Finding tan line to curve via implicit dif.

[QuarkCharmer]

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.
$$x^2+2xy-y^2+x=2$$
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?)

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

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

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' :

$$\frac{2(1)+2(2)+1}{2(2)-2(1)} = \frac{2+4+1}{4-2} = \frac{7}{2}$$

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.
$$2=\frac{7}{2}(1)+k$$
$$2-\frac{7}{2} = k$$
$$k = -\frac{3}{2}$$

So the solution is:
$$y=\frac{7}{2}x-\frac{3}{2}$$

?

 

[micromass]

Seems correct Quark!

Well done!

 

[QuarkCharmer]

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.

 

[micromass]

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! )

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.

 

[QuarkCharmer]

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 :(.