Help with vector functions

**khuezy** · November 14th 2008, 05:10 PM

Consider the vector function r(t) =< 4t + 1, t2 − 3t, cos(t) >.
(a) Find the unit tangent vector T at (1, 0, 1)

I know that T = r'(t)/|r-(t)|
and r'(t) = < 4, 2t-3, -sin(t)>

I'm not sure why the book plugs in 0 for t.
Thanks in advance

**Mathstud28** · November 14th 2008, 05:28 PM

Originally Posted by khuezy

Consider the vector function r(t) =< 4t + 1, t2 − 3t, cos(t) >.
(a) Find the unit tangent vector T at (1, 0, 1)

I know that T = r'(t)/|r-(t)|
and r'(t) = < 4, 2t-3, -sin(t)>

I'm not sure why the book plugs in 0 for t.
Thanks in advance

Ok $\bold{r}(t)=\left\langle4t+1,t^2-3t,\cos(t)\right\rangle$

So as you noted $\bold{T}(t)=\frac{\bold{T}'(t)}{|\bold{T}'(t)|}$

So we can compute this to be

$\bold{T}(t)=\frac{\langle4,2t-3,-\sin(t)\rangle}{\sqrt{16+\left(2t-3\right)^3+\sin^2(t)}}$

So $\bold{T}(1,0,1)=\frac{\langle4,-3,-\sin(1)\rangle}{\sqrt{16+9+\sin^2(1)}}$

**khuezy** · November 14th 2008, 05:35 PM

In the answer key, they did this
T(0) to get <4, -3, 0> as r'(0)
Why did they use 0?

**Chris L T521** · November 14th 2008, 05:44 PM

Originally Posted by khuezy

In the answer key, they did this
T(0) to get <4, -3, 0> as r'(0)
Why did they use 0?

$\bold r(t)=\left<4t+1,t^2-3t,\cos t\right>$ at the point $(1,0,1)$ implies that there is a t value such that $4t+1=1,~t^2-3t=0,$ and $\cos(t)=1$

It should be a little clearer why they picked $t=0$ ...

--Chris

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