How do you find parametric equations for the tangent line to the curve with the...

rzfansubs87

rzfansubs87

Answered

2022-07-06

How do you find parametric equations for the tangent line to the curve with the given parametric equations at the point ( 1 , 4 , 4 ) .
x = cos ( t ) , y = 4 e 8 t , z = 4 e - 8 t

Answer & Explanation

enfeinadag0

enfeinadag0

Expert

2022-07-07Added 16 answers

Step 1
We have parametric equations as functions of t for x, y and z so we can form the derivatives wrt t for each variable:
x = cos t   d x d t = - sin t
y = 4 e 8 t     d y d t = 32 e 8 t
z = 4 e - 8 t d z d t = - 32 e 8 t
In order to find the normal at any particular point in vector space we use the Del, or gradient operator:
f ( t ) = d x d t i ^ + d y d t j ^ + d z d t k ^
                    = - sin t i ^ + 32 e 8 t j ^ - 32 e - 8 t k ^
So for the particular point ( 1 , 4 , 4 ) we need to establish the corresponding value of t, which by inspection is t = 0 .
So for this particular point, the normal vector to the surface is given by:
f ( 0 ) = 0 i ^ + 32 j ^ - 32 k ^
= 32 j ^ - 32 k ^
= 32 ( j ^ - k ^ ) So the tangent line is a line with direction j ^ - k ^ that passes through the point ( 1 , 4 , 4 ) , which therefore has the vector equation:
r = ( 1 4 4 ) + λ ( 0 1 - 1 )
Which we can therefore parametrise (using λ ) as follows
x = 1
y = 4 + λ
z = 4 - λ

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