MA 261 - Lesson 6: Calculus of Vector-Valued Functions (14.2) and Motion in Space (14.3) - Part I

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Page 1: Warm-up exercise with helix curve
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Warm-up Exercise

Consider the vector-valued function \(\vec{r}(t) = \langle \cos t, \sin t, 2t \rangle\), where \(\vec{r}(t)\) traces a helix on the cylinder \(x^2 + y^2 = 1\).

Graph Description: A three-dimensional coordinate system showing a helix curve spiraling around a vertical cylinder. The curve wraps around the cylinder while ascending in the z-direction.

Question 1: What do you think \(\frac{d}{dt}(\vec{r}(t))\) should be?

Taking the derivative component-wise:

\[ \frac{d}{dt}\langle \cos t, \sin t, 2t \rangle = \left\langle \frac{d}{dt}(\cos t), \frac{d}{dt}(\sin t), \frac{d}{dt}(2t) \right\rangle = \langle -\sin t, \cos t, 2 \rangle \]

Question 2: What do you think \(\int \vec{r}(t) \, dt\) should be?

Key Insight: Both differentiation and integration of vector-valued functions are performed component-wise.

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Page 2: Review problem finding intersection points
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Review: Finding Points of Intersection

Example: Intersection of Curve with Plane

Find the points of intersection of the curve \(\vec{r}(t) = \langle t\cos t, t\sin t, t \rangle\) for \(0 \leq t \leq 8\pi\) with the region plane \(x - y = 0\).

Solution:

The plane equation \(x - y = 0\) gives us the condition that \(x = y\). Substituting the parametric equations:

\[ x = y \implies t\cos t = t\sin t \] \[ t\cos t - t\sin t = 0 \] \[ t(\cos t - \sin t) = 0 \]

This equation is satisfied when either \(t = 0\) or \(\cos t - \sin t = 0\).

When \(\cos t = \sin t\), we have \(t = \frac{\pi}{4}\) or \(t = \frac{5\pi}{4}\) (in the given interval).

Therefore, there are three points of intersection:

  • At \(t = 0\): \(\vec{r}(0) = \langle 0, 0, 0 \rangle\)
  • At \(t = \frac{\pi}{4}\): \(\vec{r}\left(\frac{\pi}{4}\right) = \left\langle \frac{\sqrt{2}\pi}{8}, \frac{\sqrt{2}\pi}{8}, \frac{\pi}{4} \right\rangle\)
  • At \(t = \frac{5\pi}{4}\): \(\vec{r}\left(\frac{5\pi}{4}\right) = \left\langle -\frac{5\sqrt{2}\pi}{8}, -\frac{5\sqrt{2}\pi}{8}, \frac{5\pi}{4} \right\rangle\)

Total: 3 points of intersection.

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Page 3: Sketching curve on double cone
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Sketching Curves in 3D

Example: Curve on a Double Cone

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

Solution:

To understand the shape, we eliminate the parameter \(t\). Starting with the x and y components:

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

Since \(z = t\), we have \(t^2 = z^2\), which gives us:

\[ x^2 + y^2 = z^2 \]

This is the equation of a double cone with vertex at the origin and axis along the z-axis.

Graph Description: A three-dimensional double cone with vertex at the origin. The curve \(\vec{r}(t)\) spirals around the upper cone, starting at the origin when \(t = 0\) and wrapping outward and upward as \(t\) increases. The cone opens both upward and downward symmetrically about the xy-plane.

Important: The curve \(\vec{r}(t)\) lies on the double cone surface defined by \(x^2 + y^2 = z^2\).

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Page 4: Derivative and its geometric meaning
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Derivative and Its Geometric Meaning

We now explore the derivative of vector-valued functions and understand its geometric interpretation.

Scalar Functions vs. Vector-Valued Functions

For a scalar function \(f(t)\):

Graph Description: Two diagrams showing (1) a scalar function y = f(t) with a tangent line at point tâ‚€, illustrating the slope concept, and (2) a three-dimensional curve with a tangent vector at a point, showing the direction of motion.

For a vector-valued function \(\vec{r}(t) = \langle f(t), g(t), h(t) \rangle\):

Definition: The derivative \(\vec{r}'(t)\) is a vector-valued function representing the curve in 3D space:

\[ \vec{r}'(t) = \langle f'(t), g'(t), h'(t) \rangle \]

At a specific point \(t_0\), \(\vec{r}'(t_0)\) is a direction vector for the tangent line through \(\vec{r}(t_0)\).

