- #1
flyingpig
- 2,579
- 1
Homework Statement
Find the parametric equations of the tangent to the curve defined
[tex]r = <0, 2 + t^2, 3 + 2t - t^3>[/tex] at the point [tex](0,2,3)[/tex]
The Attempt at a Solution
a) [tex]r'(0) = <0, 0, 2>[/tex]
[tex]r = <0, 2 + t^2, 3 + 2t - t^3>[/tex]
[tex]r' = <0, 2 + 2t, 2 - 3t^2>[/tex]
[tex]t = 0[/tex]
[tex]r'(0) = <0,0,2>[/tex]
[tex]r' = <0, 2 + 2t, 2 - 3t^2>[/tex]
[tex]t = 0[/tex]
[tex]r'(0) = <0,0,2>[/tex]
r = <0, 2 + t^2, 3 + 2t - t^3>
y = 2 + t^2 = 2
z = 3 + 2t - t^3 = 3
t = 0
Let y = f(z)
y' = \frac{2t}{2 - 3t^2}
y'(0) = 0
y - 2 = 0(z - 3) = 2[/tex]
Hence it is y = 2 or q(s) = <0,2,0>
y = 2 + t^2 = 2
z = 3 + 2t - t^3 = 3
t = 0
Let y = f(z)
y' = \frac{2t}{2 - 3t^2}
y'(0) = 0
y - 2 = 0(z - 3) = 2[/tex]
Hence it is y = 2 or q(s) = <0,2,0>
c) [tex]p(r) = <0,2,3> + r<0,2,0>[/tex]
[tex]r'(0) = <0, 0, 2>[/tex]
I think a) is wrong because it is a point, not a point.
I can't decide between b) and c) because they both look correct with their reasoning. I lean towards c) though
r = <0, 2 + t^2, 3 + 2t - t^3>
r' = <0, 2 + 2t, 2 - 3t^2>
t = 0
r'(0) = <0,0,2>
p(r) = <0,2,3> + r<0,2,0>
r' = <0, 2 + 2t, 2 - 3t^2>
t = 0
r'(0) = <0,0,2>
p(r) = <0,2,3> + r<0,2,0>
I can't decide between b) and c) because they both look correct with their reasoning. I lean towards c) though