Finding tan line to curve via implicit dif.

In summary, the homework statement asks for the equation of a tangent line to the curve at the given point. Through point p(1,2), x^2+2xy-y^2+x=2 can be solved for y. The derivative of y(y') is found to be 2x+2y+1=-y'(2y-2x) which gives y=\frac{7}{2}x-\frac{3}{2}.
  • #1
QuarkCharmer
1,051
3

Homework Statement


Stewart Calculus (6E) 3.6 question #26
Using implicit differentiation, find the equation of a tangent line to the curve at the given point.
[tex]x^2+2xy-y^2+x=2[/tex]
Through point p(1,2)

I just want to know if I am doing this correctly.

Homework Equations



The Attempt at a Solution


I start with the original equation, and find the derivative with respect to x. (I am assuming x right?)

[tex]x^2+2xy-y^2+x=2[/tex]
[tex]\frac{d}{dx}x^2+2xy-y^2+x=2[/tex]
[tex]2x+2(y+xy')-2yy'+1=0[/tex]
[tex]2x+2y+2xy'-2yy'+1=0[/tex]

Now I isolate the derivative of y(y') and solve for it:
[tex]2x+2y+1=y'(2y-2x)[/tex]
[tex]y'=\frac{2x+2y+1}{2y-2x}[/tex]

So, to get the derivative (slope of tangent) at that point, I put that x/y coordinate in for the derivative equation and get y' :

[tex]\frac{2(1)+2(2)+1}{2(2)-2(1)} = \frac{2+4+1}{4-2} = \frac{7}{2}[/tex]

Now, having the slope of the tangent line at that point on the original curve, I can just stick it in the equation of a line with the coordinate points.
[tex]2=\frac{7}{2}(1)+k[/tex]
[tex]2-\frac{7}{2} = k[/tex]
[tex]k = -\frac{3}{2}[/tex]

So the solution is:
[tex]y=\frac{7}{2}x-\frac{3}{2}[/tex]

?
 
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  • #2
Seems correct Quark! :smile:

Well done!
 
  • #3
Thank you.

When I check the dy/dx on my calculator (TI-89) via:
d(x^2+2xy-y^2+x=2,x)
it returns 2x+1=0 ??

and I can't figure out how to get the original equation in terms of y= +/- something to graph it with my tangent line to see if it looks right. How can I verify this solution via other means?

Also, how do I know they want the dy/dx and not a dx/dy ? There is no indication in the problem.
 
  • #4
QuarkCharmer said:
Thank you.

When I check the dy/dx on my calculator (TI-89) via:
d(x^2+2xy-y^2+x=2,x)
it returns 2x+1=0 ??

and I can't figure out how to get the original equation in terms of y= +/- something to graph it with my tangent line to see if it looks right. How can I verify this solution via other means?

Let me present you one of the best free graphing programs out there:

http://www.padowan.dk/graph/

Install it, then go to function -> add a relation (or something like that) or type F6.
This way you can add the implicit relation.

Then go to function -> add a function (or type Ins) and add the tangent line.

You'll see that the line is a perfect tangent at the curve!

(Also, your calculator is weird! :biggrin: )


Also, how do I know they want the dy/dx and not a dx/dy ? There is no indication in the problem.

This shouldn't really matter. If you try it the other way, then I'm pretty sure you'll end up with the same answer.
 
  • #5
I used that program to verify the solution, that won't do me much good on a test though!

Thanks for the help! I am sure I will post many more questions, it's the weekend :(.
 

Related to Finding tan line to curve via implicit dif.

1. What is the purpose of finding the tan line to curve?

The purpose of finding the tan line to curve is to determine the slope of the tangent line to a curve at a specific point. This can be useful in solving problems in calculus, physics, and engineering, among other fields.

2. How is the tan line to curve found via implicit differentiation?

The tan line to curve can be found by first taking the derivative of the implicit function, which will yield the slope of the tangent line. Then, the x and y values of the point of interest are plugged into the derivative to find the slope at that specific point.

3. Can implicit differentiation be used for any type of curve?

Yes, implicit differentiation can be used for any type of curve, as long as the equation is written in implicit form. This means that the equation does not explicitly express y as a function of x, but rather in terms of both x and y.

4. Is it possible to find multiple tangent lines to a curve using implicit differentiation?

Yes, it is possible to find multiple tangent lines to a curve using implicit differentiation. This is because the derivative of an implicit function is a function of both x and y, allowing us to find the slope at any point on the curve.

5. What are some real-life applications of finding the tan line to curve via implicit differentiation?

Finding the tan line to curve via implicit differentiation has many real-life applications, such as in physics for determining the acceleration of an object, in engineering for optimizing the design of a structure, and in economics for analyzing the marginal cost of production. It is also commonly used in computer graphics for creating smooth curves in animations and video games.

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