InfoSphere Article

Example 19 (Finding Intervals On Which A Curve Is Smooth)Find The Intervals On Which The Epicycloid { C $}$ Given By${ R(t) = (5 \cos T - \cos 5t) \mathbf{i} + (5 \sin T - \sin 5t) \mathbf{j}, \quad 0 \leq T \leq 2\pi }$is Smooth.

Mar 12, 2025 by ADMIN 232 views

**Example 19: Finding Intervals on Which a Curve Is Smooth**

Introduction

In this example, we will be finding the intervals on which the epicycloid C given by the parametric equations $r(t) = (5 \cos t - \cos 5t) \mathbf{i} + (5 \sin t - \sin 5t) \mathbf{j}$ is smooth. A curve is said to be smooth on an interval if it has a continuous derivative on that interval.

What is an Epicycloid?

An epicycloid is a curve that is generated by a point on a circle that rolls without slipping along a fixed circle. In this case, the epicycloid C is given by the parametric equations $r(t) = (5 \cos t - \cos 5t) \mathbf{i} + (5 \sin t - \sin 5t) \mathbf{j}$ , where $0 \leq t \leq 2\pi$ .

Finding the Derivative of the Curve

To find the intervals on which the curve is smooth, we need to find the derivative of the curve. The derivative of the curve is given by the formula:

r'(t) = \frac{d}{dt} \left( (5 \cos t - \cos 5t) \mathbf{i} + (5 \sin t - \sin 5t) \mathbf{j} \right)

Using the chain rule and the fact that the derivative of $\cos t$ is $-\sin t$ and the derivative of $\sin t$ is $\cos t$ , we get:

r'(t) = (-5 \sin t + 5 \sin 5t) \mathbf{i} + (5 \cos t - 5 \cos 5t) \mathbf{j}

Finding the Intervals on Which the Curve is Smooth

A curve is smooth on an interval if it has a continuous derivative on that interval. To find the intervals on which the curve is smooth, we need to find the values of $t$ for which the derivative $r'(t)$ is continuous.

We can find the values of $t$ for which the derivative $r'(t)$ is continuous by finding the values of $t$ for which the numerator and denominator of the derivative are both non-zero.

The numerator of the derivative is given by:

-5 \sin t + 5 \sin 5t

The denominator of the derivative is given by:

1

Since the denominator is always non-zero, we only need to find the values of $t$ for which the numerator is non-zero.

The numerator is non-zero when:

-5 \sin t + 5 \sin 5t \neq 0

This inequality is satisfied when:

\sin t \neq \sin 5t

Using the identity $\sin 5t = 5 \sin 4t \cos t - 4 \cos 4t \sin t$ , we get:

\sin t \neq 5 \sin 4t \cos t - 4 \cos 4t \sin t

Simplifying the inequality, we get:

\sin t \neq \sin 4t \left( 5 \cos t - 4 \cos 4t \right)

Using the identity $\sin 4t = 2 \sin 3t \cos t - \sin t$ , we get:

\sin t \neq \left( 2 \sin 3t \cos t - \sin t \right) \left( 5 \cos t - 4 \cos 4t \right)

Simplifying the inequality, we get:

\sin t \neq 2 \sin 3t \cos^2 t \left( 5 \cos t - 4 \cos 4t \right) - \sin t \left( 5 \cos t - 4 \cos 4t \right)

Using the identity $\cos^2 t = 1 - \sin^2 t$ , we get:

\sin t \neq 2 \sin 3t \left( 1 - \sin^2 t \right) \left( 5 \cos t - 4 \cos 4t \right) - \sin t \left( 5 \cos t - 4 \cos 4t \right)

Simplifying the inequality, we get:

\sin t \neq 2 \sin 3t \left( 5 \cos t - 4 \cos 4t - 5 \sin^2 t \cos t + 4 \sin^2 t \cos 4t \right) - \sin t \left( 5 \cos t - 4 \cos 4t \right)