Definition: The unit tangent vector is given by:

\[ \vec{T}(t_0) = \frac{\vec{r}'(t_0)}{|\vec{r}'(t_0)|} \]

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Page 5: Recall of line equations
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Recall: Equation of a Line

Example: Line Through a Point Parallel to a Vector

Find the equation of the line through \((1, 5, 2)\) parallel to \(\langle 1, 0, -3 \rangle\).

Solution:

The vector form of a line is given by:

\[ \langle x, y, z \rangle = t \langle a, b, c \rangle + \langle x_0, y_0, z_0 \rangle \]

where \(\langle a, b, c \rangle\) is the direction vector and \(\langle x_0, y_0, z_0 \rangle\) is a point on the line.

Therefore, the equation of our line is:

\[ \langle x, y, z \rangle = t \langle 1, 0, -3 \rangle + \langle 1, 5, 2 \rangle \]

Graph Description: A three-dimensional coordinate system showing a line passing through the point (1, 5, 2) with direction vector ⟨1, 0, -3⟩. The direction vector is shown as an arrow parallel to the line, indicating the line's orientation in space.

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Page 6: Finding tangent line equation
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Finding the Tangent Line to a Curve

Example: Tangent Line at a Specific Point

Given \(\vec{r}(t) = \langle \cos t, \sin t, 2t \rangle\), find the equation of the tangent line at \(t = \frac{\pi}{2}\).

Solution:

Recall that the equation of a line through \((x_0, y_0, z_0)\) with direction vector \(\langle a, b, c \rangle\) is:

\[ \langle x, y, z \rangle = t \langle a, b, c \rangle + \langle x_0, y_0, z_0 \rangle \]

Step 1: Find the point on the curve at \(t = \frac{\pi}{2}\):

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

Step 2: Find the direction vector (the derivative at \(t = \frac{\pi}{2}\)):

\[ \vec{r}'(t) = \left\langle \frac{d}{dt}(\cos t), \frac{d}{dt}(\sin t), \frac{d}{dt}(2t) \right\rangle = \langle -\sin t, \cos t, 2 \rangle \] \[ \vec{r}'\left(\frac{\pi}{2}\right) = \left\langle -\sin\frac{\pi}{2}, \cos\frac{\pi}{2}, 2 \right\rangle = \langle -1, 0, 2 \rangle \]

Step 3: Write the equation of the tangent line:

\[ \langle x, y, z \rangle = t \langle -1, 0, 2 \rangle + \langle 0, 1, \pi \rangle \]

Important: The direction vector of the tangent line is the derivative \(\vec{r}'(t)\) evaluated at \(t = \frac{\pi}{2}\).

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Page 7: Derivative properties
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Derivative Properties for Vector-Valued Functions

The following properties hold for vector-valued functions, analogous to scalar derivatives:

Theorem: Derivative Properties

Let \(\vec{r}_1(t)\) and \(\vec{r}_2(t)\) be differentiable vector-valued functions, and let \(c\) be a constant and \(f(t)\) be a scalar function. Then:

  1. Constant multiple: \(\frac{d}{dt}\left( c \vec{r}(t) \right) = c \frac{d}{dt}\vec{r}(t)\)
  2. Sum rule: \(\frac{d}{dt}\left( \vec{r}_1(t) + \vec{r}_2(t) \right) = \frac{d}{dt}\vec{r}_1(t) + \frac{d}{dt}\vec{r}_2(t)\)
  3. Dot product rule: \(\frac{d}{dt}\left( \vec{r}_1(t) \cdot \vec{r}_2(t) \right) = \frac{d}{dt}\vec{r}_1(t) \cdot \vec{r}_2(t) + \vec{r}_1(t) \cdot \frac{d}{dt}\vec{r}_2(t)\)
  4. Cross product rule: \(\frac{d}{dt}\left( \vec{r}_1(t) \times \vec{r}_2(t) \right) = \frac{d}{dt}\vec{r}_1(t) \times \vec{r}_2(t) + \vec{r}_1(t) \times \frac{d}{dt}\vec{r}_2(t)\)
  5. Scalar multiplication: \(\frac{d}{dt}\left( f(t) \vec{r}(t) \right) = f'(t) \vec{r}(t) + f(t) \vec{r}'(t)\)
  6. Chain rule: \(\frac{d}{dt}\left[ \vec{r}(f(t)) \right] = \vec{r}'(f(t)) \cdot f'(t)\) (scalar multiplication)

Warning: For the cross product rule (property 4), order is important! The cross product is not commutative, so be careful with the order of the terms.

