<p>So I have this problem and I thought I knew how to solve it but I tried and obviously I didn't. </p>
<p>So I have to show that the curve of the intersection of the surface x^2+2y-z^2+3x=1 and 2x^2+4y^2-2z^2-5y=0 lie in a plane. How do I go about doing this? </p>
<p>Thanks!</p>
<p>hmmm... 20 people viewed and no one knows?!?!</p>
<p>53 people... I know that you're all in college and I'm sure most of you have taken calc...</p>
<p>just because a lot have probably taken it doesn't mean they want to do calc in their free time</p>
<p>set the left hand equation to zero, set the two equations equal to each other, collect terms and go from there, i suppose</p>
<p>probably because the problem is multivariable calc. i'm not there yet, and i don't think a lot of people go up to multivariable calc.</p>
<p>feenotype... I tried that. Thanks for the try.</p>
<p>its been a long time since calculus, but I'd try adding 1 to the right equation on both sides. Now they're equal, so put the terms from each in a new equation and solve. Pray that when you reduce the terms you end up with something that is in a plane (eg. has only 2 of the variables {x,y,z} instead of 3). A function of 2 variables must lie in a plane.</p>
<p>no it doesnt. x^2+y^2=4, graphed in 3d, is an infinitely long cylinder (in the z-direction) with radius 2</p>
<p>I think feenotype was onto the right method... set them both equal to zero (and then set them equal to eachother).. collect/simplify and then the resulting equation should be the equation of a plane or something 2 dimensional that fits in a plane</p>
<p>Remember that a plane is defined by a point and an orthogonal vector.</p>
<p>You also might check the SOS math cyberboard for help. </p>
<p>And don't forget to cite any help that you end up using. Academic honesty requires that.</p>
<p>I did what feenotype said and got a decently messy answer. Traditionally, our teacher doesn't assign messy answers because we're not allowed to use a calculator and he wants us to understand the concept and not fret over the general math part. Well, I turned in what I think was the answer but it was pretty messy numbers. I asked a bunch of kids before class and I didn't find one who really knew what they were doing. I'm not sure if what I did was right, but hey, I tried.</p>
<p>Thanks to those who provided input.</p>
<p>If a curve lies in a 3-D plane, its equation should be z = (linear function of x and y)
i.e. z = ax + by + c (or equivalently z^2 = (ax + by + c)^2 )</p>
<p>You can try rewriting both given equations as z^2 = RHS1 and z^2 = RHS2, set RHS1=RHS2 and come up with some relationship between x & y. Then try using this relationship to re-write one of the original equations in the form z^2 = (ax + by + c)^2 . </p>
<p>Are you sure you have the original equations right? They lead to a <em>messy</em> relationship between x & y after eliminating z^2 ...</p>
<p>Unless they're printed wrong in the textbook, I have them right. I've emailed my TA and now that we've turned in the work, she'll give the answers and explanation. Thanks.</p>
<p>When you have a chance, please post the answer here. I'm curious :).</p>
<p>I think everyone is mistaken here. The result of two surfaces intersecting is NOT A PLANE, but rather a CURVE in space as the problem already indicates. So you have to find the eq of the curve and show that it lies on a plane, that is...the x, y, and z components of the curve/vector are linear equations. How do u do that? I don't know lol.</p>
<p>x^2+2y-z^2+3x=1 and
2x^2+4y^2-2z^2-5y=0</p>
<p>multiply the top equation by -2 on both sides
add the first equation to the second by elimination to get </p>
<p>-9y+4y^2-6x+0z=-2
because z=0, every point in the equation must lie in the two-dimensional xy plane</p>
<p>no.</p>
<p>the coefficient of z is 0. that means nothing. z doesnt have to be 0. In fact, like i said before, z can take on an infinite number of values once its coefficient becomes zero.</p>
<p>think about this: 0*x + y = 2 <===> y=2 (aka x can take any value at all when you graph this line in 2d).</p>
<p>im also interested in the answer.</p>
<p>I've done this problem three different time in three different sittings and I keep getting something that is not a space curve as the problem contends I'm supposed to.</p>
<p>I keep getting </p>
<p>(x-3/2)^2+4(4-7/2)^2-z^2=69/16</p>
<p>which is the quadric equation for a hyperboloid of one sheet.</p>
<p>graphing the three functions, it doesn't look like that is their intersection (though they both sort of taper off from different parts of it), but I haven't rotated the view so it might just be a deceptive angle</p>