vector problems

• Sep 11th 2008, 07:44 PM
iwonder
vector problems
1] how to find the unit tangent vector T(t)at the indicated point of the vector function r(t)=(e^(8t) cos(t)) i + (e^(8t)sin(t)) j + (e^(8t)) k ,what is $T(pi/2)$ ?

>> i am not sure, but i kind of think that it should be derivative and divided by the magnitude.. but somehow it cannot be done, please help!!!

And also 2] how to find parametric equations for the tangent line at the point $(cos(3pi/6),sin(3pi/6),(3pi/6))$ on the curve $x = cos t, y = sin t ,z = t$?

>> i totally get lost from the question.. can someone please help? :)
• Sep 12th 2008, 03:04 AM
mr fantastic
Quote:

Originally Posted by iwonder
1] how to find the unit tangent vector T(t)at the indicated point of the vector function r(t)=(e^(8t) cos(t)) i + (e^(8t)sin(t)) j + (e^(8t)) k ,what is $T(pi/2)$ ?

>> i am not sure, but i kind of think that it should be derivative and divided by the magnitude.. but somehow it cannot be done, please help!!!

[snip]

A vector tangent to the curve defined by r(t) is $T(t) = \frac{dr}{dt}$. Substitute $t = \frac{\pi}{2}$ and divide the resulting vector by its magnitude.
• Sep 12th 2008, 03:12 AM
mr fantastic
Quote:

Originally Posted by iwonder
[snip]
And also 2] how to find parametric equations for the tangent line at the point $(cos(3pi/6),sin(3pi/6),(3pi/6))$ on the curve $x = cos t, y = sin t ,z = t$?

>> i totally get lost from the question.. can someone please help? :)

A vector tangent to the curve has i, j and k components given respectively by $\frac{dx}{dt} = - \sin t$, $\frac{dy}{dt} = \cos t$, $\frac{dz}{dt} = 1$.

So at $t = \frac{3 \pi}{6} = \frac{\pi}{2}$ a vector in the direction of the tangent is ......

And a point on the tangent is obviously $\left( \cos \frac{\pi}{2} = 0, \, \sin \frac{\pi}{2} = 1, \, \frac{\pi}{2} \right)$.

So you have a vector in the direction of the line and a point on the line. You should know how to get the parametric equations of the line from these two things.