<p>Hey guys, this is not my hw, don't worry. This is one of the problems our teacher gave us as an example of one that will be on the AP test.</p>
<p>dy/dx=(4x-2xy)/(x^2+y^2+1)</p>
<p>Question: the line y=-x is tangent to the curve at point B. Find the coordinates of point B.</p>
<p>I had solved for that derivate from a previous equation (I'm fairly sure it's right), but I don't know how to do this part, our book stinks at explaining it. Anyone care to explain. Thanks.</p>
<p>Well, the slope of the tangent is -1, so that is what the derivative equals. Since the tangent line, y=-x, touches the curve, the point you are looking for must have coordinates that fit the tangent equation.</p>
<p>Hope this helps.</p>
<p>Okay... Hmmm.</p>
<p>So I set up the equation like this:</p>
<p>-1=(4x-2xy)/(x^2+y^2+1)</p>
<p>Then I tried guess and check with the points (-2,2),(-1,1),(0,0),(1,-1)...yada yada.</p>
<p>But it never reaches -1. So a set of points does not exist?</p>
<p>Thanks for the help, averagemathgeek :).</p>
<p>Well if worse comes to worse, just bust out a TI calculator and graph the two equations and find the point.</p>
<p>Yeah, I tried that. But I wasn't able to do the algebra to get it too look like: y=something because of the multiple y-values, which means I can't put it on the calculator (I have a ti-83).</p>
<p>-(4x-2xy)= x^2+y^2+1</p>
<p>-4x-1= (x-y)^2</p>
<p>x-sqrt(-4x-1)</p>
<p>i mean plug it into the derivative</p>
<p>x-sqrt(-4x-1)=y must also agree with plugging in x to your old equation (not the derivative one but the integral of this) and then getting y.</p>
<p>Oh wow, great algebra!...but I don't understand what you mean by plug it into the derivate and the integral? I don't get how you get the coordinates? </p>
<p>Sorry, I'm new at this calculus stuff. The first two chapters were really easy, but now it's getting difficult.</p>
<p>well you have two equations now:</p>
<p>x-sqrt(-4x-1)=y</p>
<p>(your original equation, that you didnt tell us)=y</p>
<p>solve them simultaneously.<br>
i think</p>
<p>Ehhhh, the original equation looked like this:</p>
<p>2y^3+6x^2*y-12x^2+6y=1</p>
<p>Is that possible? I don't even know how to do that on the calculator.</p>
<p>Wait a second, the question i'm trying to answer is: find the point on line y=-x where it's tangent to the curve. This restrics me to a limited set of points, and I tried those on the equation you found, and they don't work, or the original.</p>
<p>x-sqrt(-4x-1)=-x</p>
<p>sqrt(-4x-1)=2x</p>
<p>4x^2+4x+1=0</p>
<p>{-4+/-sqrt(0)}/8
doesnt work</p>
<p>Well, thanks for all the help Togos. I'm gonna sleep on it, and talk with my teacher before school starts. You lost me somewhere. Thanks though.</p>
<p>Okay this is how to do it:</p>
<p>dy/dx=(4x-2xy)/(x^2+y^2+1)
At B (the point you are looking for), y=-x and dy/dx=-1
-1=[4x-2x*(-x)]/[x^2+(-x)^2+1]
-2x^2-1=4x+2x^2
4x^2+4x+1=(2x+1)^2=0
x=-1/2</p>
<p>It seems that everybody missed the big clue: y=-x.</p>
<p>Hope this helps.</p>
<p>again, your way is easier to understand than mine, but as you see above my solution was also -1/2, i accidently put it DOESNT work when it does. a couple of posts up</p>
<p>x=-1/2 and y=1/2 work just fine.</p>
<p>I just saw that he/she gave us the original equation: 2y^3+6x^2*y-12x^2+6y=1. (-1/2,1/2) works in there.</p>
<p>Yeah thats right, averagemathgeek got to it before i could post it. good method.</p>