1. ## Line Integral

Hello! I need some help with this problem, please!

Let $f$ be a field and $\gamma$ a curve such that for all $t\in{\mathbb{R}}$ happens that $f(\gamma (t))=m\gamma ''(t)$ (where m is a constant).
Show that:

$\displaystyle\int_{\gamma (a)}^{\gamma (b)}fd\gamma = \displaystyle\frac{1}{2}m \left\|{\gamma ''(b)}\right\|-\displaystyle\frac{1}{2}m \left\|{\gamma ''(a)}\right\|$

Thank you!!

2. Originally Posted by Veronika
Hello! I need some help with this problem, please!

Let $f$ be a field and $\gamma$ a curve such that for all $t\in{\mathbb{R}}$ happens that $f(\gamma (t))=m\gamma ''(t)$ (where m is a constant).
Show that:

$\displaystyle\int_{\gamma (a)}^{\gamma (b)}fd\gamma = \displaystyle\frac{1}{2}m \left\|{\gamma ''(b)}\right\|-\displaystyle\frac{1}{2}m \left\|{\gamma ''(a)}\right\|$

Thank you!!
I wonder whether this formula is at all valid, since $\int_{\gamma(a)}^{\gamma(b)}f d\gamma=\int_a^bf(\gamma(t))\cdot \gamma'(t)\,dt=m\int_a^b\gamma''(t)\cdot \gamma'(t)\,dt$ and $\frac{d}{dt}\big[\parallel \gamma'(t)\parallel^2\big]=\frac{d}{dt}\big[\gamma'(t)\cdot \gamma'(t)\big]=2\gamma''(t)\cdot\gamma'(t)$.
So I would have expected the formula to look more like $\int_{\gamma(a)}^{\gamma(b)}f d\gamma\overset{?}{=}\tfrac{1}{2}m\parallel \gamma'(b)\parallel^2-\tfrac{1}{2}m\parallel \gamma'(a)\parallel^2$

3. Yes, I think is a typing error from my professor.

So, to prove the last equation should I just derive and see if the are the same? I mean, using the 1 Fundamental Theorem of Calculus.

Thanks!

4. Originally Posted by Veronika
Yes, I think is a typing error from my professor.

So, to prove the last equation should I just derive and see if the are the same? I mean, using the 1 Fundamental Theorem of Calculus.

Thanks!
Um, well, yes, I think that's the way to go. You can really just determine the derivative of $F(t) := \frac{m}{2}\parallel \gamma'(t)\parallel^2$ and show, that it is equal to $f(\gamma(t))\cdot \gamma'(t)$, given that $f(\gamma(t))=m\gamma''(t)$.

$\displaystyle\int_{\gamma (a)}^{\gamma (b)}f(\gamma (t))d\gamma =m\displaystyle\int_{a}^{b}\gamma (t)''\gamma (t)'dt$
We change $u(t)=(\gamma (t)')^2$ so that $du=2\gamma (t)'\gamma (t)''$ and
$\displaystyle\frac{m}{2}\displaystyle\int_{a}^{b}d u=\displaystyle\frac{m}{2}(u(b)-u(a))=\displaystyle\frac{m}{2}(\gamma (a)')^2-\displaystyle\frac{m}{2}(\gamma (b)')^2$