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Can you help me?

So far I have turned $y = 7-x^4$ into $\langle1, 1, 0\rangle$ and used it to make the equation $L = (0, 7, -3) + t(1, 1, 0)$. I know this is wrong, but I just don't know what, and I know it has to do with $y = 7-x^4$.

Thank you in advance.

Narasimham
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anon12345
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  • $y=7-x^4$ defines a surface, not a curve, given the implication that this is all in $\mathbb{R}^3$. – 2'5 9'2 Mar 14 '14 at 03:30
  • @alex.jordan yes, yet he is clearly asking for the curve defined by intersecting said surface with the plane $z=-3$ – John C Oct 03 '14 at 13:22

2 Answers2

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Required parametrization of curve parallel to base curve defined by first two coordinates is

$$ x= t , y = 7 -t^4 , z = -3 $$

$x,y $ relation shows that it is a fourth order parabola. Also no $x,y$ is involved in $z$. So it is a surface obtained by extruding the higher order parabola parallelly drawn along $z.$ Such independent two coordinates satisfy not a curve but the full surface, which are called cylinders.

The base parabola curve is shown in green, and the surface generated upto $z=-3$ is yellow.The upper edge $z=-3$ is the curve your question.

enter image description here

Narasimham
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When you choose your curve, $z$ can take any value, so just choose $f(t)=(t, 7-t^4,-3)$. Clearly, the curve is parallel to the xy-plane (a curve is parallel to a plane if it lies in a plane parallel to the plane)...no reparametrization is necessary and certainly we don't need to think about lines through the point $(0,7,-3)$.

John C
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Matt R
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