Find parametric equations for the tangent line to the curve with the given parametric equation s at the specified point. x=t^(2)+1, y=4sqrt(t), z=e^(t^(2)-t);(2,4,1)

Neveah Salazar

Neveah Salazar

Open question

2022-08-18

Find parametric equations for the tangent line to the curve with the given parametric equation s at the specified point.
x=t2+1,y=4t,z=et2t;(2,4,1)

Answer & Explanation

Cruz Marshall

Cruz Marshall

Beginner2022-08-19Added 6 answers

Frist we need to find the tangent vector r'(t)
We have
r(t)=t2+1i+4tj+et2tk
Differentiate to get the tangent vector
r(t)=2ti+2tj+(2t1)et2tk
Note that the point (2,4,1) corresponds to t=1
Since as we put t=1 in r(t), we get (2,4,1) as the position.
r(1)=21i+21j+(211)e121k
r'(1)=2i+2j+k
Remember that: Equation of a line passing through a point with position vector a, and parallel to the vector b is
r(t)=a+t b
Hence equation of the tangent line at (2,4,1) is
r(t)=2,4,1+t2,2,1
r(t)=2+2t,4+2t,1+t
Hence the parametric equations of the line are
x=2+2t, y=4+2t, z=1+t
Result:
The required equation for tangent vector is x=2+2t, y=4+2t, z=1+t.

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