/*! 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 17 Find parametric equations of the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find parametric equations of the line that satisfies the stated conditions. The line that is tangent to the circle \(x^{2}+y^{2}=25\) at the point \((3,-4)\).

Short Answer

Expert verified
Parametric equations: \(x = 3 + 4t\), \(y = -4 + 3t\).

Step by step solution

01

Equation of the Circle

The given circle has the equation: \(x^2 + y^2 = 25\). This implies that the circle is centered at the origin \((0, 0)\) with a radius of 5, since \(\sqrt{25} = 5\).
02

Find the Derivative of Circle's Implicit Function

To find the slope of the tangent line, first differentiate the equation of the circle implicitly with respect to \(x\). \(2x + 2y\frac{dy}{dx} = 0\) simplifies to \(\frac{dy}{dx} = -\frac{x}{y}\).
03

Compute the Slope at Point (3, -4)

Substitute the point \((3, -4)\) into the slope formula: \(\frac{dy}{dx} = -\frac{3}{-4} = \frac{3}{4}\). Thus, the slope of the tangent at this point is \(\frac{3}{4}\).
04

Equation of the Tangent Line

Using the point-slope form of the line: \(y = mx + c\), and knowing \(m = \frac{3}{4}\), and passing through \((3, -4)\), the line equation becomes \(y + 4 = \frac{3}{4}(x - 3)\). Simplifying, we get \(y = \frac{3}{4}x - \frac{9}{4} - 4\), or \(y = \frac{3}{4}x - \frac{25}{4}\).
05

Convert to Parametric Form

To express the line in parametric form, let \(x = 3 + 4t\) and \(y = -4 + 3t\). These parametric equations follow since the direction vector of the line based on the slope \(\frac{3}{4}\) is \(\langle 4, 3 \rangle\). Therefore, any point on the line can be expressed as \((3, -4) + t \langle 4, 3 \rangle\).

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 91Ó°ÊÓ!

Key Concepts

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

Tangent Line
A tangent line is a straight line that touches a curve at a single point without crossing over it. Imagine a bicycle wheel; the tangent line would be like the road underneath touching only a specific part of the wheel without going through it. In mathematics, finding the tangent line to a curve at a point means finding a line that has the same slope as the curve does at that point.
The slope of a tangent line is crucial. It represents the rate at which the curve is changing at that specific point. To better understand this, think about how steep a hill is while you’re biking - the steeper the hill, the higher the slope.
  • Tangent lines help us understand curves better.
  • They are used in various applications such as calculating angles and analyzing graphs.
For the circle in the given problem, the tangent line at the point (3, -4) is special because that's where the line balances snugly against the circle.
Circle
A circle is a set of points in a plane that are all the same distance, called the radius, from a certain point, the center. In the given problem, the circle's equation is \(x^2 + y^2 = 25\). This means that every point \((x,y)\) on this circle is exactly 5 units (since \(r = \sqrt{25} = 5\)) from the center at (0,0).
The equation of the circle is derived from the Pythagorean theorem, which relates to distances in a plane. Circles are symmetrical shapes, making them interesting in various mathematics problems.
  • Every point on the circle is equidistant from the center.
  • The circle serves as the basis for defining angles and trigonometric functions.
In this exercise, focusing on the circle helps us pinpoint where the tangent touches and provides a background for further calculations.
Derivative
The derivative is a powerful concept in calculus, representing the slope or rate of change of a function. It tells you how quickly a function is increasing or decreasing at any given point. In simpler terms, if you imagine a rolling ball, the derivative at any point would tell you the ball's speed.
For the given circle's equation \(x^2 + y^2 = 25\), using calculus allows us to derive the slope of the tangent line. Implicit differentiation gives us the derivative \(\frac{dy}{dx} = -\frac{x}{y}\), which can be used to find the slope at a specific point like (3, -4).
  • Derivatives indicate the slope of a function at a particular point.
  • They help in determining maxima, minima, and points of inflection in graphs.
This makes derivatives essential for not only tangent lines but a broad spectrum of mathematical applications.
Implicit Differentiation
Implicit differentiation is a technique used when a function isn’t written explicitly in terms of one variable. This is particularly useful with equations like circles, where both variables, x and y, are intertwined within the equation.
For the circle given as \(x^2 + y^2 = 25\), implicit differentiation helps us find \(\frac{dy}{dx}\), the derivative of y with respect to x. Here’s how it works: we differentiate both sides concerning x, treating y as a function of x. This gives us \(2x + 2y \frac{dy}{dx} = 0\). Solving for \(\frac{dy}{dx}\), we get \(\frac{dy}{dx} = -\frac{x}{y}\).
  • It is useful for handling equations where y cannot be easily isolated.
  • Allows us to find slopes of tangents to curves like circles.
Implicit differentiation is a fantastic tool in calculus that provides access to vital information about complex mathematical relationships.

One App. One Place for Learning.

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

Get started for free

Study anywhere. Anytime. Across all devices.