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14.2 Calculus of Vector-Valued Functions

Recall if \(y = f(t)\) is a scalar function, then \(y' = f'(t) = \lim_{h \to 0} \frac{f(t+h) - f(t)}{h}\).

A graph showing a curve y=f(t) with a secant line connecting points at t and t+h, illustrating the slope calculation.

If \(\vec{r}(t) = \langle x(t), y(t) \rangle\) is a vector-valued function, its derivative is as expected:

\[ \vec{r}'(t) = \lim_{h \to 0} \frac{\vec{r}(t+h) - \vec{r}(t)}{h} \]

What is the geometric interpretation?

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\[ \vec{r}'(t) = \lim_{h \to 0} \frac{\vec{r}(t+h) - \vec{r}(t)}{h} \]

Then if h is small, then \(\vec{r}'(t) \approx \frac{\vec{r}(t+h) - \vec{r}(t)}{h}\) or \(\vec{r}(t+h) - \vec{r}(t) \approx h\vec{r}'(t)\).

Two diagrams showing a curve with vectors r(t) and r(t+h) originating from the origin, and a difference vector u(t) connecting them.

Notice \(\vec{r}(t) + \vec{u}(t) = \vec{r}(t+h)\) or \(\vec{r}(t+h) - \vec{r}(t) = \vec{u}(t)\).

As \(h \to 0\), h is small, so \(\vec{r}(t+h) - \vec{r}(t) \approx h\vec{r}'(t) = \vec{u}(t)\). \(\vec{u}(t)\) will end up following the slope of \(\vec{r}(t)\) at t, which is the tangent vector.

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\( \vec{r}'(t) = \vec{u}(t) \) which is tangent vector to \( \vec{r}(t) \)

so \( \vec{r}'(t) \) is also tangent to \( \vec{r}(t) \)

back to \( \vec{r}(t) = \langle x(t), y(t) \rangle \)

to calculate \( \vec{r}'(t) \), we do:

\[ \vec{r}'(t) = \lim_{h \to 0} \frac{\vec{r}(t+h) - \vec{r}(t)}{h} \]
\[ \vec{r}'(t) = \lim_{h \to 0} \left\langle \frac{x(t+h) - x(t)}{h}, \frac{y(t+h) - y(t)}{h} \right\rangle \]
\[ = \left\langle \lim_{h \to 0} \frac{x(t+h) - x(t)}{h}, \lim_{h \to 0} \frac{y(t+h) - y(t)}{h} \right\rangle \]
\[ \vec{r}'(t) = \langle x'(t), y'(t) \rangle \]

Same is true if \( \vec{r}(t) \) has more components.

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Example

\( \vec{r}(t) = \langle t \cos t, t \sin t, t \rangle \)

\( x = t \cos t, \quad y = t \sin t, \quad z = t \)

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

So, \( x^2 + y^2 = z^2 \) (cone, \( z \)-axis as symmetry axis).

Cross sections are circles that get bigger as \( z \) increases.

A 3D coordinate system showing a spiral curve winding around the z-axis, representing the vector function r(t).

\( \vec{r}(t) = \langle t \cos t, t \sin t, t \rangle \)

\( \vec{r}'(t) = \langle -t \sin t + \cos t, t \cos t + \sin t, 1 \rangle \)

at \( t = \pi/2 \):

\( \vec{r}'(\pi/2) = \langle -\pi/2, 1, 1 \rangle \) vector tangent to \( \vec{r}(t) \) at \( t = \pi/2 \)

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Tangent Vectors and Unit Tangent Vectors

\[ |\vec{r}'(t)| = \sqrt{(-t\sin t + \cos t)^2 + (t\cos t + \sin t)^2 + (1)^2} = \dots = \sqrt{t^2+2} \]

This tells us the magnitude grows as t increases.

The tangent vector \(\vec{r}'(t)\) in general varies in magnitude.

Later in the course, the unit tangent vector becomes important:

\[ \vec{T} = \frac{\vec{r}'(t)}{|\vec{r}'(t)|} \]

Unit vector in direction of the tangent vector

For the spiral,

\[ \vec{T} = \left\langle \frac{-t\sin t + \cos t}{\sqrt{t^2+2}}, \frac{t\cos t + \sin t}{\sqrt{t^2+2}}, \frac{1}{\sqrt{t^2+2}} \right\rangle \]
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Differentiation Rules for Vectors

Many differentiation rules stay the same (or mostly the same).

For example, the product rule for dot product:

\[ \frac{d}{dt} [\vec{u}(t) \cdot \vec{v}(t)] = \vec{u}'(t) \cdot \vec{v}(t) + \vec{u}(t) \cdot \vec{v}'(t) = \frac{d}{dt} [\vec{v}(t) \cdot \vec{u}(t)] \]

Order doesn't matter for dot product.

\[ \frac{d}{dt} [\vec{u}(t) \times \vec{v}(t)] = \vec{u}'(t) \times \vec{v}(t) + \vec{u}(t) \times \vec{v}'(t) \]

Keep the order as written because order is important in cross product.

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Integrals work similarly, too

\[ \int \vec{r}'(t) dt = \vec{r}(t) + \vec{C} \]

Vector constant of integration

\[ \int_{a}^{b} \vec{r}'(t) dt = \vec{r}(b) - \vec{r}(a) \]

This is the displacement vector from \( t=a \) to \( t=b \).

A vector diagram showing a curve with position vectors r(a) and r(b) originating from the origin, with the displacement vector r(b)-r(a) connecting them.
Figure 1: Geometric representation of displacement.

The sum of infinitely many \( \vec{r}'(t) \) along \( t=a \) to \( t=b \) gives us \( \vec{r}(b) - \vec{r}(a) \).

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14.3 Motions in space (part 1)

If \( \vec{r}(t) \) describes the position of an object, then \( \vec{r}'(t) \) is the velocity vector.

And \( |\vec{r}'(t)| \) is the speed.

And \( \vec{r}''(t) \) is the acceleration vector.

Example: If \( \vec{a}(t) = \langle 1, t, t^2 \rangle \) is the acceleration (\( t \ge 0 \), find the velocity such that \( \vec{v}(0) = \langle 1, 2, 3 \rangle \).

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Vector Integration

Given the acceleration vector function:

\[ \vec{a}(t) = \langle 1, t, t^2 \rangle = \vec{v}'(t) \]

To find the velocity vector function \(\vec{v}(t)\), we integrate \(\vec{a}(t)\):

\[ \vec{v}(t) = \int \vec{a}(t) dt = \int \langle 1, t, t^2 \rangle dt \]
\[ = \langle \int 1 dt, \int t dt, \int t^2 dt \rangle \]
\[ = \langle t + C_1, \frac{1}{2}t^2 + C_2, \frac{1}{3}t^3 + C_3 \rangle \]
\[ \vec{v}(t) = \langle t, \frac{1}{2}t^2, \frac{1}{3}t^3 \rangle + \langle C_1, C_2, C_3 \rangle \]

This is the vector constant of integration \(\vec{C}\).

To find \(\vec{C}\), we use the initial condition \(\vec{v}(0) = \langle 1, 2, 3 \rangle\):

\[ \langle 1, 2, 3 \rangle = \langle 0, 0, 0 \rangle + \langle C_1, C_2, C_3 \rangle \]

Therefore, the velocity vector is:

\[ \vec{v}(t) = \langle t, \frac{1}{2}t^2, \frac{1}{3}t^3 \rangle + \langle 1, 2, 3 \rangle \]
\[ = \langle t+1, \frac{1}{2}t^2+2, \frac{1}{3}t^3+3 \rangle \]
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