Question

Please help me with the following problem. I am asked to find a vector equation of the line tangent to the graph of r(t) at the point on the curve.
r(t)=t^2i-1/(t+1)j+(4-t^2)k
P(4,1,0)

Answer 1

\[r'(t)=2ti-\frac{ 1 }{ (t+1)^{2} }j-2tk\]

Answer 2

I found the the derivative of r(t), where do I go from here?

Answer 3

I got an answer but it wasn't quite right according to the book....
here is my answer:
(4i+j)+t(2i-1/(t+1)^2j-2k)

Answer 4

r=(4i+j)+t(how do I obtain these values?)

Answer 5

I think the gradient (the derivative wrt to i,j,k) is perpendicular to the curve

Answer 6

P(4,1,0)
r(t)=t^2i-1/(t+1)j+(4-t^2)k

as Loser said, find the t that gives you point P.
we can choose any of the dimensions, so choose J
-1/(t+1) = 1
-1 = t+1
-2 = t

as a check, if you use t=-2 in r(t) you get 4i + 1j +0 k

Answer 7

(4i+j)+t(2i-1/(t+1)^2j-2k) with t= -2

Answer 8

ok i see

Answer 9

so you don't need to take the derivative of r(t) then?

Answer 10

isn't (2i-1/(t+1)^2j-2k) r'(t) ?

Answer 11

no i don't think so....wouldn't there be a t-variable in both the i and k component?

Answer 12

I copy and pasted your post, I did not check it.

Answer 13

I'm not sure if we are doing this right....the answer in the book is...
r=(4i+j)+t(-4+j+4k)

Answer 14

the derivative i posted has t's in the i and k component

Answer 15

ok, r'(t)= 2 t i - 1/(t+1)^2j -2 t k

evaluate this at t= -2 (this gives you the slope at point P)
you get -4 i + 1 j + 4 k

now you can make an equation using the point P and that vector

Answer 16

Ok, how did you obtain a positive 1 in the j component for the t-terms

Answer 17

if you plug-in (-2) into the derivative of the j-component don't you get (-1) in return?

Answer 18

I guess we have to be careful about minus signs!
r(t)=t^2i-1/(t+1)j+(4-t^2)k

looking at the j component:
\[ \frac{d}{dt} -(t+1)^{-1} = (t+1)^{-2}\]

notice it is positive. now when we put in (-2 + 1)^-2 = (-1)^-2 = 1

Answer 19

ok then i just messed up on taking that derivative, ok thanks...