/*! 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 15 Let \(P_{1}=\left(x_{1}, y_{1}\r... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 15

Let \(P_{1}=\left(x_{1}, y_{1}\right)\) and \(P_{2}=\left(x_{2}, y_{2}\right)\) be two points on the curve \(y=\) \(a x^{2}+b x+c .\) If \(P_{3}=\left(x_{3}, y_{3}\right)\) lies on the arc \(P_{1} P_{2}\) and the tangent to the curve at \(P_{3}\) is parallel to the chord \(P_{1} P_{2}\), show that \(x_{3}=\left(x_{1}+x_{2}\right) / 2\).

Short Answer

Expert verified
To show that \(x_3 = \frac{x_1 + x_2}{2}\), when the tangent to the curve \(y = ax^2 + bx + c\) at point \(P_3(x_3, y_3)\) is parallel to the chord \(P_1P_2\), we found the slopes of the tangent and the chord, equated them, and solved the resulting equation for \(x_3\). After simplifying, we derived the desired result \(x_3 = \frac{x_1 + x_2}{2}\).

Step by step solution

01

Find the slope of the tangent at P3

Since P3 is on the given curve, we can write y3 = ax3^2 + bx3 + c. The derivative with respect to x of the given curve, y' = 2ax + b, is the slope of the tangent at any point on the curve. So, the slope of the tangent at P3 is 2ax3 + b.
02

Find the slope of the chord P1P2

The slope of a line segment between two points (x1, y1) and (x2, y2) is given by (y2 - y1)/(x2 - x1). Therefore, the slope of the chord P1P2 is (y2 - y1)/(x2 - x1).
03

Equate the slopes of the tangent and the chord

According to the problem statement, the tangent to the curve at P3 is parallel to the chord P1P2. Two lines are parallel if their slopes are equal. Thus, we can equate the slopes derived in steps 1 and 2: 2ax3 + b = (y2 - y1)/(x2 - x1)
04

Solve the equation for x3

To show that x3 = (x1 + x2)/2, we'll solve the above equation for x3: 2ax3 + b = (y2 - y1)/(x2 - x1) Now, substitute the expressions for y1 and y2 using the given curve: 2ax3 + b = (ax2^2 + bx2 + c - (ax1^2 + bx1 + c))/(x2 - x1) Simplifying, we get: 2ax3 + b = (ax2^2 - ax1^2 + bx2 - bx1)/(x2 - x1) Multiply both sides by (x2 - x1): 2ax3(x2 - x1) + b(x2 - x1) = ax2^2 - ax1^2 + bx2 - bx1 Expanding, we get: 2ax3x2 - 2ax3x1 + bx2 - bx1 = ax2^2 - ax1^2 + bx2 - bx1 Subtract bx2 - bx1 from both sides: 2ax3x2 - 2ax3x1 = ax2^2 - ax1^2 Factoring the right-hand side: 2ax3(x2 - x1) = a(x2 + x1)(x2 - x1) Now, divide both sides by a(x2 - x1): 2x3 = x2 + x1 And finally, divide both sides by 2: x3 = (x1 + x2)/2 This is the required result.

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

Let \(f:\left[-\frac{1}{2}, \frac{1}{2}\right] \rightarrow \mathbb{R}\) be given by $$ f(x)=\left\\{\begin{array}{ll} \sqrt{2 x-x^{2}} & \text { if } 0 \leq x \leq \frac{1}{2} \\ \sqrt{-2 x-x^{2}} & \text { if }-\frac{1}{2} \leq x \leq 0 . \end{array}\right. $$ Show that \(f\left(\frac{1}{2}\right)=f\left(-\frac{1}{2}\right)\) but \(f^{\prime}(x) \neq 0\) for all \(x\) with \(0<|x|<\frac{1}{2}\). Does this contradict Rolle's Theorem? Justify your answer.

