/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 Consider the following parametri... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 63

Consider the following parametric curves. a. Determine \(d y / d x\) in terms of \(t\) and evaluate it at the given value of \(t.\) b. Make a sketch of the curve showing the tangent line at the point corresponding to the given value of \(t.\) $$x=t+1 / t, y=t-1 / t ; t=1$$

Short Answer

Expert verified
Question: Determine the equation of the tangent line to the curve defined by the parametric equations \(x = t + \frac{1}{t}\) and \(y = t - \frac{1}{t}\) at the point corresponding to \(t = 1\), and provide a rough sketch of the curve and tangent line. Answer: The tangent line to the curve at the point corresponding to \(t = 1\) is vertical and passes through the point (2, 0). Since it is vertical, its equation is \(x = 2\).

Step by step solution

01

Finding \(dx/dt\) and \(dy/dt\)

First, we need to find the derivatives of \(x\) and \(y\) with respect to \(t\). Given the parametric equations: $$x = t + \frac{1}{t}$$ $$y = t - \frac{1}{t}$$ Now, differentiate both equations with respect to \(t\): $$\frac{dx}{dt} = 1 - \frac{1}{t^2}$$ $$\frac{dy}{dt} = 1 + \frac{1}{t^2}$$
02

Finding \(\frac{dy}{dx}\)

Now, we will use the chain rule to find the derivative \(\frac{dy}{dx}\). Recall, that \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\). Now we can substitute the expressions found in Step 1 into this equation: $$\frac{dy}{dx} = \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}}$$
03

Evaluating \(\frac{dy}{dx}\) at \(t=1\)

Now that we have found the expression for \(\frac{dy}{dx}\), we need to evaluate it at \(t=1\). So, we replace \(t\) with \(1\): $$\frac{dy}{dx}\Big|_{t=1} = \frac{1 + \frac{1}{1^2}}{1 - \frac{1}{1^2}} = \frac{1+1}{1-1}$$ However, this expression evaluates to an undefined value as the denominator becomes zero, which means the tangent is going to be vertical.
04

Find the point corresponding to \(t=1\)

Now, we need to find the coordinate of the point on the graph corresponding to \(t=1\). Substitute \(t=1\) into the parametric equations to find the point \((x,y)\): $$x = 1 + \frac{1}{1} = 2$$ $$y = 1 - \frac{1}{1} = 0$$ So, our point is \((2, 0)\).
05

Sketch the curve and the tangent line

Since the tangent line is vertical, the line will pass through the point \((2,0)\) with an undefined slope. It's going to be parallel to the \(y\)-axis. To sketch the curve, we can generate some points using different values of \(t\), and then plot the points on the Cartesian plane together with the tangent line passing through \((2,0)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative of Parametric Equations
In calculus, parametric equations allow us to describe a set of related quantities as functions of an independent variable, often represented by \( t \). To find the rate of change of the dependent variable \( y \) with respect to \( x \), one must first determine their derivatives with respect to \( t \). This involves differentiating each parametric equation separately.
  • From the parametric equations \( x = t + \frac{1}{t} \) and \( y = t - \frac{1}{t} \), their derivatives with respect to \( t \) are \( \frac{dx}{dt} = 1 - \frac{1}{t^2} \) and \( \frac{dy}{dt} = 1 + \frac{1}{t^2} \).
  • These derivatives express how \( x \) and \( y \) change with the parameter \( t \).
Finding the ratio \( \frac{dy}{dx} \) is crucial for understanding how the curve behaves as \( t \) varies.
Chain Rule in Calculus
The chain rule is a fundamental theorem in calculus used to determine the derivative of composite functions. In the context of parametric equations, it simplifies finding \( \frac{dy}{dx} \) by expressing it in terms of \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \).
  • This is done through the relation \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \).
  • For our parametric curve, substituting the derivatives found earlier gives \( \frac{dy}{dx} = \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}} \).
The chain rule here links the change in \( y \) with respect to \( x \) indirectly using \( t \), making complex problems much more tractable. Its usage is crucial especially when direct differentiation with respect to \( x \) is not feasible.
Tangent Line to Curve
A tangent line to a curve at a given point is a straight line that just barely "touches" the curve at that point. Mathematically, it's defined by the derivative \( \frac{dy}{dx} \) at a particular point, representing the slope of the tangent.
  • In the parametric context, it describes the instantaneous direction of the curve.
  • For \( t = 1 \), substituting into our expression for \( \frac{dy}{dx} \) results in an undefined slope due to division by zero, signifying a vertical tangent line.
Thus, in this case, the tangent line is aligned with the y-axis through the point \( (2, 0) \), indicating a vertical orientation.
Evaluating Derivatives
Evaluating derivatives means substituting specific values into our derivative expression to find the slope or different characteristics of the curve at those points. This helps us understand the curve鈥檚 shape and behavior locally.
  • Substituting \( t = 1 \) into the parametric equations \( x = t + \frac{1}{t} \) and \( y = t - \frac{1}{t} \) yields the point \((2, 0)\).
  • Plugging the same value into the derivative expression \( \frac{dy}{dx} = \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}} \) leads to an undefined result.
This specific outcome shows that at \( t = 1 \), the slope is undefined, confirming a vertical tangent, and highlights how derivative evaluation can offer insights into the nature of curves.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find an equation of the line tangent to the hyperbola \(x^{2} / a^{2}-y^{2} / b^{2}=1\) at the point \(\left(x_{0}, y_{0}\right)\)

Show that the polar equation $$r^{2}-2 r r_{0} \cos \left(\theta-\theta_{0}\right)=R^{2}-r_{0}^{2}$$ describes a circle of radius \(R\) whose center has polar coordinates \(\left(r_{0}, \theta_{0}\right)\).

Find an equation of the following curves, assuming the center is at the origin. Sketch a graph labeling the vertices, foci, asymptotes (if they exist), and directrices. Use a graphing utility to check your work. An ellipse with vertices (卤9,0) and eccentricity \(\frac{1}{3}\)

A simplified model assumes that the orbits of Earth and Mars are circular with radii of 2 and \(3,\) respectively, and that Earth completes one orbit in one year while Mars takes two years. When \(t=0,\) Earth is at (2,0) and Mars is at (3,0) both orbit the Sun (at (0,0) ) in the counterclockwise direction. The position of Mars relative to Earth is given by the parametric equations \(x=(3-4 \cos \pi t) \cos \pi t+2, \quad y=(3-4 \cos \pi t) \sin \pi t\) a. Graph the parametric equations, for \(0 \leq t \leq 2\) b. Letting \(r=(3-4 \cos \pi t),\) explain why the path of Mars relative to Earth is a lima莽on (Exercise 89).

A trochoid is the path followed by a point \(b\) units from the center of a wheel of radius \(a\) as the wheel rolls along the \(x\) -axis. Its parametric description is \(x=a t-b \sin t, y=a-b \cos t .\) Choose specific values of \(a\) and \(b,\) and use a graphing utility to plot different trochoids. In particular, explore the difference between the cases \(a>b\) and \(a
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.