Chapter 11: Problem 17
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 \((3, -4) + t \langle 4, 3 \rangle\).
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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.
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.
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.
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.
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).
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.
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 \(\frac{dy}{dx} = -\frac{x}{y}\).
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.