Question

Find the tangent line of the curve defined by the vector equation at the point t=0. u(t)=(1+t^3) i+te^(-t) j+e^(2t) k

Answer 1

the tangent line of a vector defines function is the derivatives of each component

Answer 2

<3t^2, -t e^-t + e^-t, 2e^2t> if i see it right

Answer 3

at t=0 we got:

<0,0,2> right? plug that into the point to get the line

Answer 4

x = 1+t^3 ; x' = 3t^2
y = t e^(-t) ; y' = -t e^(-t) + e^(-t)
z = e^(2t) ; z' = 2 e^(2t)

at t=0 we get the point (1,0,1)

the tangent vector at t=0 is: n<0,1,2>

Answer 5

the line equation in then:

x=1
y=1n
z=1 +2n

where n is the scalar for the tangent vector

Answer 6

do youagree?

Answer 7

but the point should be <0,0,1>

Answer 8

hmm;

x = 1+t^3 = 1 + (0) =1

y = t e^(-t) = 0(e^...) = 0

z = e^(2t) = e^0 = 1

point (1,0,1) ; how do we get x=o?

Answer 9

ohhh, you are right