Troubleshooting Mathematica's Simplify Function for Finding the Norm of a Vector

[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?

 

[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 $t\in\mathbb{R}$

 

[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...

 

[D H]

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.

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?

PowerExpand[Simplify[]] gives me 12Cosh[2Re[t]] and that can't be it...
Someone think they know what's going on here?

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).

 

[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 I am 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 this…$$\sqrt {72 + 36e^{-4t} + 36e^{4t} }$$