Finding the Tangent Line at a Point on a Curve

In summary, the conversation discusses finding the tangent line to the curve xe^Y+ye^x=1 at the point (0,1). It includes finding the derivative, setting it equal to 0, and attempting to use natural logarithms to separate the variables. The conversation concludes with the realization that the derivative was already found and can be used to find the slope of the tangent line.
  • #1
Ron Powers
3
0
I need to find the tangent line to the curve xe^Y+ye^x=1 at the point (0,1).
I took the derivative and found to be:
dy/dx=-(ye^x-e^y)/(xe^y+ye^x)

I set that equal to 0 so:
0=-ye^x-e^y

I have tried using a natural log to get y on one side and x on the other, but so far no good. How can I separate the two variables, or was there a mistake in my derivative that I am just not catching?
 
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  • #2
Homework-type questions must be posted in the Homework & Coursework section.
 
  • #3
Ron Powers said:
I need to find the tangent line to the curve xe^Y+ye^x=1 at the point (0,1).
I took the derivative and found to be:
dy/dx=-(ye^x-e^y)/(xe^y+ye^x)

I set that equal to 0 so:
0=-ye^x-e^y
Why?

You have dy/dx, so evaluate it at the point (0, 1) to get the slope of the tangent line at that point.
Ron Powers said:
I have tried using a natural log to get y on one side and x on the other, but so far no good. How can I separate the two variables, or was there a mistake in my derivative that I am just not catching?
 
  • #4
Well I end up having y=-x+lny, and unless I am mistaken I cannot get an equation for the tangent line with that equation. I can find the slope, but I can't really find y=mx+b using that, can I?
 
  • #5
Never mind, I figured it out. I have to take the natural log of the equation first and then find the derivative.
 
  • #6
Ron Powers said:
Well I end up having y=-x+lny, and unless I am mistaken I cannot get an equation for the tangent line with that equation. I can find the slope, but I can't really find y=mx+b using that, can I?

You have an expression for dy/dx in post 1. All you need to do to find the slope of the tangent line at the point (0, 1) is to substitute 0 for x and 1 for 1 in what you have for dy/dx. When you have the slope of a line and a point on the line, it's not hard to find the equation of the line.

Ron Powers said:
Never mind, I figured it out. I have to take the natural log of the equation first and then find the derivative.
Why? You already found the derivative in post 1 (assuming that your work was correct - I didn't check it).
 

FAQ: Finding the Tangent Line at a Point on a Curve

What is a tangent line?

A tangent line is a straight line that touches a curve at only one point, and has the same slope as the curve at that point.

How do you find the equation of a tangent line?

To find the equation of a tangent line, you need to first find the slope of the curve at the point of tangency. This can be done by taking the derivative of the curve at that point. Then, using the slope and the coordinates of the point of tangency, you can use the point-slope form of a line to find the equation of the tangent line.

Why is finding a tangent line important?

Finding a tangent line allows us to approximate the behavior of a curve at a specific point. This is useful in many applications, such as optimization problems, physics, and engineering.

Can a curve have more than one tangent line?

No, a curve can only have one tangent line at a given point. This is because a tangent line must have the same slope as the curve at that point, and a curve can only have one slope at a given point.

What is the difference between a tangent line and a secant line?

A tangent line touches a curve at only one point, while a secant line intersects a curve at two points. Additionally, a tangent line has the same slope as the curve at the point of tangency, while a secant line has an average slope between the two points of intersection.

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