Problem 16 - Calculus Early Transcendentals

Question

Reparametrize the curve $$\mathbf{r}(t)=\left(\frac{2}{t^{2}+1}-1\right) \mathbf{i}+\frac{2 t}{t^{2}+1} \mathbf{j}$$ with respect to arc length measured from the point \((1,0)\) in the direction of increasing \(t\) . Express the reparametrization in its simplest form. What can you conclude about the curve?

Step-by-Step Solution

Answer

The curve is a circle.

Step 1: Compute the speed of the curve

The arc length for a parameterized curve \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} \) is given by integrating the speed, which is \( \sqrt{(x'(t))^2 + (y'(t))^2} \), over the desired interval. First, find the derivatives \( x'(t) \) and \( y'(t) \). For \( x(t) = \frac{2}{t^2 + 1} - 1 \), the derivative is \( x'(t) = \frac{-4t}{(t^2 + 1)^2} \). For \( y(t) = \frac{2t}{t^2 + 1} \), the derivative is \( y'(t) = \frac{2(t^2 + 1) - 4t^2}{(t^2 + 1)^2} = \frac{2 - 2t^2}{(t^2 + 1)^2} \).

Step 2: Calculate the magnitude of the velocity

The magnitude of the velocity (speed) is \( \sqrt{(x'(t))^2 + (y'(t))^2} \). Substitute the expressions for \( x'(t) \) and \( y'(t) \): \[ \text{Speed} = \sqrt{ \left( \frac{-4t}{(t^2 + 1)^2} \right)^2 + \left( \frac{2 - 2t^2}{(t^2 + 1)^2} \right)^2 } \].Simplifying this, we find \( \text{Speed} = \frac{2}{t^2+1} \).

Step 3: Calculate the arc length function

To find the arc length \( s(t) \), integrate the speed function from \( t = 0 \) to \( t \). \[ s(t) = \int_0^t \frac{2}{u^2+1} \, du = [2 \tan^{-1} u]_0^t = 2 \tan^{-1}(t) \].

Step 4: Solve for \( t \) in terms of \( s \)

The function we have from the arc length \( s(t) = 2 \tan^{-1}(t) \) allows us to express \( t \) in terms of \( s \): \[ \tan^{-1}(t) = \frac{s}{2} \]Therefore, \( t = \tan\left(\frac{s}{2}\right) \).

Step 5: Reparametrize the curve using \( s \)

Now substitute \( t = \tan\left( \frac{s}{2} \right) \) back into the original parameterization \( \mathbf{r}(t) \):- For \( x(t) \), substitute \( t = \tan\left( \frac{s}{2} \right) \):\[ x(s) = \frac{2}{\tan^2\left(\frac{s}{2}\right) + 1} - 1 \]- For \( y(t) \), substitute \( t = \tan\left( \frac{s}{2} \right) \):\[ y(s) = \frac{2\tan\left(\frac{s}{2}\right)}{\tan^2\left(\frac{s}{2}\right) + 1} \].This gives the reparametrized curve as \( \mathbf{r}(s) = x(s) \mathbf{i} + y(s) \mathbf{j} \).

Step 6: Conclusion about the curve

The curve is a circle, because the parametrization leads us to forms that satisfy the equation of the unit circle after simplification. As one can verify, \( x^2(s) + y^2(s) = 1 \), indicating a circle.

Key Concepts

Arc Length

The concept of arc length is fundamental in calculus, especially when working with curves. In simple terms, the arc length is the distance along the curve from one point to another. It's a bit like measuring out a string that follows the path of the curve and then straightening it to see how long it is.

To find the arc length of a parametric curve, we need to calculate the integral of the curve's speed over the interval of interest. If we have a curve given by a vector function \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} \), the speed \( v(t) \) at any point is the magnitude of the derivative \( \mathbf{r}'(t) \). The speed is given by

• \( v(t) = \sqrt{(x'(t))^2 + (y'(t))^2} \).

By integrating this speed, we obtain the total arc length \( s(t) \) from a starting point up to some value \( t \). It's given as:

• \( s(t) = \int_{t_0}^t v(u) \, du \).

The arc length is particularly useful when we seek a new way to describe the curve, like measuring the curve from a start point and using it as a new parameter.

Parametric Equations

Parametric equations are a powerful way to describe mathematical curves. Unlike regular functions where \( y \) is usually expressed directly in terms of \( x \), parametric equations provide a pair of equations to describe both \( x \) and \( y \) as functions of a third variable, typically \( t \), which is the parameter.

So, we might have:

• \( x = f(t) \)
• \( y = g(t) \)

The parameter \( t \) usually represents 'time' or another dimension along which the curve progresses. By varying \( t \), one can trace the path of the curve in a plane. This allows for a more flexible and comprehensive representation of curves that might loop or twist in ways not easily represented by simple \( y = f(x) \) equations.

A noteworthy feature is the ability to reparametrize these curves with respect to a different measure, such as arc length, as we did in the original exercise. This reparametrization provides fresh insights and interpretations of the same geometric path.

Circle in Calculus

Circles are a classic and elegant subject in calculus, appearing often in a variety of forms and problems. When defining a circle in terms of parametric equations, things become very insightful.

For the unit circle, the typical parameterization is:

• \( x(t) = \cos(t) \)
• \( y(t) = \sin(t) \)

This is because \( \cos^2(t) + \sin^2(t) = 1 \), which is the equation for a circle of radius 1 centered at the origin.

The original problem concludes that a given curve is a circle because after simplification, its reparameterized form satisfies \( x^2(s) + y^2(s) = 1 \). In calculus, finding that a parametrized equation satisfies this equation often reveals the hidden circle. Recognizing a circle in equations can significantly simplify understanding its properties and behavior.