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

umatisi6ar

umatisi6ar

Answered question

2023-03-08

How to find parametric equations for the tangent line to the curvewith the given parametric equations x = 7 t 2 - 4 and y = 7 t 2 + 4 and z = 6 t + 5 and (3,11,11)?

Answer & Explanation

Veronella6nk

Veronella6nk

Beginner2023-03-09Added 6 answers

Answer:
x = 3 + 14 t
y = 11 + 14 t
z = 11 + 6 t
The point (3,11,11) is for t = 1 , as you can see substituting it in the three equations of the curve.
Let us now look for the generic vector tangent to the curve:
x = 14 t
y = 14 t
z = 6
Therefore, for t = 1 it is: v ( 14 , 14 , 6 )
Thus, remembering that given a point P ( x P , y P , z P ) and a direction v ( a , b , c ) the line that passes from that point with that direction is:
x = x P + a t
y = y P + b t
z = z P + c t
Then the tangent is:
x = 3 + 14 t
y = 11 + 14 t
z = 11 + 6 t
N.B. The direction (14,14,6) is the same that (7,7,3), so the line could be written also:
x = 3 + 7 t
y = 11 + 7 t
z = 11 + 3 t

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