Problem 21 - Thomas Calculus

Question

Find an equation for the circle of curvature of the curve \(\mathbf{r}(t)=t \mathbf{i}+(\sin t) \mathbf{j}\) at the point \((\pi / 2,1) .\) (The curve parametrizes the graph of \(y=\sin x\) in the \(x y\) -plane.)

Step-by-Step Solution

Answer

\((x - \frac{\pi}{2})^2 + y^2 = 1\)

Step 1: Confirm Point on Curve

We need to confirm that the point (\(\frac{\pi}{2}, 1\)) lies on the curve \(\mathbf{r}(t)=t \mathbf{i}+\sin(t) \mathbf{j}\). Evaluate \(\mathbf{r}(t)\) at \(t = \frac{\pi}{2}\):\[ \mathbf{r}\left(\frac{\pi}{2}\right) = \frac{\pi}{2} \mathbf{i} + \sin\left(\frac{\pi}{2}\right) \mathbf{j} = \frac{\pi}{2} \mathbf{i} + 1 \mathbf{j} = \left(\frac{\pi}{2}, 1\right), \]which confirms the point is on the curve.

Step 2: Compute First Derivative

Compute the first derivative \(\mathbf{r}'(t)\) to find the velocity vector:\[ \mathbf{r}'(t) = \frac{d}{dt}[t \mathbf{i} + \sin t \mathbf{j}] = \mathbf{i} + \cos t \mathbf{j}. \]Evaluate at \(t = \frac{\pi}{2}\):\[ \mathbf{r}'\left(\frac{\pi}{2}\right) = \mathbf{i} + \cos\left(\frac{\pi}{2}\right) \mathbf{j} = \mathbf{i}. \]

Step 3: Compute Second Derivative

Compute the second derivative \(\mathbf{r}''(t)\), the acceleration vector:\[ \mathbf{r}''(t) = \frac{d}{dt}[\mathbf{i} + \cos t \mathbf{j}] = -\sin t \mathbf{j}. \]Evaluate at \(t = \frac{\pi}{2}\):\[ \mathbf{r}''\left(\frac{\pi}{2}\right) = -\sin\left(\frac{\pi}{2}\right) \mathbf{j} = -\mathbf{j}. \]

Step 4: Compute Curvature

The formula for curvature \(\kappa\) is \(\kappa = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}\). First, compute the cross product.\[ \mathbf{r}'(t) \times \mathbf{r}''(t) = |\mathbf{i} \mathbf{j} \mathbf{k} ewline 1 \quad \cos t \quad 0 ewline 0 \quad -\sin t \quad 0| = 0 \mathbf{i} + 0 \mathbf{j} - \sin t \mathbf{k} = -\sin t \mathbf{k}. \]At \(t = \frac{\pi}{2}\), this gives us \(-\mathbf{k}\).\[ |\mathbf{r}'\left(\frac{\pi}{2}\right) \times \mathbf{r}''\left(\frac{\pi}{2}\right)| = 1. \]\[ |\mathbf{r}'\left(\frac{\pi}{2}\right)| = 1. \]Thus, \[ \kappa = \frac{1}{1^3} = 1. \]

Step 5: Find Radius and Center of Circle of Curvature

The radius \(R\) of the circle of curvature is \(\frac{1}{\kappa} = 1\). The unit normal vector is \(\mathbf{N}(t) = \frac{\mathbf{r}''(t)}{|\mathbf{r}''(t)|}\), calculated at \(t = \frac{\pi}{2}\):\[ \mathbf{N}\left(\frac{\pi}{2}\right) = \frac{-\mathbf{j}}{1} = -\mathbf{j}. \]The center of the circle of curvature is\[ \mathbf{C} = \mathbf{r}\left(\frac{\pi}{2}\right) + R \mathbf{N}\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2}, 1\right) + 1(-\mathbf{j}) = \left(\frac{\pi}{2}, 0\right). \]

Step 6: Write Equation of Circle of Curvature

The equation of a circle is \((x - h)^2 + (y - k)^2 = R^2\). With center \(\left(\frac{\pi}{2}, 0\right)\) and \(R = 1\), the equation is\[ \left(x - \frac{\pi}{2}\right)^2 + (y - 0)^2 = 1. \]Simplifying gives\[ \left(x - \frac{\pi}{2}\right)^2 + y^2 = 1. \]

Key Concepts

Parametric Equations

Parametric equations are a way of defining a path or curve in terms of parameters. In this exercise, the curve is given by the parametric equation \(\mathbf{r}(t) = t \mathbf{i} + \sin t \mathbf{j}\). This setup allows us to express the curve as a function of a single variable \(t\), known as the parameter.

• \(\mathbf{i}\) and \(\mathbf{j}\) represent the unit vectors along the x-axis and y-axis, respectively.
• The expression \(t \mathbf{i}\) gives the x-coordinate as the parameter \(t\).
• The expression \(\sin t \mathbf{j}\) provides the y-coordinate, linking it to the sine of the parameter \(t\).

This parametric setup is particularly useful when dealing with curves that cannot be easily defined by a single equation involving only \(x\) and \(y\). In this exercise, we transform information about the path into another form that can be easily manipulated to find derivatives and curvature, as seen in the following steps.

Curvature Calculation

Curvature is a measure of how sharply a curve bends at a particular point. For a parametrically defined curve, as in our example, the curvature \(\kappa\) can be calculated using the formula:
\[ \kappa = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3} \]

• \(\mathbf{r}'(t)\) is the first derivative of the position vector, representing the velocity of the curve.
• \(\mathbf{r}''(t)\) is the second derivative, representing the acceleration of the curve.
• The cross product \(\mathbf{r}'(t) \times \mathbf{r}''(t)\) gives a measure of the curve's twist, and its magnitude provides curvature information.

In this exercise, at the point \(t = \frac{\pi}{2}\), the curvative \(\kappa\) equals 1. This indicates that the circle best fitting the curvature at this point has a radius of 1, known as the radius of curvature.

Derivatives in Calculus

Derivatives are a fundamental tool in calculus used to understand how a function changes. In this exercise, derivatives help us analyze the motion and change of the curve over the parameter \(t\).

• The first derivative \(\mathbf{r}'(t)\) gives the velocity of the curve, describing the direction and speed of movement at each value of \(t\).
• In our exercise: \(\mathbf{r}'(t) = \mathbf{i} + \cos t \mathbf{j}\).

This indicates the rate of change for each component along the axes.

• The second derivative \(\mathbf{r}''(t)\) gives the acceleration, or how the velocity itself is changing over time.
• For this problem, \(\mathbf{r}''(t) = -\sin t \mathbf{j}\).

Derivatives thus provide valuable insights into the geometry and dynamics of parametric curves, helping in tasks like calculating the circle of curvature by relating them to concepts like velocity and acceleration.