Question

Find the equation of the osculating circle to the curve at the indicated \(t\) -value. \(\vec{r}(t)=\left\langle t, t^{2}\right\rangle,\) at \(t=0\)

Short answer

The osculating circle is given by the equation \(x^2 + (y - \frac{1}{2})^2 = \frac{1}{4}\).

Step-by-step solution

Find the Derivative

First, find the derivative of the vector function \(\vec{r}(t) = \langle t, t^2 \rangle\). Differentiate each component with respect to \(t\): \(\vec{r}\,'(t) = \left\langle \frac{d}{dt}(t), \frac{d}{dt}(t^2) \right\rangle = \langle 1, 2t \rangle\).

Calculate the Second Derivative

Differentiate \(\vec{r}\,'(t) = \langle 1, 2t \rangle\) to find the second derivative:\(\vec{r}\,''(t) = \left\langle \frac{d}{dt}(1), \frac{d}{dt}(2t) \right\rangle = \langle 0, 2 \rangle\).

Evaluate the Derivatives at t=0

Evaluate the first and second derivatives at \(t = 0\):\(\vec{r}\,'(0) = \langle 1, 0 \rangle\) and \(\vec{r}\,''(0) = \langle 0, 2 \rangle\).

Compute the Curvature

The formula for curvature \(\kappa\) at a point is:\[ \kappa = \frac{\| \vec{r}\,'(t) \times \vec{r}\,''(t) \|}{\| \vec{r}\,'(t) \|^3} \]Since we are in 2D, we use:\[ \kappa = \frac{|x' y'' - y' x''|}{\left((x')^2 + (y')^2\right)^{3/2}} \]Using \(\vec{r}\,'(0) = \langle 1, 0 \rangle\) and \(\vec{r}\,''(0) = \langle 0, 2 \rangle\), we get:\[ \kappa = \frac{|1 \cdot 2 - 0 \cdot 0|}{(1^2 + 0^2)^{3/2}} = \frac{2}{1} = 2 \].

Calculate the Osculating Circle Radius

The radius \(R\) of the osculating circle is the reciprocal of the curvature:\[ R = \frac{1}{\kappa} = \frac{1}{2} \].

Determine the Center of the Osculating Circle

The center of the osculating circle \(C\) is found using:\[ C = \, \vec{r}(t) + \frac{\vec{r}\,'(t) \times \vec{k}}{\| \vec{r}\,'(t) \|} \cdot R \]where \(\vec{k} \) is the unit vector perpendicular to the velocity vector \(\vec{r}\,'(t)\) in the plane. Here, we rotate the velocity vector by 90 degrees counter-clockwise, so \(\vec{k} = \langle 0, 1 \rangle\).At \(t = 0\), \(\vec{r}(0) = \langle 0, 0 \rangle\), \(\vec{r}\,'(0) = \langle 1, 0 \rangle\). The perpendicular vector is \(\langle 0, 1 \rangle\), and the center of the circle is\[ C = \langle 0, 0 \rangle + \frac{\langle 0, 1 \rangle}{1} \cdot \frac{1}{2} = \langle 0, \frac{1}{2} \rangle \].

Write the Equation of the Osculating Circle

The equation of the circle with center \( C(ap, a)\) and radius \(R(a,b)\) is:\[ (x - 0)^2 + \left(y - \frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^2 \]Simplifying, we get:\[ x^2 + \left(y - \frac{1}{2}\right)^2 = \frac{1}{4} \].

Key concepts

Curvature

Curvature helps us understand how much a curve deviates from being a straight line. It essentially measures how "bent" the curve is at a particular point. When a curve bends more sharply, it has a higher curvature. In the context of our curve \[ \vec{r}(t) = \langle t, t^2 \rangle \]at the point where \( t = 0 \), the formula used to calculate the curvature is:\[ \kappa = \frac{|x' y'' - y' x''|}{((x')^2 + (y')^2)^{3/2}} \]This formula is particularly suited for 2D curves and uses derivatives of the curve's components.

• The term \( x' y'' - y' x'' \) in the numerator involves the derivatives of the component functions.
• \( x' \) and \( y'' \) represent the first derivative of the first component and the second derivative of the second component respectively.
• In the denominator, \( (x')^2 + (y')^2 \) is raised to the power \( 3/2 \), emphasizing the speed of change.

By plugging in the values, we found the curvature \( \kappa = 2 \), which shows us the rate at which the curve bends at \( t = 0 \).

Vector Calculus

Vector calculus is a mathematical tool that deals with vector fields. It explores how vectors change and interact in different directions. Here's a quick overview of the key aspects relevant in this problem:

• **Vector function**: In our example, the vector function \( \vec{r}(t) = \langle t, t^2 \rangle \) describes a path traced by a point in the plane.
• **Derivatives**: The change in the vector function, or its derivative, shows the direction and speed of movement along the curve.
• The vector parameter \( t \) commonly acts as a time parameter, tracing the curve as \( t \) increases.

In calculating the osculating circle, vector calculus allows us to determine precise characteristics of the curve, like its speed, direction, and how it bends through space. The concepts involved, such as derivatives and cross-products, are essential in determining quantities like velocity, acceleration, and curvature, which further lead to defining the osculating circle.

Derivative

Derivatives play a crucial role in analyzing and understanding curves. They provide information about the rate at which things change. In the problem of finding the osculating circle, derivatives help us determine both the velocity and acceleration along the curve.First, the derivative of the vector function:\[ \vec{r}'(t) = \langle 1, 2t \rangle \]This tells us the velocity at which the particle moves along the curve. At \( t = 0 \), the velocity is \( \langle 1, 0 \rangle \), showing a horizontal motion.Next, the second derivative:\[ \vec{r}''(t) = \langle 0, 2 \rangle \]This represents the acceleration, indicating the rate of change of velocity. Its evaluation at \( t = 0 \) yields \( \langle 0, 2 \rangle \), suggesting an upward acceleration of the particle.

• The velocity vector \( \vec{r}'(t) \) helps in identifying the direction and speed at any point.
• The acceleration vector \( \vec{r}''(t) \) gives insight into how the velocity changes.

These derivatives are essential for calculating the elements needed to find the osculating circle, such as curvature and circle radius. Understanding derivatives enables us to explore how intricate paths and motions can be and how they evolve.