MHB 13.4.7 Find the curvature of the curve r(t).

[karush]

$\textit{$13.4.7$ Find the curvature of the curve $r(t).$}$
\begin{align*}\displaystyle
r(t)&=(5+9 \cos 8t) i - (4 + 9 \sin 8t)j + 6k\\
v&= -72\sin(8t)+72\cos(8t)\\
|v|&=\sqrt{(-72\sin(8t))^2 +(72\cos(8t))^2}\\
&=5184\\
\frac{v}{|v|}&=\frac{-72\sin(8t)+72\cos(8t)}{5184}\\
K&=\frac{1}{|v|}\left|\frac{d\textbf{T}}{dr}\right|
\end{align*}ok the answer is $\displaystyle k=\frac{1}{9}$
but got lost in this process

 

[karush]

Find the curvature of the curve

$\tiny{13.4.7}$
$\textit{Find the curvature of the curve $r(t).$}$
\begin{align*}\displaystyle
r(t)&=(5+9 \cos 8t) \textbf{i}
- (4+9\sin 8t)\textbf{j}
+ 6\textbf{k}\\
T(t)&= \frac{r'(t)}{||r'(t)||}
=\frac{-72\sin(8t)i-72\cos(8t)j+6}{\sqrt{72^2 +6^2}}\\
\end{align*}

ok i can't seem to get this first srep correct
the answer is
$ K= \color{red}{\frac{1}{9}}$

 

[MarkFL]

I've merged the two threads so that we don't have a double posting of the same problem. We discourage multiple topics of the same problem because not only is it redundant, but it causes a potential duplication of effort on the part of our helpers, whose time is valuable. Any time we can save on the part of the volunteer helpers would, I hope you can see, help you in the long run!

 

[MarkFL]

I tried to delete the first one but there is no option to so.

Yes, we only want our staff to be able to delete threads. Unfortunately, there would be some who would delete their threads after getting help to reduce the chances that their professor might find the thread. I know this doesn't apply to you, nor to the majority actually, but it is a case (one of many in society) in which the morally bereft minority ruins things for everyone else.

There have been no replies for help.

It has only been a bit more than a day since you posted the original, and it is the weekend. Sometimes it may take a few days to get a reply, and we ask for patience in that regard. :)

 

[MarkFL]

We are given:

$$r(t)=\left(5+9\cos(8t)\right)\hat{\imath}-\left(4+9\sin(8t)\right)\hat{\jmath}+(6)\hat{k}$$

And so:

$$r'(t)=-72\left(\sin(8t)\hat{\imath}+\cos(8t)\hat{\jmath}\right)$$

$$|r'(t)|=72$$

And so:

$$T(t)=\frac{r'(t)}{|r'(t)|}=-\left(\sin(8t)\hat{\imath}+\cos(8t)\hat{\jmath}\right)$$

Hence:

$$T'(t)=8\left(-\cos(8t)\hat{\imath}+\sin(8t)\hat{\jmath}\right)\implies |T'(t)|=8$$

Thus:

$$\kappa=\frac{|T'(t)|}{|r'(t)|}=\frac{8}{72}=\frac{1}{9}$$

 

[karush]

$\displaystyle T'(t)=8\left(-\cos(8t)\hat{\imath}+\sin(8t)\hat{\jmath}\right)\implies |T'(t)|=8$

I didn't ? This step

 

[MarkFL]

$\displaystyle T'(t)=8\left(-\cos(8t)\hat{\imath}+\sin(8t)\hat{\jmath}\right)\implies |T'(t)|=8$

I didn't ? This step

I am assuming you didn't understand that step? Please use complete English for clarity. :)

If that's indeed what you intended to say:

$$T'(t)=8\left(-\cos(8t)\hat{\imath}+\sin(8t)\hat{\jmath}\right)$$

$$|T'(t)|=8\sqrt{\cos^2(8t)+\sin^2(8t)}=8\cdot1=8$$

 

[MarkFL]

Ok

? = I don't understand

Yes, I assumed that was the case, but it is better to just use good ol' fashioned English to say what you actually mean. This way is it clear, and no guesswork is involved.

