Problem 145 - Calculus Volume 3

Question

Find the curvature at each point \((x, y)\) on the hyperbola \(\mathbf{r}(t)=\langle a \cosh (t), b \sinh (t)\rangle\).

Step-by-Step Solution

Answer

The curvature of the hyperbola at each point \((x, y)\) is \(\frac{ab \cosh(2t)}{(a^2 \sinh^2(t) + b^2 \cosh^2(t))^{3/2}}\).

Step 1: Differentiate the Position Vector

First, we need to find the derivative of the position vector \( \mathbf{r}(t) \). To do this, differentiate each component with respect to \( t \):\[ \mathbf{r}'(t) = \langle a \sinh(t), b \cosh(t) \rangle \]

Step 2: Compute the Magnitude of the Velocity Vector

Calculate the magnitude of \( \mathbf{r}'(t) \):\[\|\mathbf{r}'(t)\| = \sqrt{ (a \sinh(t))^2 + (b \cosh(t))^2}\]Simplify it using \( \sinh^2(t) + \cosh^2(t) = \cosh(2t) \):\[\|\mathbf{r}'(t)\| = \sqrt{ a^2 \sinh^2(t) + b^2 \cosh^2(t) }\]

Step 3: Differentiate the Velocity Vector

Find the second derivative \( \mathbf{r}''(t) \) by differentiating \( \mathbf{r}'(t) \):\[ \mathbf{r}''(t) = \langle a \cosh(t), b \sinh(t) \rangle \]

Step 4: Compute the Cross Product

Since we are working in 2D, we consider the cross product equivalent for a planar curve to calculate the curvature denominator:\[\|\mathbf{r}'(t) \times \mathbf{r}''(t)\| = a b |\cosh^2(t) + \sinh^2(t)| = a b \cosh(2t)\]

Step 5: Curvature Formula for 2D Curves

The curvature \( \kappa(t) \) is given by:\[\kappa(t) = \frac{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3}\]Substitute the values:\[\kappa(t) = \frac{a b \cosh(2t)}{(a^2 \sinh^2(t) + b^2 \cosh^2(t))^{3/2}}\]

Step 6: Find the Value of Curvature at Each Point

For any point \((x, y)\) on the hyperbola, substitute the corresponding \(t\) value from the given parametric equations into the curvature formula to obtain the specific curvature at that point.

Key Concepts

hyperbola

A hyperbola is a type of conic section that occurs when a plane intersects both halves of a double cone. It results in an open curve with two disconnected branches, often resembling two mirrored arcs. Unlike ellipses, where the sum of distances from focal points to any point on the curve is constant, in a hyperbola, the difference of these distances is constant.

Key features of a hyperbola include:

• Two focal points that determine the curve's shape.
• An axis of symmetry, often referred to as the transverse axis.
• Vertices, which are the points on the curve closest to each other.

These elements define its geometric properties and help in plotting the hyperbola on a graph. Understanding hyperbolas is essential for solving problems involving curves and intersections in various mathematical and engineering fields.

parametric equations

Parametric equations provide a way to express a set of related quantities as explicit functions of one or more independent parameters. In the context of curves such as hyperbolas, they define each point on the curve as a function of a third variable, often denoted as 't'.

This method allows for a more flexible and descriptive representation of complex curves:

• Parametric forms can simplify calculations, especially when describing motion or changing quantities over time.
• They can be easily differentiated to find velocities and accelerations if derived from motion context.

For example, a hyperbola can be described through parametric equations like \( \mathbf{r}(t) = \langle a \cosh (t), b \sinh (t) \rangle \), making it easier to analyze its properties over a range of values. Parametric equations are fundamental in calculus for developing integral theorems and curve plotting.

position vector

In mathematics, particularly vector calculus, a position vector represents the position of a point relative to an origin in space. For a given point on a curve or surface, the position vector offers a convenient and systematic way to describe its location. The position vector is typically expressed in unit vectors along the axes, such as \( \langle x(t), y(t) \rangle \).

Some important aspects of position vectors include:

• They provide a straightforward way to describe points in both two-dimensional and three-dimensional spaces.
• The components of the position vector can be functions of one or more variables, such as time, allowing for the study of dynamic systems.

By analyzing the derivatives of position vectors, one can derive information about the object's velocity and acceleration, which is crucial in physics and engineering for understanding the dynamics of systems and objects.

derivative

The derivative is a fundamental concept in calculus that measures how a function changes as its input changes. In simple terms, it provides the rate of change or the slope of the function at a given point. For curves and surfaces described by position vectors, derivatives are tools for understanding motion dynamics.

Key elements include:

• First Derivative: Represents the velocity of a system, showing how one state changes over time or another variable.
• Second Derivative: Indicates acceleration, reflecting the rate of velocity change, crucial for applications in motion analysis.

In our example with position vectors of a hyperbola, deriving these vectors helps to determine key properties, such as curvature, by expressing how the direction and magnitude of vectors evolve. This understanding aids not only in theoretical pursuits but also in practical applications spanning physics and engineering domains. Proper comprehension of derivatives is essential for advanced calculus operations and problem-solving techniques.