<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).
Wow, does no one know how to do this? I figured I'll post my work, because maybe someone can tell me if I'm approaching it correctly or not.</p>
<p>The point exists where t=-1.
x'=2t, y'=3t^2, z'=5e^(t^3+1)+15t^3e^(t^3+1)
x=2<em>-1=-2, y=3</em>(-1)^2=3, z=-10
-2x+3y-10z= -2<em>-3 + 3</em>0 + -10*-5
-2x+3y-10z=56</p>