<p>Hey. I was wondering if anyone knows how to solve this problem:</p>
<p>Given that the curve C is defined by x=t^2-4, y=t^3+1,z=5te^(t^3+1), write an equation (in rectangular form and with integral coefficients and constants) for the normal plane to C at P (-3,0,5). Thanks!</p>
<p>Edit: The point is (-3,0,-5) not (-3,0,5).</p>
<p>Michael Viscardi could solve this.</p>
<p>hahaha^^^</p>
<p>yes, he probably could, but i bet he would have a hard time trying to explain it....kiddin....</p>
<p>Here you go: Motion in Space Solution PDF</p>
<p>"I formed a theorem which characterizes all such domains for which the resulting solution is rational namely in terms of their Riemann maps and their Bergman kernels," he said.
holy hell that dude is smart.....looks like a tool though...</p>
<p>He's actually seriously the nicest kid ever. So modest and sweet. By talking to him you'd only get the slightest hint of what a genius he is. He doesn't actively brag or show it.</p>
<p>^^ go ahead and ask the question</p>
<p>Find the tangent to C at that point (which is easy, assuming you can differentiate each component of C). That gives you a vector tangent to C, so use that vector as the normal to the plane you're constructing. You know the normal to the plane and a point that it passes through, so you can get the equation of the plane in any form you want.</p>
<p>Since this math problem seems to be fooling everyone, thought it would be a good idea to do this. </p>
<p>Maybe there are a few math errors, but I think you can see how it works with this explanation. It really isn't that hard.</p>
<ol>
<li><p>Find the t that gets (-3, 0, -5). Use y because its the easiest to solve for.
0 = t³ + 1
-1 = t³
-1 = t</p></li>
<li><p>Find the slope of x, y, and z by taking derivatives. Z requires the use of the product and chain rules.
x' = 2t - 4 = -2 - 4 = -6
y' = 3t² + 1 = 3 + 1 = 4
z' = 5e^(t³+1) + 5te^(t³+1)(3t²) = (5 + 15t³)(e^(t³+1)) = (5 - 15)(e^(-1+1)) = -10(e^0) = -10</p></li>
<li><p>Find what is perpendicular using the definition of the dot product. Make sure (-3, 0, 5) is included as well.
P = (x, y, z)
((x, y, z) - (-3, 0, -5)) · (-6, 4, -10) = 0</p></li>
<li><p>Use the distributive property to simplify. Divide both sides by 2 if you want.
(x, y, z) · (-6, 4, -10) + (3, 0, 5) · (-6, 4, -10) = 0
-6x + 4y - 10z -18 - 50 = 0
-6x + 4y - 10z = 68
-3x + 2y - 5z = 34</p></li>
</ol>
<p>floyisgod.. I think you'll find that you are wrong.. I've already posted a full solution above. <a href="http://www.rumchev.com/Motion%20in%20Space.pdf%5B/url%5D">http://www.rumchev.com/Motion%20in%20Space.pdf</a></p>
<p>lol take a look at this guy</p>
<p>I assure you he may be winning scholarship cash but sure aint winning any dates :]</p>
<p>Hey!! Come on now, Michael's awesome :)</p>
<p>vrumchev, i did the problem and i got 2x - 3y + 10z + 56 = 0. I believe in your notes you forgot to multiply your -10z by -1.</p>
<p>
[quote]
I assure you he may be winning scholarship cash but sure aint winning any dates :]
[/quote]
</p>
<p>Ha ha, yeah. </p>
<p>btw: pwn3d.</p>
