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

misurrosne

misurrosne

Answered question

2022-06-23

How do you find parametric equations for the tangent line to the curve with the given parametric equations at the specified point x = 1 + 10 t , y = t 5 - t , and z = t 5 + t ; (11, 0, 2)?

Answer & Explanation

nuvolor8

nuvolor8

Beginner2022-06-24Added 32 answers

Step 1
Find the value of t that creates the point (11,0,2)
Substitute 11 for x:
11 = 1 + 10 t
10 t = 10
t = 1
check that t = 1 works for the y and z values:
1 5 - 1 = 0
1 5 + 1 = 2
Step 2
These check, t = 1
Find the tangent vector:
d x d t = 5 t
d y d t = 5 t 4 - 1
d z d t = 5 t 4 + 1
The tangent vector for all points, v ¯ ( t ) , is:
v ¯ ( t ) = { 5 t } i ^ + { 5 t 4 - 1 } j ^ + { 5 t 4 + 1 } k ^
We are interested in the tangent vector at the given point:
v ¯ ( 1 ) = 5 i ^ + 4 j ^ + 6 k ^
The vector equation of the tangent line is
( x , y , z ) = ( 11 , 0 , 2 ) + s ( 5 i ^ + 4 j ^ + 6 k ^ )
The parametric equations for this line are:
x = 11 + 5 s
y = 4 s
z = 2 + 6 s

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