Thread: find an equation to r(t)

Given : r(t) = t^2 i + t j +3t^3 k

Find an equation for the tangent line to r(t) at (1,1,3).

Thank you very much.

Hello, Jenny20!

Given: .r(t) .= .t²i + tj +3t³k

Find an equation for the tangent line to r(t) at (1,1,3).

The tangent vector is the derivative of r(t):
. . r'(t) .= .2ti + j + 9t²k

At t = 1: .r'(1) .= .2i + j + 9k .= .< 2,1,9 >

The parametric equations of the tangent line are:
. . x .= .1 + 2p
. . y .= .1 + p . . for some parameter p
. . z .= .3 + 9p

Hi Soroban,

Thank you very much.

What about this part? How should I solve it?

Given : rsub1(t) =t^2i + tj + 3t^3k
and rsub2(t) = (t-1)i + (1/4)t^2j + (5-t)k.

Find cos theta , where theta is an acute angle between the tangent lines to rsub1 and rsub2 at (1,1,3).

Hello again, Jenny!

Given: .r_1(t) .= .t²i + tj + 3t³k
. .and: .r_2(u) .= .(u-1)i + ¼u²j + (5 - u)k

Find cos θ, where θ is an acute angle between the tangent lines to r_1 and r_2 at (1,1,3).

I changed the parameter in r_2 to u to avoid confusion.

When t = 1: .r_1(1) = (1,1,3)

When u = 2: .r_2(2) = (1,1,3)


Tangent to r_1: .r'_1(t) .= .2ti + j + 9t²k
. . When t = 1: .r'_1(1) .= .2i + j + 9k

Tangent to r_2: .r'_2(t) .= .i + ½uj - k
. . When u = 2: .r'_2(2) .= .i + j - k


The angle θ between two vectors u and v is given by:

. . . . . . . . . . | u·v |
. . cos θ . = . ---------
. . . . . . . . . . |u| |v|


The numerator is: .|u·v| .= .|<2,1,9>.<1,1,-1>| .= .|2 + 1 -9| .= .6

The denominator is: .√(2² + 1² + 9²) .= .√86
. . and: .√[1² + 1² + (-1)²] .= .√3

. . . . . . . . . . . . . . . . . . . . 6
And we have: . cos θ . = . ------- . = . 0.373543684
. . . . . . . . . . . . . . . . . . .√258

Therefore: .θ .≈ .68°