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.