Let \(C\) be an algebraic plane curve, that is, let \(C\) be implicitly defined by \(F(x, y)=0\), where \(F(x, y)\) is a nonzero polynomial in two variables \(x\) and \(y\) with coefficients in \(\mathbb{R} .\) Let the (total) degree of \(F(x, y)\) be \(n .\) Let \(P=\left(x_{0}, y_{0}\right)\) be a point on \(C\), so that \(F\left(x_{0}, y_{0}\right)=0 .\) (i) If we let \(X:=x-c\) and \(Y:=y-d\) and define \(g(X, Y):=f(x, y)\), then show that \(g(X, Y)\) is a polynomial in \(X\) and \(Y\) with \(g(0,0)=0\). Deduce that there is a unique \(m \in \mathbb{N}\) such that \(m \leq n\) and $$ g(X, Y)=g_{m}(X, Y)+g_{m+1}(X, Y)+\cdots+g_{n}(X, Y) $$ where \(g_{i}(X, Y)\) is either the zero polynomial or a nonzero homogeneous polynomial of degree \(i\), for \(m \leq i \leq n\), and \(g_{m}(X, Y) \neq 0 .\) We denote the integer \(m\) by mult \(_{P}(C)\), and call it the multiplicity of \(C\) at the point \(P\). (ii) Show that a tangent to the curve \(C\) at the point \(P\) is defined (as far as calculus is concerned) if and only if mult \(_{P}(C)=1\). Moreover, if \(\operatorname{mult}_{P}(C)=1\), then there are \(\alpha_{1}, \beta_{1} \in \mathbb{R}\) such that \(g_{1}(X, Y)=\) \(\alpha_{1} X+\beta_{1} Y\), and then the line \(\alpha_{1}(x-c)+\beta_{1}(y-d)=0\) is the tangent to \(C\) at \(P\). (iii) Show that if \(F(x, y)=y-f(x)\) for some polynomial \(f(x)\) in one variable \(x\), then for the corresponding curve \(C\) given by \(F(x, y)=0\) we have mult \(_{P}(C)=1\) for every \(P\) on \(C\). (iv) Determine the integer \(m=\operatorname{mult}_{P}(C)\) and a factorization of \(g_{m}(X, Y)\) when \(P=(0,0)\) and \(C\) is the curve implicitly defined by \(F(x, y):=\) \(y^{2}-x^{2}-x^{3}=0\), or by \(F(x, y):=y^{2}-x^{3}=0\) [Note: In view of Exercise 70 of Chapter 1, the initial form \(g_{m}(X, Y)\) factors as a product of homogeneous linear polynomials, that is, $$ g_{m}(X, Y)=\prod_{i=1}^{m}\left(\alpha_{i} X+\beta_{i} Y\right) \text { for some } \alpha_{i}, \beta_{i} \in \mathbb{C}, 1 \leq i \leq m $$ In the algebraic approach to tangents, the \(m\) (complex) lines given by \(\alpha_{i}(x-c)+\beta_{i}(y-d)=0\) for \(i=1, \ldots, m\), are called the tangent lines to the curve \(C\) at the point \(P .\) ]

Let \(I\) be an interval and \(f: I \rightarrow \mathbb{R}\) be continuous on \(I\) and differentiable at every interior point of \(I\). If there is a constant \(\alpha\) such that \(\left|f^{\prime}(x)\right| \leq \alpha\) for all interior points \(x\) of \(I\), then show that \(f\) is uniformly continuous on \(I\). Is the converse true? In other words, is it true that if \(f: I \rightarrow \mathbb{R}\) is uniformly continuous on \(I\) and differentiable at every interior point of \(I\), then there is a constant \(\alpha\) such that \(\left|f^{\prime}(x)\right| \leq \alpha\) for all interior points \(x\) of \(I ?\)

Let \(f:[a, b] \rightarrow \mathbb{R}\) be such that \(f^{\prime}\) is continuous on \([a, b]\) and \(f^{\prime \prime}\) exists on \((a, b)\). Show that there is \(c \in(a, b)\) such that $$ f^{\prime \prime}(c)[f(b)-f(a)]=f^{\prime}(c)\left[f^{\prime}(b)-f^{\prime}(a)\right] $$

Let \(f(x)\) be a polynomial. A real number \(c\) is called a root of \(f(x)\) of multiplicity \(m\) if \(f(x)=(x-c)^{m} g(x)\) for some polynomial \(g(x)\) such that \(g(c) \neq 0\). (i) Let \(f(x)\) have \(r\) roots (counting multiplicities) in an open interval \((a, b)\). Show that the polynomial \(f^{\prime}(x)\) has at least \(r-1\) roots in \((a, b)\). Also, give an example where \(f^{\prime}(x)\) has more than \(r-1\) roots in \((a, b)\). More generally, for \(k \in \mathbb{N}\), show that the polynomial \(f^{(k)}(x)\) has at least \(r-k\) roots in \((a, b)\). (ii) If \(f^{(k)}(x)\) has \(s\) roots in \((a, b)\), what can you conclude about the number of roots of \(f(x)\) in \((a, b) ?\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

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.