Find equation of line that is perpendicular to the tangent line to the curve

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Homework Help Overview

The problem involves finding the equation of a line that is perpendicular to the tangent line of the curve defined by y=(3x+1)/(4x-2) at the point (1,2). The subject area pertains to calculus, specifically the concepts of derivatives and slopes of tangent and normal lines.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the process of finding the derivative to determine the slope of the tangent line at a specific point. There is confusion regarding the necessity of setting the derivative equal to zero and the relationship between the slopes of the tangent and normal lines.

Discussion Status

The discussion has progressed with some participants providing guidance on how to find the slope of the tangent line by substituting x=1 into the derivative. There is acknowledgment of the relationship between the slopes of the tangent and normal lines, with some participants confirming their understanding of the negative reciprocal relationship.

Contextual Notes

Participants are navigating through the steps of differentiation and the implications of the results, with some expressing confusion about the initial approach to the problem. There is an emphasis on clarifying the definitions and relationships involved in the problem.

Kinetica
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Homework Statement



Find the equation of the line that is perpendicular to the tangent line to the curve, y=(3x+1)/(4x-2) at the point (1,2)

Homework Equations





The Attempt at a Solution


I am absolutely confused with this problem. I tried taking a derivative of the equation. And I got y'=-10/(4x-2)2
I couldn't set it equal to 0, it will not work. What have I done wrong?
 
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There's no need to solve -10/(4x-2)2 = 0 because you're not supposed to find when/if the tangent line to the curve is ever horizontal. The problem wants you to first find the slope of the tangent line at the point (1, 2); you need to find the derivative when x = 1.
 
Excuse me, what exactly do you mean by that?
 
y'(x) is the slope of the tangent line. Substitute x=1 into your equation for y'(x) to find the slope of the tangent line. What's y'(1)? Now what's the relation between the slope of the tangent line and a perpendicular to it?
 
Dick said:
y'(x) is the slope of the tangent line. Substitute x=1 into your equation for y'(x) to find the slope of the tangent line. What's y'(1)? Now what's the relation between the slope of the tangent line and a perpendicular to it?

OK, here we go:

y'(1)=-10/4

Normal line is reciprocal of the tangent, which is 4/10.
The normal line equation is:

y-2=0.4(x-1)
y=0.4x+1.6
 
Kinetica said:
OK, here we go:

y'(1)=-10/4

Normal line is reciprocal of the tangent, which is 4/10.
The normal line equation is:

y-2=0.4(x-1)
y=0.4x+1.6

Uh, NEGATIVE reciprocal of the tangent. Works for me. Seem ok to you?
 
Dick said:
Uh, NEGATIVE reciprocal of the tangent. Works for me. Seem ok to you?

That's what I exactly meant :))
Thank you SO much for everything.
 

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