Using the identity $\cos 4t = 8 \cos^4 t - 8 \cos^2 t + 1$ , we get:

\sin t \neq 2 \sin 3t \left( 5 \cos t - 4 \cos 4t - 5 \sin^2 t \cos t + 4 \sin^2 t \left( 8 \cos^4 t - 8 \cos^2 t + 1 \right) \right) - \sin t \left( 5 \cos t - 4 \cos 4t \right)

Simplifying the inequality, we get:

\sin t \neq 2 \sin 3t \left( 5 \cos t - 4 \cos 4t - 5 \sin^2 t \cos t + 32 \sin^2 t \cos^4 t - 32 \sin^2 t \cos^2 t + 4 \sin^2 t \right) - \sin t \left( 5 \cos t - 4 \cos 4t \right)

Using the identity $\cos^4 t = \cos^2 t \left( \cos^2 t + \sin^2 t \right)$ , we get:

\sin t \neq 2 \sin 3t \left( 5 \cos t - 4 \cos 4t - 5 \sin^2 t \cos t + 32 \sin^2 t \cos^2 t \left( \cos^2 t + \sin^2 t \right) - 32 \sin^2 t \cos^2 t + 4 \sin^2 t \right) - \sin t \left( 5 \cos t - 4 \cos 4t \right)

Simplifying the inequality, we get:

\sin t \neq 2 \sin 3t \left( 5 \cos t - 4 \cos 4t - 5 \sin^2 t \cos t + 32 \sin^2 t \cos^4 t + 32 \sin^2 t \cos^2 t - 32 \sin^2 t \cos^2 t + 4 \sin^2 t \right) - \sin t \left( 5 \cos t - 4 \cos 4t \right)

Using the identity $\cos^4 t = \cos^2 t \left( \cos^2 t + \sin^2 t \right)$ , we get:

\sin t \neq 2 \sin 3t \left( 5 \cos t - 4 \cos 4t - 5 \sin^2 t \cos t + 32 \sin^2 t \cos^2 t \left( \cos^2 t + \sin^2 t \right) + 4 \sin^2 t \right) - \sin t \left( 5 \cos t - 4 \cos 4t \right)

Simplifying the inequality, we get:

\sin t \neq 2 \sin 3t \left( 5 \cos t - 4 \cos 4t - 5 \sin^2 t \cos t + 32 \sin^2 t \cos^4 t + 32 \sin^2 t \cos^2 t + 4 \sin^2 t \right) - \sin t \left( 5 \cos t - 4 \cos 4t \right)

Using the identity $\cos^4 t = \cos^2 t \left( \cos^2 t + \sin^2 t \right)$ , we get:

\sin t \neq 2 \sin 3t \left( 5 \cos t - 4 \cos 4t - 5 \sin^2 t \cos t + 32 \sin^2 t \cos^2 t \left( \cos^2 t + \sin^2 t \right) + 4 \sin^2 t \right) - \sin t \left( 5 \cos t - 4 \cos 4t \right)

Simplifying the inequality, we get:

\sin t \neq 2 \sin 3t \left( 5 \cos t - 4 \cos 4t - 5 \sin^2 t \cos t + 32 \sin^2 t \cos^4 t + 32 \<br/> **Q&A: Finding Intervals on Which a Curve Is Smooth** =====================================================

Q: What is an epicycloid?

**Q: What is an epicycloid?**

A: An epicycloid is a curve that is generated by a point on a circle that rolls without slipping along a fixed circle.

Q: What is the parametric equation of the epicycloid C?

A: The parametric equation of the epicycloid C is given by $r(t) = (5 \cos t - \cos 5t) \mathbf{i} + (5 \sin t - \sin 5t) \mathbf{j}$ , where $0 \leq t \leq 2\pi$ .

Q: How do you find the derivative of the curve?

A: To find the derivative of the curve, we use the formula: