cc77e3…: “Owing to the windless calculation”

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The position vector for the curve $C$ in the $(x,y,z)$ -space is given by $\vec {r} (t) = \vector {2t+1, t^2, 3t}$ . Find values for the constants $a$ and $b$ so the curve $C$ lies on the surface $x^2-4y +az = b$ or explain why no such constants exist.

Are there values for $a$ and $b$ for which the curve lies on the surface?

yes no

The values of $a$ and $b$ for which the curve lies on the surface are $a=\answer {-4/3 }$ and $b=\answer { 1 }$ .

Since $\vec {r} (t) = \vector {2t+1, t^2, 3t}$ , we analyze $x(t)^2-4y(t) + az(t) = b.$

$\begin{align*} x(t)^2-4y(t) + z(t) & = \left(\answer{2t+1}\right)^2-4\left(\answer{t^2}\right) +a \left(\answer{3t}\right) \\ &= 4t^2+4t+1 -4t^2+3at \\ & = (4+3a)t+1 \end{align*}$

Note that we need $(4+3a)t+1 = b$ for all $t$ , which can only occur if $4+3a=0$ and $b=1$ .