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.

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- February 21st 2007, 08:17 PMJenny20find 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. - February 21st 2007, 09:00 PMSoroban
Hello, Jenny20!

Quote:

Given: .**r**(t) .= .t²**i**+ t**j**+3t³**k**

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) .= .2t**i**+**j**+ 9t²**k**

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

The parametric equations of the tangent line are:

. . x .= .1 + 2p

. . y .= .1 + p . . for some parameter*p*

. . z .= .3 + 9p

- February 21st 2007, 09:07 PMJenny20
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). - February 21st 2007, 10:16 PMSoroban
Hello again, Jenny!

Quote:

Given: .**r_1**(t) .= .t²**i**+ t**j**+ 3t³**k**

. .and: .**r_2**(u) .= .(u-1)**i**+ ¼u²**j**+ (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) .= .2t**i**+**j**+ 9t²**k**

. . When t = 1: .r'_1(1) .= .2**i**+**j**+ 9**k**

Tangent to r_2: .r'_2(t) .= .**i**+ ½u**j**-**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°