The cycloid x=r(theta - sin(theta)), y=r(1 - cos(theta)) 

Find an equation of the tangent to the cycloid at the point where theta = pi/3

To find the equation of the tangent line to the cycloid given by the parametric equations

\[
x = r(\theta - \sin(\theta)), \quad y = r(1 - \cos(\theta)),
\]

at the point where \(\theta = \frac{\pi}{3}\), we need to follow these steps:

### 1. Compute the derivatives \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\)
The slope of the tangent line at any point on the curve is given by \(\frac{dy}{dx}\), which can be written as:

\[
\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}.
\]

First, find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\):

- For \(x = r(\theta - \sin(\theta))\),
  \[
  \frac{dx}{d\theta} = r(1 - \cos(\theta)).
  \]

- For \(y = r(1 - \cos(\theta))\),
  \[
  \frac{dy}{d\theta} = r\sin(\theta).
  \]

### 2. Evaluate the derivatives at \(\theta = \frac{\pi}{3}\)
Substitute \(\theta = \frac{\pi}{3}\) into the derivatives:

- \[
  \frac{dx}{d\theta} \bigg|_{\theta = \frac{\pi}{3}} = r\left(1 - \cos\left(\frac{\pi}{3}\right)\right) = r\left(1 - \frac{1}{2}\right) = \frac{r}{2}.
  \]

- \[
  \frac{dy}{d\theta} \bigg|_{\theta = \frac{\pi}{3}} = r\sin\left(\frac{\pi}{3}\right) = r \cdot \frac{\sqrt{3}}{2}.
  \]

Now, compute the slope \(\frac{dy}{dx}\) at \(\theta = \frac{\pi}{3}\):

\[
\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\frac{r\sqrt{3}}{2}}{\frac{r}{2}} = \sqrt{3}.
\]

So, the slope of the tangent line at \(\theta = \frac{\pi}{3}\) is \(\sqrt{3}\).

### 3. Find the point on the curve at \(\theta = \frac{\pi}{3}\)
Substitute \(\theta = \frac{\pi}{3}\) into the parametric equations to get the point on the curve:

- For \(x\):
  \[
  x = r\left(\frac{\pi}{3} - \sin\left(\frac{\pi}{3}\right)\right) = r\left(\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right).
  \]

- For \(y\):
  \[
  y = r\left(1 - \cos\left(\frac{\pi}{3}\right)\right) = r\left(1 - \frac{1}{2}\right) = \frac{r}{2}.
  \]

Thus, the point is \(\left(r\left(\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right), \frac{r}{2}\right)\).

### 4. Equation of the tangent line
The equation of a line with slope \(m\) through the point \((x_0, y_0)\) is given by:

\[
y - y_0 = m(x - x_0).
\]

Substitute \(m = \sqrt{3}\), \(x_0 = r\left(\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right)\), and \(y_0 = \frac{r}{2}\):

\[
y - \frac{r}{2} = \sqrt{3}\left(x - r\left(\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right)\right).
\]

This is the equation of the tangent line to the cycloid at \(\theta = \frac{\pi}{3}\).

Would you like more details or have any questions?

Here are 5 related questions to explore:

1. How do you find the tangent to a parametric curve in general?
2. What happens to the slope of the tangent line for different values of \(\theta\)?
3. How do you interpret the cycloid geometrically?
4. What is the normal line to the cycloid at \(\theta = \frac{\pi}{3}\)?
5. How do you find the arc length of a cycloid?

**Tip:** When dealing with parametric equations, always find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\) first to compute the slope \(\frac{dy}{dx}\) correctly.

The cycloid x=r(theta - sin(theta)), y=r(1 - cos(theta)) Find an equation of the tangent to the cycloid at the point where theta = pi/3

Discover how to find the equation of the tangent line to a cycloid at θ = π/3 using parametric equations and derivatives. This problem explores key calculus concepts such as finding the slope of the tangent line and computing derivatives of parametric curves. Suitable for students in Grades 11-12.

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