GRAPHS

• Graph 1 — A line and a circle with position, velocity, and acceleration vectors.
• Graph 2 — The graph of the parabola \(y = x^2\) with position, velocity, and acceleration vectors, and osculating circle.

Background & Notes — Exercise 1

Suppose the vector function \(\vec{r}(t)\) is a position function — representing the location \(\vec{r}\) of an object moving through space, as a function of time \(t\).

The velocity \(\vec{v}(t)\) of the object is the derivative (instantaneous rate of change) of position \(vec{r}(t)\).

The magnitude of velocity is speed \(v(t)\).

The acceleration \(\vec{a}(t)\) is the derivative of velocity.

For a curve in 3-space:

\[\begin{align*}
\vec{r}(t) &= \big< x(t), y(t), z(t) \big>\\
\vec{v}(t) = \vec{r}'(t) &= \big< x'(t), y'(t), z'(t) \big>\\
v(t) = |\vec{v}(t)| &= \sqrt{ \big[x'(t)\big]^2 + \big[y'(t)\big]^2 + \big[z'(t)\big]^2}\\
\vec{a}(t) = \vec{r}''(t) &= \big< x''(t), y''(t), z''(t) \big>
\end{align*}\]

The curvature of a curve \(\vec{r}(t)\) can be computed using the velocity, speed, and acceleration:

\[\kappa(t) = \frac{|\vec{v}(t) \times \vec{a}(t)|}{| \vec{v}(t)|^3} = \frac{|\vec{r}'(t) \times \vec{r}''(t) |}{| \vec{r}'(t)|^3}\]

EXERCISE 1 — Lines & Circles

Assumptions

Let \(\vec{r}_1(t)\), \(\vec{r}_1(t)\), \(\vec{r}_1(t)\) be the position functions:

\[\begin{align*}
\vec{r}_1(t) &= \big< 3t - 4, t - 2 \big>\\
\vec{r}_2(t) &= \big< 3 \cos t, 3 \sin t \big>
\end{align*}\]

Instructions

1. Motion & Curvature — Lines

Compute the velocity \(\vec{v}(t)\), speed \(v(t)\), acceleration \(\vec{a}(t)\), and curvature \(\kappa(t)\) for the line \(\vec{r}_1(t) = \big< 3t - 4, t - 2 \big>\).

Note: On a line, velocity, speed, acceleration, and curvature are all constant.

Go to graph 1 and identify the line \(\vec{r}_1(t)\) (red). Use the t-slider to move the position vector (blue) and velocity vector (black) over the line.

Questions:

• What is the relationship between the velocity vector and the line?
• Does the curvature \(\kappa(t)\) you computed seem to give a good description of how “curvy” the line is?

2. Motion & Curvature — Circles

Compute the velocity \(\vec{v}(t)\), speed \(v(t)\), acceleration \(\vec{a}(t)\), and curvature \(\kappa(t)\) for the circle \(\vec{r}_2(t) = \big< 3 \cos t, 3 \sin t \big>\).

Note: On a circle, only the speed and curvature are constant.

Go to graph 1 and identify the circle \(\vec{r}_2(t)\) (blue). Use the t-slider to move the position vector (blue), velocity vector (black), and acceleration vector (green) over the circle.

Questions:

• Compare the curvature \(\kappa(t)\) for this circle to the radius of the circle. What’s the relationship between them?
• Does the curvature \(\kappa(t)\) you computed seem to give a good description of how “curvy” the circle is?

3. Uniform Circular Motion

Uniform circular motion refers to an object traveling around a circle at constant speed.

Adapt the vector function \(\vec{r}_2(t) = \big< 3 \cos t, 3 \sin t \big>\) to create a vector function for the motion of an object traveling around a circle of radius \(r = 3\) with constant speed \(v(t) = 6\).