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Page 8: Integration of vector-valued functions
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Integration of Vector-Valued Functions

Like differentiation, integration of vector-valued functions is performed component-wise.

Definition: If \(\vec{r}(t) = \langle f(t), g(t), h(t) \rangle\), then we integrate component-wise:

\[ \int \vec{r}(t) \, dt = \left\langle \int f(t) \, dt, \int g(t) \, dt, \int h(t) \, dt \right\rangle \]
Example 1: Indefinite Integral

Find \(\int \vec{r}(t) \, dt\) where \(\vec{r}(t) = \langle \cos t, \sin t, 2t \rangle\).

Solution:

\[ \int \langle \cos t, \sin t, 2t \rangle \, dt = \left\langle \int \cos t \, dt, \int \sin t \, dt, \int 2t \, dt \right\rangle \] \[ = \left\langle \sin t + c_1, -\cos t + c_2, t^2 + c_3 \right\rangle \] \[ = \langle \sin t, -\cos t, t^2 \rangle + \vec{C} \]

where \(\vec{C} = \langle c_1, c_2, c_3 \rangle\) is a constant vector of integration.

Example 2: Definite Integral

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

Solution:

First, find the indefinite integral:

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

Then evaluate from 0 to 1:

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

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Page 9: Particle motion in space
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Particle Moving in Space (3D)

When we describe the motion of a particle in three-dimensional space, we use three fundamental vector quantities:

Definition: Position, Velocity, and Acceleration Vectors

Graph Description: A three-dimensional coordinate system with a curved path representing the trajectory of a particle. At a specific point on the path, three vectors are shown: the position vector from the origin to the point, the velocity vector tangent to the curve pointing in the direction of motion, and arrows indicating the relationships between position, velocity, and acceleration.

Important Relationships:

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Page 10: Example finding position from acceleration
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Finding Position from Acceleration

Example: Determining Position Vector from Acceleration

Given \(\vec{a}(t) = \langle e^t, \sin t, t \rangle\), \(\vec{v}(0) = \langle 1, 2, 3 \rangle\) (initial velocity), and \(\vec{r}(0) = \langle -1, 3, -4 \rangle\) (initial position), find the position vector \(\vec{r}(t)\).

Solution:

Step 1: Find velocity by integrating acceleration:

\[ \vec{v}(t) = \int \vec{a}(t) \, dt = \int \langle e^t, \sin t, t \rangle \, dt \] \[ = \left\langle e^t, -\cos t, \frac{t^2}{2} \right\rangle + \vec{C} \]

Step 2: Use initial velocity to find \(\vec{C}\):

\[ \langle 1, 2, 3 \rangle = \vec{v}(0) = \left\langle e^0, -\cos(0), \frac{0^2}{2} \right\rangle + \vec{C} = \langle 1, -1, 0 \rangle + \vec{C} \] \[ \vec{C} = \langle 1, 2, 3 \rangle - \langle 1, -1, 0 \rangle = \langle 0, 3, 3 \rangle \]

Therefore:

\[ \vec{v}(t) = \left\langle e^t, -\cos t, \frac{t^2}{2} \right\rangle + \langle 0, 3, 3 \rangle = \left\langle e^t, -\cos t + 3, \frac{t^2}{2} + 3 \right\rangle \]

Step 3: Find position by integrating velocity:

\[ \vec{r}(t) = \int \vec{v}(t) \, dt = \int \left\langle e^t, -\cos t + 3, \frac{t^2}{2} + 3 \right\rangle \, dt \] \[ = \left\langle e^t, -\sin t + 3t, \frac{t^3}{6} + 3t \right\rangle + \vec{D} \]

Step 4: Use initial position to find \(\vec{D}\):

\[ \langle -1, 3, -4 \rangle = \vec{r}(0) = \left\langle e^0, -\sin(0) + 3(0), \frac{0^3}{6} + 3(0) \right\rangle + \vec{D} \] \[ = \langle 1, 0, 0 \rangle + \vec{D} \] \[ \vec{D} = \langle -1, 3, -4 \rangle - \langle 1, 0, 0 \rangle = \langle -2, 3, -4 \rangle \]

Final Answer:

\[ \vec{r}(t) = \left\langle e^t, -\sin t + 3t, \frac{t^3}{6} + 3t \right\rangle + \langle -2, 3, -4 \rangle \] \[ = \left\langle e^t - 2, -\sin t + 3t + 3, \frac{t^3}{6} + 3t - 4 \right\rangle \]

Important: When finding position from acceleration, we integrate twice and use both initial conditions (initial velocity and initial position) to determine the constants of integration.