1. ## vector valued functions

r(t) = <3t, (t-1)>

find
a) Dt[r(t) dot u(t)]
b) Dt[u(t) - 2r(t)]
c) Dt[||r(t)||], t > 0
d) Dt[r(t) x u(t)]

i don't understand what i am have to do for these problems

2. Originally Posted by chris25
r(t) = <3t, (t-1)>

find
a) Dt[r(t) dot u(t)]
b) Dt[u(t) - 2r(t)]
c) Dt[||r(t)||], t > 0
d) Dt[r(t) x u(t)]

i don't understand what i am have to do for these problems
a) $\frac{d}{dt}\bigg[\vec{r}(t)\cdot \hat{u}(t)\bigg]$

First of all, you're given $\vec{r}(t)=\bigg<3t,t-1\bigg>$

The unit vector $\hat{u}(t)$ is defined as $\hat{u}(t)=\frac{\vec{r}(t)}{\Vert\vec{r}(t)\Vert}$.

Find $\Vert\vec{r}(t)\Vert$:

$\Vert\vec{r}(t)\Vert=\sqrt{(3t)^2+(t-1)^2}=\sqrt{9t^2+t^2-2t+1}=\sqrt{10t^2-2t+1}$.

Thus,
$\hat{u}(t)=\frac{\bigg<3t,(t-1)\bigg>}{\sqrt{10t^2-2t+1}}=\bigg<\frac{3t}{\sqrt{10t^2-2t+1}},\frac{t-1}{\sqrt{10t^2-2t+1}}\bigg>$

Thus,

$\vec{r}(t)\cdot\hat{u}(t)=\frac{9t^2}{\sqrt{10t^2-2t+1}}+\frac{(t-1)^2}{\sqrt{10t^2-2t+1}}=\frac{10t^2-2t+1}{\sqrt{10t^2-2t+1}}=\sqrt{10t^2-2t+1}$

Now find $\frac{d}{dt}\bigg[\vec{r}(t)\cdot\hat{u}(t)\bigg]$

$\frac{d}{dt}\bigg[\vec{r}(t)\cdot\hat{u}(t)\bigg]=\frac{1}{2}\bigg[10t^2-2t+1\bigg]^{-\frac{1}{2}}\cdot\bigg(20t^2-2\bigg)=\color{red}\boxed{\frac{10t^2-1}{\sqrt{10t^2-2t+1}}}$

Hope this helps. Try to do the others.

--Chris

3. Originally Posted by chris25
r(t) = <3t, (t-1)>

find
a) Dt[r(t) dot u(t)]
b) Dt[u(t) - 2r(t)]
c) Dt[||r(t)||], t > 0
d) Dt[r(t) x u(t)]

i don't understand what i am have to do for these problems
For the second one as Chris pointed out

$\hat{u}(t)=\frac{\left\langle{3t,(t-1)}\right\rangle}{\sqrt{10t^2-2t+1}}=\left\langle\frac{3t}{\sqrt{10t^2-2t+1}},\frac{t-1}{\sqrt{10t^2-2t+1}}\right\rangle$

So $\bold{\hat{u}(t)}-2\bold{r(t)}$ $=\left\langle\frac{3t}{\sqrt{10t^2-2t+1}}-6t,\frac{t-1}{\sqrt{10t^2-2t+1}}-2t-2\right\rangle$

So $\frac{d}{dt}\bigg[\bold{\hat{u}(t)}-2\bold{r(t)}\bigg]=\left\langle\frac{-3(t-1)}{(10t^2-2t+1)^{\frac{3}{2}}}-6,\frac{9t}{(10t^2-2t+1)^{\frac{3}{2}}}-2\right\rangle$

Next we have

$||\bold{r(t)}||=\sqrt{(3t)^2+(t-1)^2}=\sqrt{10t^2-2t+1}$

so

$\frac{d}{dt}\bigg[\bold{r(t)}\bigg]=\frac{10t-1}{\sqrt{10t^2-2t+1}}$

Lastly we have

$\frac{d}{dt}\bigg[\bold{\hat{u}(t)}\text{ x }\bold{r(t)}\bigg]=\bold{\hat{u}(t)}'\text{ x }\bold{r(t)}+\bold{\hat{u}(t)}\text{ x }\bold{r(t)}$

Which actually doesnt make sense, because you have a vector in $\mathbb{R}^2$ and the cross product is only defined in $\mathbb{R}^3$

Off topic , how do you represent cross product, the representation I used isn't very nice looking. I tried \cross, but to no avail

4. Originally Posted by Mathstud28
Off topic , how do you represent cross product, the representation I used isn't very nice looking. I tried \cross, but to no avail

Use \times :

i.e. $\vec{r}(t)\times\hat{u}(t)$