Challenge — Show that, on the circle:

\[\vec{r}_2(t) = \big< 3 \cos t, 3 \sin t \big>\]

• Velocity \(\vec{v}(t)\) is orthogonal to position \(\vec{r}_2(t)\).
• Acceleration \(\vec{a}(t)\) is orthogonal to velocity \(\vec{v}(t)\).

Big Ideas — Lines & Circles

On the line \(\vec{r}(t) = t \vec{v} + \vec{r}_0\):

• \(\vec{v}(t) = \vec{v}\) — Velocity is constant.
• \(v(t) = |\vec{v}|\) — Speed is constant.
• \(\vec{v}(t) = \vec{0}\) — Acceleration is constant (equals zero).
• \(\kappa(t) = 0\) — Curvature is constant (equals zero).

On the (uniform) circle \(\vec{r}(t) = \big< r \cos c t, r \sin c t \big>\):

• \(r\) is the radius of the circle.
• \(v(t) = cr\) — Speed is constant.
• \(\kappa(t) = 1/r\) — Curvature is constant (equals the reciprocal of the radius).
• \(\vec{v}(t) \perp \vec{r}(t)\), \(\vec{a}(t) \perp \vec{v}(t)\), \(\vec{a}(t) \| \vec{r}(t)\) — Velocity is orthogonal to position, acceleration is orthogonal to velocity, and acceleration is parallel to position.

Background & Notes — Exercise 2

The graph of a function \(y = f(x)\) in the \(xy\)-plane can be parametrized using the vector function:

\[\vec{r}(x) = \big< t, f(x) \big>\]

Adding a third component \(z(x) = 0\), we think about the graph in the \(xy\)-plane (2-dimensional) sitting inside 3-space (3-dimensional):

\[\vec{r}(x) = \big< x, f(x), 0 \big>\]

This allows us to use the cross product to compute the curvature of the graph.

EXERCISE 2 — Curvature & Graphs of Functions

Assumptions

Suppose \(y = f(x)\) is the graph a function with continuous first and second derivatives.

Let \(\vec{r}(x)\) be the parametrization for this graph:

\[\vec{r}(x) = \big< x, f(x), 0 \big>\]

Instructions

1. Compute

Use the equation:

\[\kappa(x) = \frac{|\vec{r}'(x) \times \vec{r}''(x)|}{| \vec{r}'(x)|^3}\]

to show that the curvature function \(\kappa(x)\) for this graph is:

\[\kappa(x) = \frac{|f''(x)|}{\left(1 + \big[f'(x)\big]^2\right)^{3/2}}\]

Note: On this parabola, only the acceleration is constant.

2. Compute & Identify

Use the equation:

\[\kappa(x) = \frac{|f''(x)|}{\sqrt{1 + \big[f'(x)\big]^2}}\]

to compute the curvature function \(\kappa(x)\) for the graph of the parabola \(y = x^2\).

Use the curvature function \(\kappa(x)\) to identify the maximum value of curvature on the parabola, and the value of \(x\) at which it occurs.

3. Graph — The osculating circle.

Go to graph 2 and use the t-slider to move the position (blue), velocity (black), and acceleration (green) across the graph.

The black circle that moves with the vectors is the osculating circle. This is the “best circular approximation” to the curve.

Move the t-slider to \(t = 0\), and determine the radius of the osculating circle.

When \(x = 0\), the graph passes through the origin. Compare the curvature \(\kappa(0)\) to the radius of the osculating circle at the origin. What is the relationship between the curvature \(\kappa(0)\) and the radius of the osculating circle?

Big Ideas — Curvature & Graphs of Functions

The curvature function of the graph of a twice-differentiable function \(y = f(x)\) is:

\[\kappa(x) = \frac{|f''(x)|}{\left(1 + \big[f'(x)\big]^2\right)^{3/2}}\]

The osculating circle is the “best circular approximation” of a curve.

The curvature \(\kappa\) at a point on a curve is the reciprocal of the radius \(R\) of the osculating circle at that point:

\[\kappa = 1/R\]