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

  • Thread starter Kinetica
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  • #1
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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?
 

Answers and Replies

  • #2
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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.
 
  • #3
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Excuse me, what exactly do you mean by that?
 
  • #4
Dick
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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?
 
  • #5
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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
 
  • #6
Dick
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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?
 
  • #7
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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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