How do you find parametric equations for the tangent line

Sandra Allison 2021-12-16 Answered
How do you find parametric equations for the tangent line to the curve with the given parametric equations x=7t24 and y=7t2+4 and z=6t+5 and (3,11,11)?
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Expert Answer

Bob Huerta
Answered 2021-12-17 Author has 41 answers

The answer is:
x=3+14t
y=11+14t
z=11+6t
The point (3,11,11) is for 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'=14
y'=14
z=6
So, for t=1 it is v(14,14,6)
So, remembering that given a point P(xp,yp,zp) and a direction v(a,b,c) the line that passes from that point with that direction is:
x=xp+α
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 simpler!

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Mary Goodson
Answered 2021-12-18 Author has 37 answers
I got it. thanks

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