Question

For the following exercises, find the equation of the tangent line to the given curve. Graph both the function and its tangent line. $$ x=\ln (t), y=t^{2}-1, t=1 $$

Short answer

The equation of the tangent line is \( y = 2x \).

Step-by-step solution

Identify the parametric equations

The parametric equations for the given problem are: \[ x = \ln(t) \] and \[ y = t^2 - 1 \]. The value of the parameter is given as \( t = 1 \).

Differentiate the parametric equations

Find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). The derivative of \( x = \ln(t) \) with respect to \( t \) is: \[ \frac{dx}{dt} = \frac{1}{t} \]. The derivative of \( y = t^2 - 1 \) with respect to \( t \) is: \[ \frac{dy}{dt} = 2t \].

Find the slope of the tangent line

The slope of the tangent line to the curve at a given \( t \) is given by the formula: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \]. Evaluate this at \( t = 1 \): \[ \frac{dy}{dx} = \frac{2 \cdot 1}{\frac{1}{1}} = 2 \].

Find the coordinates of the tangent point

Substitute \( t = 1 \) into the parametric equations to find the coordinates of the point: \[ x = \ln(1) = 0 \] and \[ y = 1^2 - 1 = 0 \]. Therefore, the point is \((0, 0)\).

Write the equation of the tangent line

Using the point-slope form of a line \( y - y_1 = m(x - x_1) \), with \((x_1, y_1) = (0, 0)\) and slope \( m = 2 \), the equation is: \[ y - 0 = 2(x - 0) \] which simplifies to \[ y = 2x \].

Graph the curve and tangent line

Graph the curve by plotting the parametric equations for an appropriate range of \( t \), and graph the line \( y = 2x \) to show the tangent at the point \((0, 0)\).

Key concepts

Parametric Equations

Parametric equations are a great tool in mathematics. They define a set of functions that describe a curve, using a parameter, often denoted by "t". Instead of expressing variables in terms of each other, parametric equations use a third parameter. This can make certain problems easier to solve or understand.
For example, the parametric equations given in the problem are:

• \( x = \ln(t) \)
• \( y = t^2 - 1 \)

Here, "t" is the parameter that runs through values like a clock, generating points on the curve.
At any given "t", you can find specific values for "x" and "y", which gives you a point on the curve. This representation is powerful, especially for complex curves that cannot be easily described in the traditional \( y = f(x) \) format.

Tangent Line

A tangent line is a straight line that touches a curve at only one point. This meeting point gives the tangent line the same slope as the curve where they meet.
In the context of parametric curves, finding a tangent line involves differentiating both parametric equations and determining the slope of the line using both the derivatives at a specific value of t.
The tangent line provides a simple linear approximation of the curve close to the point of tangency. It is useful in understanding the behavior of the curve at specific points and is foundational in calculus for analyzing curves.

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the rate at which things change. In the context of parametric equations, it involves calculating derivatives with respect to the parameter, "t".
For the given equations:

• \( \frac{dx}{dt} = \frac{1}{t} \)
• \( \frac{dy}{dt} = 2t \)

Differentiation allows us to find the slope of the tangent line to the parametric curve. By dividing \( \frac{dy}{dt} \) by \( \frac{dx}{dt} \), we arrive at the slope of the tangent line to the curve. At \( t = 1 \), the differentiation gives us a slope of 2, vital for constructing the tangent line at that point.

Point-Slope Form

The point-slope form is an algebraic equation written as \( y - y_1 = m(x - x_1) \), and it is useful in defining the equation of a line given a point and a slope.
When constructing the tangent line to a curve, we use this form to plug in the point through which our tangent passes \((x_1, y_1)\) and its slope "m".
In the exercise, at point \((0, 0)\) with slope 2, we use the point-slope form to derive the tangent line equation:

• Plugging in the values, the equation becomes: \( y - 0 = 2(x - 0) \).
• Which simplifies to: \( y = 2x \).

This method is precise, straightforward, and critical for modeling linear relationships in calculus.