{r}(t)\).

Given information:

Initial position: \(\vec{r}(0) = \langle 0, 0, 49 \rangle\) m
Initial velocity: \(\vec{v}(0) = \langle 2, 4, 14.7 \rangle\) m/s
Acceleration due to gravity: \(\vec{a}(t) = \langle 0, 0, -9.8 \rangle\) m/s²

Solution:

Step 1: Find velocity by integrating acceleration:

\[ \vec{v}(t) = \int \vec{a}(t) \, dt + \vec{C} = \int \langle 0, 0, -9.8 \rangle \, dt + \vec{C} \] \[ = \langle 0, 0, -9.8t \rangle + \vec{C} \]

Using the initial condition \(\vec{v}(0) = \langle 2, 4, 14.7 \rangle\):

\[ \langle 2, 4, 14.7 \rangle = \langle 0, 0, 0 \rangle + \vec{C} \implies \vec{C} = \langle 2, 4, 14.7 \rangle \]

Therefore:

\[ \vec{v}(t) = \langle 2, 4, -9.8t + 14.7 \rangle \]

Step 2: Find position by integrating velocity:

\[ \vec{r}(t) = \int \vec{v}(t) \, dt + \vec{D} = \int \langle 2, 4, -9.8t + 14.7 \rangle \, dt + \vec{D} \] \[ = \left\langle 2t, 4t, -4.9t^2 + 14.7t \right\rangle + \vec{D} \]

Using the initial condition \(\vec{r}(0) = \langle 0, 0, 49 \rangle\):

\[ \langle 0, 0, 49 \rangle = \langle 0, 0, 0 \rangle + \vec{D} \implies \vec{D} = \langle 0, 0, 49 \rangle \]

Final Answer:

\[ \vec{r}(t) = \left\langle 2t, 4t, -4.9t^2 + 14.7t + 49 \right\rangle \]

Graph Description: A parabolic trajectory showing a ball launched from a height of 49 m, following a curved path under the influence of gravity before returning to ground level.

Continuation: When Does the Ball Hit the Ground?

Question: When does the ball hit the ground?

Using the position vector \(\vec{r}(t) = \left\langle 2t, 4t, -4.9t^2 + 14.7t + 49 \right\rangle\), find when the ball reaches the ground (where \(z = 0\)).

Solution:

The ball hits the ground when the z-component equals 0:

\[ z(t) = -4.9t^2 + 14.7t + 49 = 0 \]

Factoring out -4.9:

\[ -4.9(t^2 - 3t - 10) = 0 \] \[ -4.9(t - 5)(t + 2) = 0 \]

This gives \(t = 5\) or \(t = -2\).

Since time must be positive, \(t = 5\) seconds.

At \(t = 5\), the position is:

\[ \vec{r}(5) = \langle 2(5), 4(5), -4.9(5)^2 + 14.7(5) + 49 \rangle = \langle 10, 20, 0 \rangle \]

Graph Description: A parabolic trajectory showing the complete path of the ball from launch at 49 m height to landing at ground level, with the landing point marked.

Continuation: When Does the Ball Reach Maximum Height?

Question: When does the ball attain its peak height?

Find the time when the ball reaches maximum height and calculate that height.

Solution:

At maximum height, the vertical component of velocity equals zero:

\[ \vec{v}(t) = \langle 2, 4, -9.8t + 14.7 \rangle \]

Setting the z-component equal to zero:

\[ -9.8t + 14.7 = 0 \] \[ t = \frac{14.7}{9.8} = \frac{3}{2} = 1.5 \text{ seconds} \]

The maximum height is the z-component at \(t = 1.5\):

\[ \vec{r}(t) = \left\langle 2t, 4t, -4.9t^2 + 14.7t + 49 \right\rangle \] \[ \text{Max height} = z(1.5) = -4.9(1.5)^2 + 14.7(1.5) + 49 \] \[ = -4.9(2.25) + 22.05 + 49 \] \[ = -11.025 + 22.05 + 49 = 60.025 \text{ m} \]

Key Concept: Maximum height occurs when the vertical component of velocity equals zero, which is when the ball momentarily stops rising before beginning to fall.