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14.3 Motions in space (continued)

\(\vec{r}_1(t) = \langle \cos t, \sin t \rangle \quad 0 \le t \le 2\pi\)

\(\vec{r}_2(t) = \langle \sin t, \cos t \rangle \quad 0 \le t \le 2\pi\)

shape? circle

note for \(\vec{r}_1\), \(x(t) = \cos t\), \(y(t) = \sin t\) \(\cos^2 t + \sin^2 t = 1 \rightarrow x^2 + y^2 = 1\) (circle radius 1)

note for \(\vec{r}_2\), \(x(t) = \sin t\), \(y(t) = \cos t\) \(\cos^2 t + \sin^2 t = 1 \rightarrow y^2 + x^2 = 1\)

Figure: A coordinate plane showing a circle centered at the origin with radius 1. The path is traced counter-clockwise, starting at t=0, 2pi on the positive x-axis, moving through t=pi/2 on the positive y-axis, t=pi on the negative x-axis, and t=3pi/2 on the negative y-axis.

Figure: A coordinate plane showing a circle centered at the origin with radius 1. The path is traced clockwise, starting at t=0, 2pi on the positive y-axis, moving through t=pi/2 on the positive x-axis, t=pi on the negative y-axis, and t=3pi/2 on the negative x-axis.

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we can also find direction from velocity

\(\vec{r}_1 = \langle \cos t, \sin t \rangle \quad 0 \le t \le 2\pi\)

\(\vec{r}_1' = \vec{v}_1 = \langle -\sin t, \cos t \rangle \quad 0 \le t \le 2\pi\)

\(\vec{v}_1(0) = \langle 0, 1 \rangle\) vector pointing "up"

\(\vec{v}_1(\frac{\pi}{2}) = \langle -1, 0 \rangle\) vector pointing "left"

note \(\vec{r}_1 \cdot \vec{v}_1 = \langle \cos t, \sin t \rangle \cdot \langle -\sin t, \cos t \rangle = 0\)

so position and velocity vectors are orthogonal for all \(t\)

this is true for all circles

in general, if \(\vec{r}(t)\) is a circle, \(\vec{r}(t) = \langle x(t), y(t) \rangle\)

then \(x^2 + y^2 = R^2\) (R: some constant (radius))

and \(\vec{r} \cdot \vec{r}' = 0\)

Figure: A coordinate plane showing a circle with two tangent vectors: one pointing up at the top of the circle and one pointing left at the right side of the circle.

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Vector Analysis of Circular Motion

Given the vector function:

\[ \vec{r}_3(t) = \langle \sin t + 2\sqrt{6} \cos t, \quad 2\sqrt{6} \sin t - \cos t \rangle, \quad 0 \le t \le 2\pi \]

We analyze the components:

\[ x^2 = \sin^2 t + 4\sqrt{6} \sin t \cos t + 24 \cos^2 t \] \[ y^2 = 24 \sin^2 t - 4\sqrt{6} \sin t \cos t + \cos^2 t \] \[ x^2 + y^2 = 25 \cos^2 t + 25 \sin^2 t = 25 \]

This represents a circle with radius 5.

Calculating the derivative:

\[ \vec{r}_3' = \langle \cos t - 2\sqrt{6} \sin t, \quad 2\sqrt{6} \cos t + \sin t \rangle \]

Note that the dot product is zero:

\[ \vec{r}_3 \cdot \vec{r}_3' = 0 \]

This is a circle with a shifted starting location, because:

\[ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \]

Figure: A coordinate plane showing a circle of radius 5 centered at the origin, with a point marked at t=0 at an angle of -11.5 degrees.

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Frequency and Speed in Circular Motion

\[ \vec{r}_1 = \langle \cos t, \sin t \rangle, \quad 0 \le t \le 2\pi \] \[ \vec{r}_4 = \langle \cos(2t), \sin(2t) \rangle, \quad 0 \le t \le B \]

The "2" is the frequency, where the period is defined as:

\[ \text{period} = \frac{2\pi}{\text{freq}} \]

For the second function, the identity holds:

\[ \cos^2(2t) + \sin^2(2t) = 1 \]

Both represent a circle of radius 1.

Figure: A unit circle graph for r1 showing t=0, 2pi at (1,0) and t=pi/2 at (0,1).

Figure: A unit circle graph for r4 showing t=0, pi at (1,0) and t=pi/4, 3pi/4 at (0,1) and (0,-1).

So, \(B = \pi\).

