How do you find parametric equations for the tangent line to the curve with the given parametric equations $x = 7 t^{2} - 4$ $x=7t^2-4$ and $y = 7 t^{2} + 4$ $y=7t^2+4$ and $z = 6 t + 5$ $z=6t+5$ and (3,11,11)?

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Answer 1 · 2015-02-22T18:12:09

The answer is:

$x = 3 + 14 t$ $x=3+14t$
$y = 11 + 14 t$ $y=11+14t$
$z = 11 + 6 t$ $z=11+6t$

The point $(3, 11, 11)$ $(3,11,11)$ is for $t = 1$ $t=1$ , as you can see substituting it in the three equations of the curve.

Now let's search the generic vector tangent to the curve:

$x'=14t$
$y'=14t$
$z'=6$

So, for $t=1$ it is: $vecv(14,14,6)$ .

So, remembering that given a point $P(x_P,y_P,z_P)$ and a direction $vecv(a,b,c)$ the line that passes from that point with that direction is:

$x=x_P+at$
$y=y_P+bt$
$z=z_P+ct$

so the tangent is:

$x=3+14t$
$y=11+14t$
$z=11+6t$

N.B. The direction $(14,14,6)$ is the same that $(7,7,3)$ , so the line could be written also:

$x=3+7t$
$y=11+7t$
$z=11+3t$

that's simplier!

How do you find parametric equations for the tangent line to the curve with the given parametric equations x=7t^2-4x=7t2−4x=7t^2-4 and y=7t^2+4y=7t2+4y=7t^2+4 and z=6t+5z=6t+5z=6t+5 and (3,11,11)?