# Mathematica can't simplify

1. Nov 27, 2005

### Pengwuino

I have a function here...

r(t) ={6Sqrt[2]t, 3Exp[2t], 3Exp[-2t]}

I needed to find the norm of r'[t]... yet when Mathematica tries to simplify it, it can't do it with simplify, powerexpand[simplify[]], fullsimplify... and this function IS suppose to be able to simplify. I then started running through every basic algebraic manipulation tool and came up empty...

I restarted mathematica thinking the kernel had just given up on me but that didn't work.

PowerExpand[Simplify[]] gives me 12Cosh[2Re[t]] and that can't be it...

Someone think they know what's going on here?

2. Nov 27, 2005

### benorin

Sure it's right

$$r(t) =\left< 6\sqrt{2}t, 3e^{2t}, 3e^{-2t}\right>$$
$$\Rightarrow r^{\prime}(t) =\left< 6\sqrt{2}, 6e^{2t}, -6e^{-2t}\right>$$
$$\Rightarrow \left| r^{\prime}(t)\right| = \left| \left< 6\sqrt{2}, 6e^{2t}, -6e^{-2t}\right> \right| = \sqrt{6^{2}2+6^{2}e^{4t}+6^{2}e^{-4t}}$$
$$= 6\sqrt{2+e^{4t}+e^{-4t}}=6\sqrt{\left(e^{2t}+e^{-2t}\right) ^2}=12\left(\frac{e^{2t}+e^{-2t}}{2}\right) = 12\cosh(2t)$$

assuming $t\in\mathbb{R}$

3. Nov 27, 2005

### Pengwuino

Crap... then I need to figure out how to simplify it into something that isn't a hyperbolic function....

I need to find tangential and normal and binormal vectors and I'm not sure how i'm going to be able to do it with that equation...

4. Nov 27, 2005

### D H

Staff Emeritus
I feel your pain. I don't have access to Mathematica, but I do use Matlab's Symbolic Toolbox, which is based on the Maple engine. Sometimes it is soooo stupid ... Forcing things sometimes helps, sometimes not.

On the other hand, this is simple expression to simplify by hand.
Assuming $t$ is real,
\begin{align*} \mathbf{r}(t) &=\begin{bmatrix}6\sqrt 2 t & 3\exp(2t) & 3\exp(-2t)\end{bmatrix} \\ \dot{\mathbf r}(t) &= \begin{bmatrix}6\sqrt 2 & 6\exp(2t) & -6\exp(-2t)\end{bmatrix} \\ \lVert{\dot{\mathbf r}(t)}\rVert^2 &= \dot{\mathbf r}(t) \cdot \dot{\mathbf r}(t) \\ &= 36 \exp(4t) + 72 + 36 \exp(-4t) \\ &= \left(6\left(\exp(2t) + \exp(-2t)\right)\right)^2 \\ &= \left(12 \cosh(2t)\right)^2 \\ \intertext{thus} \lVert{\dot{\mathbf r}(t)}\rVert &= 12 \cosh(2t)\right \end{align*}
Why use Mathematica?

Wait a sec, I thought you said Mathematica couldn't simplify it. Now you said it does, and it came up with the right answer to boot. (Mathematica's answer is correct even if $t$ is complex).

Last edited: Nov 27, 2005
5. Nov 27, 2005

### Pengwuino

Well I thought it was screwing up, the unsimplified form had these Re[t]'s in it as the magnitude of r'[t] when it seems like it only should have had e's unless im going into really complicated things here.

I got this...

$$\sqrt {72 + 36e^{ - 4{\mathop{\rm Re}\nolimits} [t]} + 36e^{4{\mathop{\rm Re}\nolimits} [t]} }$$

When I thought I would get thisâ€¦

$$\sqrt {72 + 36e^{-4t} + 36e^{4t} }$$