Comparing speeds:

\[ \vec{r}_1' = \langle -\sin t, \cos t \rangle, \quad |\vec{r}_1'| = 1 \quad (\text{speed of } \vec{r}_1) \] \[ \vec{r}_4' = \langle -2 \sin 2t, 2 \cos 2t \rangle, \quad |\vec{r}_4'| = 2 \quad (\text{twice the speed}) \]

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Given the position vector:

\[ \vec{r} = \langle 3 \cos t, \sin t \rangle, \quad 0 \le t \le 2\pi \]

This represents an ellipse:

\[ x = 3 \cos t, \quad y = \sin t \implies \frac{x}{3} = \cos t, \quad y = \sin t \implies \left(\frac{x}{3}\right)^2 + y^2 = 1 \implies \frac{x^2}{9} + y^2 = 1 \]

Figure: A coordinate graph showing an ellipse centered at the origin with x-intercepts at 3 and -3, and y-intercepts at 1 and -1.

Calculating velocity and orthogonality:

\[ \vec{v} = \vec{r}' = \langle -3 \sin t, \cos t \rangle \] \[ |\vec{v}|^2 = 9 \sin^2 t + \cos^2 t = 1 + 8 \sin^2 t \] \[ \vec{r} \cdot \vec{v} = -9 \cos t \sin t + \cos t \sin t = -8 \cos t \sin t \neq 0 \text{ for all } t \]

Also, \(|\vec{v}| \neq \text{constant}\). The velocity is no longer ALWAYS orthogonal to position.

Figure: A coordinate graph of an ellipse with vectors drawn from the origin to the curve, showing velocity vectors at specific points.

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Projectile Motion

For sake of calculation, let \(g = 10 \text{ m/s}^2\).

An object is launched from ground with initial velocity:

\[ \vec{v}(0) = \langle 1.25, 2.5, 5 \rangle \text{ m/s} \]

Under only gravitational acceleration:

\[ \vec{a}(t) = \langle 0, 0, -10 \rangle \]

Assume starting at origin \(\vec{r}(0) = \langle 0, 0, 0 \rangle\).

Find:

time of flight
range (distance covered)
max height

Figure: A 3D coordinate system showing a parabolic trajectory starting at the origin and landing on the ground plane.

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Kinematics Calculation

\[ \vec{a}(t) = \langle 0, 0, -10 \rangle \]

\[ \vec{v}(t) = \int \vec{a}(t) \, dt = \langle 0, 0, -10t \rangle + \vec{C} = \langle C_1, C_2, -10t + C_3 \rangle \]

Use \( \vec{v}(0) = \langle 1.25, 2.5, 5 \rangle \) to find \( C_1, C_2, C_3 \).

\[ \vec{v}(0) = \langle C_1, C_2, 0 + C_3 \rangle = \langle 1.25, 2.5, 5 \rangle \]

So, \( C_1 = 1.25 \), \( C_2 = 2.5 \), \( C_3 = 5 \).

So, \( \vec{v}(t) = \langle 1.25, 2.5, -10t + 5 \rangle \)

\[ \vec{r}(t) = \int \vec{v}(t) \, dt = \langle 1.25t + d_1, 2.5t + d_2, -5t^2 + 5t + d_3 \rangle \]

Use \( \vec{r}(0) = \langle 0, 0, 0 \rangle \) to find \( d_1, d_2, d_3 \). We see \( d_1 = 0, d_2 = 0, d_3 = 0 \).

So, \( \vec{r}(t) = \langle 1.25t, 2.5t, -5t^2 + 5t \rangle \)

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Flight Analysis

Time of Flight

z-component of position: \( -5t^2 + 5t = 0 \) (ground)

\[ -5t(t - 1) = 0 \implies t = 0, t = 1 \]

So time of flight is 1 second.

Range

Figure: A 3D coordinate system showing the displacement vector r(1) from the origin to the landing point.

\[ \vec{r}(1) = \langle 1.25, 2.5, 0 \rangle \]

\( |\vec{r}(1)| \approx 2.8 \text{ meters} \)

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Max height?

When the vertical component of \(\vec{v}\) is 0:

\[ \vec{v}_z = -10t + 5 = 0 \rightarrow t = \frac{1}{2} \]

Height: \(z\) of position at \(t = \frac{1}{2}\)

1.25 meters

What if \(\vec{v}(0)\) is doubled?

New \(\vec{v}(0) = \langle 2.5, 5, 10 \rangle\), same \(\vec{r}(0) = \langle 0, 0, 0 \rangle\)

\[ \vec{a}(t) = \langle 0, 0, -10 \rangle \] \[ \vec{v}(t) = \int \vec{a}(t) dt = \langle C_1, C_2, -10t + C_3 \rangle \] \[ = \langle 2.5, 5, -10t + 10 \rangle \] \[ \vec{r}(t) = \int \vec{v}(t) dt = \dots = \langle 2.5t, 5t, -5t^2 + 10t \rangle \]

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Time of flight and range

Time of flight: \(-5t^2 + 10t = 0\)

\[ -5t(t - 2) = 0 \rightarrow t = 0, t = 2 \]

doubled

Range: \(|\vec{r}(2)| = \dots = 11.18\)

quadrupled