/*! 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 57 Vertical tangent lines a. Dete... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 57

Vertical tangent lines a. Determine the points where the curve \(x+y^{2}-y=1\) has a vertical tangent line (see Exercise 53 ). b. Does the curve have any horizontal tangent lines? Explain.

Short Answer

Expert verified
Question: Find the point on the curve \(x+y^2-y=1\) where there is a vertical tangent line and determine if there are any horizontal tangent lines. Answer: The point with a vertical tangent line is \((x, y) = \left(\frac{1}{4}, \frac{1}{2}\right)\). The curve does not have any horizontal tangent lines.

Step by step solution

01

Differentiate implicitly with respect to x

Using the given equation, \(x+y^2-y=1\), we will perform implicit differentiation with respect to x: $$\frac{d}{dx}(x+y^2-y) = \frac{d}{dx}(1)$$ To differentiate each term, we will use the chain rule when there is a need to differentiate \(y\) with respect to \(x\). Differentiating both sides, we have: $$1 + (2y\frac{dy}{dx} - \frac{dy}{dx})=0$$ Now, let's solve for \(dy/dx\).
02

Solve for the derivative dy/dx

Rearrange the equation above to collect terms with \(dy/dx\) on one side: $$\frac{dy}{dx}(2y-1)=-1$$ Now, divide by \((2y-1)\): $$\frac{dy}{dx} = \frac{-1}{2y-1}$$
03

Find vertical tangent points

To determine when there is a vertical tangent line, we need to look for places where the denominator of the derivative is zero, as this will cause the tangent to be undefined, which corresponds to a vertical tangent: $$2y-1=0$$ And solving for \(y\): $$y=\frac{1}{2}$$ Now using the value of \(y\) found above, we substitute it back into the original equation to find the corresponding value of \(x\): $$x + \left(\frac{1}{2}\right)^2 - \frac{1}{2} = 1$$ Solving for \(x\), we get: $$x=\frac{1}{4}$$ So, the point with a vertical tangent line is \((x, y) = \left(\frac{1}{4}, \frac{1}{2}\right)\).
04

Evaluate the possibility of horizontal tangent lines

Horizontal tangent lines are those whose derivative, \(dy/dx\), is equal to zero. However, in the equation we derived for \(dy/dx=\frac{-1}{2y-1}\), the numerator is always \(-1\) and it will never be equal to zero. This implies that there are no horizontal tangent lines on this curve.
05

Conclusion

For the given curve \(x+y^2-y=1\), there is one point with a vertical tangent line: \((x, y) = \left(\frac{1}{4}, \frac{1}{2}\right)\). The curve does not have any horizontal tangent lines, as per the explanation above.

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!

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

Once Kate's kite reaches a height of \(50 \mathrm{ft}\) (above her hands), it rises no higher but drifts due east in a wind blowing \(5 \mathrm{ft} / \mathrm{s} .\) How fast is the string running through Kate's hands at the moment that she has released \(120 \mathrm{ft}\) of string?

Prove that \(\frac{d^{2 n}}{d x^{2 n}}(\sin x)=(-1)^{n} \sin x\) and \(\frac{d^{2 n}}{d x^{2 n}}(\cos x)=(-1)^{n} \cos x\)

The following limits equal the derivative of a function \(f\) at a point a. a. Find one possible \(f\) and \(a\) b. Evaluate the limit. $$\lim _{h \rightarrow 0} \frac{\cos \left(\frac{\pi}{6}+h\right)-\frac{\sqrt{3}}{2}}{h}$$

An airliner passes over an airport at noon traveling \(500 \mathrm{mi} / \mathrm{hr}\) due west. At \(1: 00 \mathrm{p} . \mathrm{m} .,\) another airliner passes over the same airport at the same elevation traveling due north at \(550 \mathrm{mi} / \mathrm{hr} .\) Assuming both airliners maintain their (equal) elevations, how fast is the distance between them changing at 2: 30 p.m.?

The following limits equal the derivative of a function \(f\) at a point a. a. Find one possible \(f\) and \(a\) b. Evaluate the limit. $$\lim _{h \rightarrow 0} \frac{\sin \left(\frac{\pi}{6}+h\right)-\frac{1}{2}}{h}$$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Decision Maths

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.