To find the equation of a tangent line from parametric equations, we look at a curve defined by two equations: one for the $x$-coordinate, $x(t)$, and one for the $y$-coordinate, $y(t)$. The variable $t$ acts as a parameter that helps us understand how the curve works. We want to find the slope of the tangent line at a certain value of $t$, so we can write the equation of that line.

Step 1: Calculate the Derivatives

First, we need to find the derivatives of $x(t)$ and $y(t)$. Derivatives help us see how $x$ and $y$ change:

$\frac{dx}{dt} \quad \text{and} \quad \frac{dy}{dt}.$

Step 2: Find the Slope of the Tangent Line

Next, we find the slope of the tangent line at a specific point on the curve. We use this formula:

$m = \frac{dy/dt}{dx/dt},$

where $m$ is the slope. This formula tells us how much $y$ changes compared to how much $x$ changes when $t$ changes.

Step 3: Evaluate at a Specific Point

Now, we look at the derivatives we just calculated at a particular value of $t$, which we’ll call $t_0$. This gives us the slope at that point:

$m(t_0) = \frac{y'(t_0)}{x'(t_0)}.$

At the same time, we find the coordinates of the point on the curve:

$P(x(t_0), y(t_0)),$

so we have the $(x, y)$ coordinates we need for our tangent line.

Step 4: Use the Point-Slope Form of a Linear Equation

With the slope and a point on the tangent line, we can use the point-slope form of a linear equation. It looks like this:

$y - y_1 = m(x - x_1),$

where $(x_1, y_1)$ is the point we found using $t_0$. Plugging in our values, we get:

$y - y(t_0) = m(t_0)(x - x(t_0)).$

This equation gives us the tangent line to the curve at the point for $t_0$.

Example

Let’s go through a simple example with these parametric equations:

$x(t) = t^2 \quad \text{and} \quad y(t) = t^3.$

1. Find the derivatives:

   • $\frac{dx}{dt} = 2t$
   • $\frac{dy}{dt} = 3t^2$

2. Calculate the slope at $t_0 = 1$:

   • $m(1) = \frac{3(1^2)}{2(1)} = \frac{3}{2}.$

3. Find the point on the curve:

   • $P(1^2, 1^3) = P(1, 1).$

4. Write the tangent line equation: Using the point-slope form:

   • $y - 1 = \frac{3}{2}(x - 1).$

   If we simplify this, we find:

   • $y = \frac{3}{2}x - \frac{1}{2}.$

Now we have the equation of the tangent line from our parametric equations!

Summary

To sum it up, to get the equation of a tangent line from parametric equations, we follow these steps:

1. Calculate the derivatives to find the slope.
2. Evaluate those derivatives at a specific value of $t$.
3. Use the point-slope form of a linear equation to write the tangent line.

By doing this, we can understand how the curve behaves at certain points!