How do I find the shortest distance between the curves y=x^2 and y=ln(x)? Note: I've already asked this question once, but whoever answered it thought that the 2 points should have had the same value of x. When visualizing this problem with a graphing calculator, it's clear to see that the 2 points on the curves for which the distance is the shortest do not have the same value of x. Hence ,the distance function should consist of 2 variables. I believe that in order to solve this problem there should be a use of partial derivatives. May someone help me get this right this time? Thanks
How do I find the shortest distance between the curves y=x^2 and y=ln(x)? Note: I've already asked this question once, but whoever answered it thought that the 2 points should have had the same value of x. When visualizing this problem with a graphing calculator, it's clear to see that the 2 points on the curves for which the distance is the shortest do not have the same value of x. Hence ,the distance function should consist of 2 variables. I believe that in order to solve this problem there should be a use of partial derivatives. May someone help me get this right this time? Thanks
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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How do I find the shortest distance between the curves y=x^2 and y=ln(x)?
Note: I've already asked this question once, but whoever answered it thought that the 2 points should have had the same value of x.
When visualizing this problem with a graphing calculator, it's clear to see that the 2 points on the curves for which the distance is the shortest do not have the same value of x. Hence ,the distance function should consist of 2 variables. I believe that in order to solve this problem there should be a use of partial derivatives.
May someone help me get this right this time?
Thanks
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