Textbook Question
In Exercises 41–58, find dy/dt.
y = (t⁻³/⁴ sin(t))⁴/³
Ch. 3 - Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 3, Problem 3.7.49
Normal lines to a parabola Show that if it is possible to draw three normal lines from the point (a, 0) to the parabola x = y² shown in the accompanying diagram, then a must be greater than 1/2. One of the normal lines is the x-axis. For what value of a are the other two normal lines perpendicular?
Verified step by step guidance1
Step 1: Begin by understanding the geometry of the problem. The parabola is given as x = y², and the point (a, 0) lies on the x-axis. A normal line to the parabola is perpendicular to the tangent line at a given point on the curve. The goal is to show that three normal lines can be drawn from (a, 0) to the parabola only if a > 1/2, and to find the value of a for which the other two normal lines are perpendicular.
Step 2: Find the slope of the tangent line to the parabola at any point (x, y). The derivative of x = y² with respect to y is dx/dy = 2y. The slope of the tangent line is therefore 2y. The slope of the normal line, being perpendicular to the tangent line, is -1/(2y).
Step 3: Write the equation of the normal line passing through a point (x₀, y₀) on the parabola. The equation is given by: x - x₀ = (-1/(2y₀))(y - y₀). Substitute x₀ = y₀² (from the parabola equation) into this normal line equation.
Step 4: To find the intersection of the normal line with the point (a, 0), substitute x = a and y = 0 into the normal line equation derived in Step 3. This will yield a relationship between a and y₀. Solve this equation to determine the values of y₀ for which the normal line passes through (a, 0).
Step 5: Analyze the resulting equation to determine the conditions under which three solutions for y₀ exist. This corresponds to three normal lines passing through (a, 0). Show that this is possible only if a > 1/2. Then, determine the specific value of a for which the other two normal lines are perpendicular by examining the slopes of these lines and ensuring their product equals -1.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Normal Line to a Curve
A normal line to a curve at a given point is a line perpendicular to the tangent line at that point. For a curve defined by a function, the slope of the normal line is the negative reciprocal of the slope of the tangent line. In the context of the parabola x = y², finding the normal line involves determining the derivative to get the tangent slope and then using it to find the normal slope.
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Slopes of Tangent Lines
Parabola and its Properties
A parabola is a symmetric curve defined by a quadratic equation. In this case, the parabola is given by x = y², which opens to the right. Key properties include its vertex, axis of symmetry, and the fact that it is symmetric about the x-axis. Understanding these properties helps in analyzing how lines, such as normals, interact with the parabola.
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7:42Properties of Parabolas
Perpendicular Lines
Two lines are perpendicular if the product of their slopes is -1. This concept is crucial when determining conditions under which two normal lines to a curve are perpendicular. In the problem, it involves setting up equations based on the slopes of the normal lines and solving for the parameter a to find when the lines are perpendicular.
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05:13Slopes of Tangent Lines
Related Practice
Textbook Question
Show that the line y = mx + b is its own tangent line at any point (x₀, mx₀ + b).
Textbook Question
In Exercises 83–88, find equations for the lines that are tangent, and the lines that are normal, to the curve at the given point.
(y - x)² = 2x + 4, (6, 2)
Textbook Question
In Exercises 41–58, find dy/dt.
y = tan²(sin³(t))
Textbook Question
Differential Estimates of Change
In Exercises 35–40, write a differential formula that estimates the given change in volume or surface area.
The change in the surface area S = 6x² of a cube when the edge lengths change from x₀ to x₀ + dx
Textbook Question
Implicit Differentiation
In Exercises 43–50, find by implicit differentiation.
xy + 2x + 3y = 1