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

In summary, the problem asks for the equation of a line perpendicular to the tangent line to the curve y=(3x+1)/(4x-2) at the point (1,2). To find this, the first step is to find the slope of the tangent line, which is done by taking the derivative of the curve and substituting x=1. The slope of the tangent line is -10/4 and the slope of the perpendicular line is the negative reciprocal, 4/10. The equation of the perpendicular line is then found using the point-slope formula.
  • #1
Kinetica
88
0

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

Question 1: What is a tangent line and why is it important?

A tangent line is a line that touches a curve at only one point and has the same slope as the curve at that point. It is important because it helps us understand the rate of change of a curve at a specific point and can be used to approximate the curve's behavior near that point.

Question 2: How do you find the slope of a tangent line?

The slope of a tangent line can be found using the derivative of the curve at the desired point. The derivative represents the rate of change of the curve and when evaluated at a specific point, it gives the slope of the tangent line at that point.

Question 3: What is a perpendicular line and how is it related to a tangent line?

A perpendicular line is a line that forms a 90-degree angle with another line. In the context of a tangent line, a perpendicular line would intersect the tangent line at a right angle. This is important because a perpendicular line to a tangent line will have a slope that is the negative reciprocal of the tangent line's slope.

Question 4: How do you find the equation of a line that is perpendicular to a tangent line?

To find the equation of a line that is perpendicular to a tangent line, you will first need to find the slope of the tangent line using the derivative. Then, take the negative reciprocal of that slope and use it as the slope of the perpendicular line. Next, use the point of tangency and the new slope to find the equation of the perpendicular line using the point-slope form.

Question 5: Can there be more than one line perpendicular to a tangent line?

Yes, there can be more than one line perpendicular to a tangent line. This is because a tangent line can intersect a curve at multiple points, and each point will have a different slope and therefore a different perpendicular line.

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