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Magazine Chapter 1: Problem 334
For the following exercises, find the equation of the tangent line to the given curve. Graph both the function and its tangent line. \(x=\ln (t), \quad y=t^{2}-1, \quad t=1\)
Short Answer
Expert verified
The equation of the tangent line is \(y = 2x\).
Step by step solution
01
Understand the Problem
We need to find the equation of the tangent line to the curve given by the parametric equations \(x = \ln(t)\) and \(y = t^2 - 1\) at the parameter value \(t = 1\).
02
Calculate Derivatives
To find the equation of the tangent line at \(t = 1\), we first need the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). Compute these as follows:\[ \frac{dx}{dt} = \frac{1}{t} \]\[ \frac{dy}{dt} = 2t \]
03
Evaluate Derivatives at t = 1
Substitute \(t = 1\) into \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) to find the point's slope on the curve:\[ \frac{dx}{dt}\bigg|_{t=1} = 1 \]\[ \frac{dy}{dt}\bigg|_{t=1} = 2 \]
04
Find Slope of Tangent Line
The slope of the tangent line is given by \(\frac{dy}{dx}\):\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t}{1/t} = 2t^2 \]At \(t = 1\), the slope is \(2 \times 1^2 = 2\).
05
Compute the Point of Tangency
Now find the point at \(t = 1\) using the parametric equations:\[ x = \ln(1) = 0 \]\[ y = 1^2 - 1 = 0 \]The point of tangency is \((0, 0)\).
06
Write the Equation of the Tangent Line
Using the point-slope form of a line \((y - y_1) = m(x - x_1)\), where \((x_1, y_1) = (0, 0)\) and the slope \(m = 2\):\[ y - 0 = 2(x - 0) \Rightarrow y = 2x \]
07
Graphing the Function and Tangent Line
Graph the curve represented by the parametric equations and the line \(y = 2x\) on the same axes. The curve will be a parabola opening upwards, and the tangent line is a straight line passing through the origin with slope 2.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parametric Equations
Parametric equations are a way to define a curve or a path in terms of a third variable, often referred to as a parameter.
The parameter, usually denoted as 't', allows us to express both the x and y coordinates of a point along the curve.
This method provides a flexible way to describe complex curves where a single equation in terms of x and y might be insufficient or impractical.In this exercise, the parametric equations are given by:
The parameter, usually denoted as 't', allows us to express both the x and y coordinates of a point along the curve.
This method provides a flexible way to describe complex curves where a single equation in terms of x and y might be insufficient or impractical.In this exercise, the parametric equations are given by:
- \( x = \ln(t) \)
- \( y = t^2 - 1 \)
Slope of the Tangent Line
To determine the slope of a tangent line to a parametric curve at a particular point, we need to understand how the curve itself changes.This involves calculating the slope at a specific parameter value, which helps in understanding how steep the curve is at that point. For parametric equations, the slope of the tangent line is computed as:
- \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \)
- Derivative for x: \( \frac{dx}{dt} = 1 \)
- Derivative for y: \( \frac{dy}{dt} = 2 \)
- Thus, slope \( \frac{dy}{dx} = 2 \times 1 = 2 \)
Point-Slope Form of a Line
The point-slope form is a mathematical way of expressing a linear equation. It's particularly useful for describing lines given a point and a slope.The general form of the equation is:
This formula is handy when you need to find the equation of a tangent line since you usually have both a point and a slope from previous calculations.In this exercise, we used the point \((0, 0)\) and slope \(2\) to compute:
- \( y - y_1 = m(x - x_1) \)
This formula is handy when you need to find the equation of a tangent line since you usually have both a point and a slope from previous calculations.In this exercise, we used the point \((0, 0)\) and slope \(2\) to compute:
- Plug into point-slope: \( y - 0 = 2(x - 0) \)
- Simplify to obtain: \( y = 2x \)
Derivative Calculation
The derivative calculation is a critical part of finding the tangent line to a parametric curve.
We need to calculate the derivatives of both the x and y parametric equations with respect to the parameter 't'.
These derivatives reveal how each coordinate changes and are essential in determining the slope of the tangent.For this problem:
Thus, derivative calculation is the foundational step in determining how the curve behaves at a point.
We need to calculate the derivatives of both the x and y parametric equations with respect to the parameter 't'.
These derivatives reveal how each coordinate changes and are essential in determining the slope of the tangent.For this problem:
- Derivatives help us find how x and y change with 't', \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
- \( \frac{dx}{dt} = \frac{1}{t} \), so at \( t=1 \), \( \frac{dx}{dt} = 1 \).
- \( \frac{dy}{dt} = 2t \), so at \( t=1 \), \( \frac{dy}{dt} = 2 \).
Thus, derivative calculation is the foundational step in determining how the curve behaves at a point.