Now, we do allow certain standard abbreviations/acronyms that are in common usage within the math community (such as IBP and FTOC), but text speak and other non-standard abbreviations are discouraged.

 

[karush]

why are you putting a hat on top of i,j,k ?
other texts books had a small arrow?

 

[I like Serena]

why are you putting a hat on top of i,j,k ?
other texts books had and small arrow?

It's a matter of preference:
$$\hat\imath=\boldsymbol{\hat\imath} = \vec i = \mathbf i = i = \mathbf{\hat x}$$
These are all the same depending on the textbook we're using.My own preference is actually $\mathbf{\hat x}$, since that's the only one that truly communicates what we're talking about: a vector of unit length in the x-direction.
That's as opposed to $i$, which is sometimes in the x-direction and sometimes in the y-direction. (Nerd)

 

[karush]

ok here is my plagiarized version;)
I don't like hats so removed
also still don't know how you can put a radical on that and get a 1
English is my mother tongue so you have to be patience

$\tiny{13x.26}$
$\textrm{Find the curvature of the curve}$
\begin{align*} \displaystyle
r(t)&=(5+9\cos(8t))\textbf{i}-(4+9\sin(8t)\textbf{j}+(6)\textbf{k}\\
r'(t)&=-72\sin(8t)\textbf{i}+\cos(8t)\textbf{j}\\
|r'(t)|&=72\\
T(t)&=\frac{r'(t)}{|r'(t)|}\\
&=-(\sin(8t)\textbf{i}+\cos(8t)\textbf{j}\\
T'(t)&=8\left(-\cos(8t)\hat{\imath}+\sin(8t)\hat{\jmath}\right)\\
|T'(t)|&=8\sqrt{\cos^2(8t)+\sin^2(8t)}\\
&=8\cdot1=8\\
\kappa&=\frac{|T'(t)|}{|r'(t)|}=\frac{8}{72}=\color{red}{\frac{1}{9}}
\end{align*}

 

[MarkFL]

...also still don't know how you can put a radical on that and get a 1...

It's a Pythagorean identity:

$$\sin^2(\theta)+\cos^2(\theta)=1$$

 

[HallsofIvy]

$\textit{$13.4.7$ Find the curvature of the curve $r(t).$}$
\displaystyle
r(t)&=(5+9 \cos 8t) i - (4 + 9 \sin 8t)j + 6k\\
v&= -72\sin(8t)+72\cos(8t)\\
|v|&=\sqrt{(-72\sin(8t))^2 +(72\cos(8t))^2}\\
&=5184\\

This is an error. $|v|= \sqrt{(-72\sin(8t)^2+ (72\cos(8t))^2}= \sqrt{72^2}= 72$
You seem to have forgotten the square root!

\frac{v}{|v|}&=\frac{-72\sin(8t)+72\cos(8t)}{5184}\\
K&=\frac{1}{|v|}\left|\frac{d\textbf{T}}{dr}\right|
ok the answer is $\displaystyle k=\frac{1}{9}$
but got lost in this process

 

[karush]

It's a Pythagorean identity:

$$\sin^2(\theta)+\cos^2(\theta)=1$$

I know but how can you just add in squares to make it work

it was basically

$\sin\left({\theta}\right)+\cos\left({\theta}\right)$

then it was magically

$\sin^2\left({\theta}\right)+\cos^2\left({\theta}\right)$

 

[MarkFL]

I know but how can you just add in squares to make it work

it was basically

$\sin\left({\theta}\right)+\cos\left({\theta}\right)$

then it was magically

$\sin^2\left({\theta}\right)+\cos^2\left({\theta}\right)$

To find the magnitude of a vector, we essentially use the $n$-space distance formula, which is the square root of the sum of the squares of the components. For example, suppose we have:

$$\vec{v}=\left\langle v_x,v_y,v_z \right\rangle$$

Then:

$$\left|\vec{v}\right|=\sqrt{v_x^2+v_y^2+v_z^2}$$