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title: '[Notes](#fn1): Problem Class, 27 Oct 2020'

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# [Notes](#fn1): Problem Class, 27 Oct 2020
> Group 11 -- Problem Class, MATH40001/40009 IUM
> Stergios M. Antonakoudis (stergios@imperial.ac.uk)

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Welcome to IUM Problem Class Group 11. \
Today's class: <u>**Problem Sheet 4 of Part III.**</u>

[^1]: <small> *These notes[^1] are typed up and created during and for the purposes of the discussion in the online session of Problem Class Group 11 for IUM on MS Teams, similarly to the use of a 'whiteboard' in an actual class.* </small>


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## Questions

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## Examples

### Chain Rule

Let $f(s)= s^2$ and $g(t)= \sin(t)$.
What is the derivative of $h(s)= (g \circ f)(s)$?
We can write $h(s)= g(f(s))= \sin(s^2)$.

Note: $f'(s)=2s$ and $g'(t)= \cos(t)$.

==**Chain rule**==: $h'(s)= g'(f(s))\cdot f'(s)$
*I.e.* $h'(s)= \cos(s^2) \cdot 2s = 2s\cos(s^2)$.

Note: we could have also used $G(t)=(\sin(t), \exp(t) -1 )$,
where $G: \mathbb{R} \rightarrow \mathbb{R}^2$.
In that case, we have by the chain rule that 
$\frac{d}{ds} (G \circ f)(s) = f'(s)G'(f(s))$ 
$\frac{d}{ds} (G \circ f)(s) = \frac{d}{ds} (\sin(s^2), \exp(s^2)-1) = (2s\cos(s^2), 2s\exp(s^2))$.

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## Problems

### ==Problem 7==

Show that the curvature of the curve given by the vector function $r(t)$ is given by $\kappa(t)= \frac{|r'(t)\times r''(t)|}{|r'(t)|^3}$.

### ==Solution 7==

Recall that $T(t)= r'/|r'|$ and $|r'|= ds/dt$, where $T$ is the unit tangent vector and $s$ is the arc-length parameter along the curve, respectively.

Note that ==$\kappa (t) = |\frac{dT}{ds}(t)| =|T'|/|r'|$==, using the chain rule and the above two equations/definitions.

Let's write $r' = |r'|T = \frac{ds}{dt} T$ and $r''= \frac{d^2 s}{dt^2}T +\frac{ds}{dt}T'$.

Using $T \times T =0$, we can compute $r' \times r''= (\text{scalar terms})\; T\times T +(\frac{ds}{dt})^2\; T \times T' = (\frac{ds}{dt})^2\; T \times T'$

==**$T$ and $T'$ are orthogonal vectors (!)**== ( ==$|T(t)|=1\;\; \forall t$== ! )
<u>*Proof*</u>
$1= |T(t)|^2 = T(t) \cdot T(t) \;\; \forall t$; hence, by differentiating both sides, 
$0 = T'(t) \cdot T(t) + T(t) \cdot T'(t) = 2\;\; T(t) \cdot T'(t)$.
<u>*QED*</u>

Hence, $|r' \times r''| = (\frac{ds}{dt})^2\; |T \times T'| = (\frac{ds}{dt})^2\; |T|\cdot |T'|$
Hence, $|r' \times r''|= (\frac{ds}{dt})^2\; |r'|\kappa(t)$, using $\kappa (t)=|T'|/|r'|$.

Since $ds/dt = |r'|$, we conclude that $\kappa(t)=\frac{|r' \times r''|}{|r'|^3}$.

**QED**










