Skip to main content
Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 4, Problem 4.5.25

Minimum distance Find the point P on the line y = 3x that is closest to the point (50, 0). What is the least distance between P and (50, 0)?

Verified step by step guidance
1
Identify the line equation y = 3x and the point (50, 0) from which we need to find the minimum distance to a point P on the line.
Express the coordinates of point P on the line as (x, 3x) since any point on the line y = 3x can be represented in this form.
Use the distance formula to express the distance D between point P(x, 3x) and the point (50, 0): D = sqrt((x - 50)^2 + (3x - 0)^2).
Simplify the distance formula: D = sqrt((x - 50)^2 + 9x^2). This simplifies to D = sqrt(10x^2 - 100x + 2500).
To find the minimum distance, minimize the expression under the square root, 10x^2 - 100x + 2500, by finding its derivative, setting it to zero, and solving for x. This will give the x-coordinate of point P that minimizes the distance.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
9m

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Distance Formula

The distance formula calculates the distance between two points in a Cartesian plane. For points (x1, y1) and (x2, y2), the distance d is given by d = √((x2 - x1)² + (y2 - y1)²). This formula is essential for determining how far the point P on the line is from the point (50, 0).
Recommended video:
5:04
Converting between Degrees & Radians

Line Equation

The equation of a line describes the relationship between x and y coordinates of points on that line. In this case, the line is given by y = 3x, which indicates that for every unit increase in x, y increases by three units. Understanding this equation helps in identifying the coordinates of point P that lies on the line.
Recommended video:
Guided course
05:14
Equations of Tangent Lines

Optimization

Optimization in calculus involves finding the maximum or minimum values of a function. In this problem, we need to minimize the distance from point P on the line to the point (50, 0). This typically involves using techniques such as taking derivatives and setting them to zero to find critical points.
Recommended video:
10:13
Intro to Applied Optimization: Maximizing Area