The questions are the following:
1.Find the length of the curve $y =(e^x + e^{-x})/2$ for 0 ≤ x ≤ b. Hint:
\begin{equation} \left(\frac{e^{x} -e^{-x}}{2}\right)^{2} +1=\left(\frac{e^{x} +e^{-x}}{2}\right)^{2} \end{equation}
2.Find an integral for the length of the curve given parametrically in Exercise 1 for 1 ≤ t ≤ 2. Simplify the integrand as much as possible but do not evaluate.
The answers given by the professor are:
1.$ \begin{equation*} y'=\frac{e^{x} -e^{-x}}{2} \end{equation*}$, so the hint says $1+(y')^2=y^2$ and $ds=\sqrt{1+( y')^{2}} \ dx=ydx$. Thus,
\begin{equation} Arclength=\frac{1}{2}\int ^{b}_{0}\left( e^{x} +e^{-x}\right) dx=\frac{e^{b} -e^{-b}}{2} \end{equation}
2.$dx/dt =1 − 1/t^2, dy/dt=1+1/t^2$. Thus
\begin{equation} ds=\sqrt{\left(\frac{dx}{dt}\right)^{2} +\left(\frac{dy}{dt}\right)^{2}} dt=\sqrt{2+\frac{2}{t^{4}}} \ dt \end{equation} and
\begin{equation} Arclength=\frac{1}{2}\int ^{2}_{1}\sqrt{2+\frac{2}{t^{4}}} \ dt \end{equation}
I don´t understand why from $1+(y')^2 = y^2$, it can be deduced that $ds=\sqrt{1+( y')^{2}} \ dx=ydx$. I neither understand where this conclusion, $dx/dt =1 − 1/t2, dy/dt=1+1/t^2$, comes from.
Here's a short justification for the arc length formula:
Given a parametric curve $\mathbf{r}(t)=(x(t),y(t))$ the arc length formula between two times $t_1$ and $t_2$ is $$s=\int_{t_1}^{t_2} \sqrt{\dot{x}(t)^2+\dot{y}(t)^2}~\mathrm{d}t$$ Why? Well the physical interpretation is this - the distance that a particle covers is simply the speed multiplied by the amount of time. The velocity of the particle is the time derivative of its position vector, that is, $\dot{\mathbf{r}}(t)=(\dot{x}(t),\dot{y}(t)).$ The speed of the particle is the magnitude of this velocity vector, that is $\Vert\dot{\mathbf{r}}(t)\Vert=\sqrt{\dot{x}(t)^2+\dot{y}(t)^2}$. In the special case that the path of the particle follows a trajectory where the $x$ and $y$ coordinates are related by some equation $y=f(x)$, one can use the parameterization $x=t~;~y=f(t)$ to obtain $$s=\int_{t_1}^{t_2}\sqrt{\left(\frac{\mathrm{d}}{\mathrm{d}t}(t)\right)^2+\left(\frac{\mathrm{d}}{\mathrm{d}t}f(t)\right)^2}~\mathrm{d}t$$ Taking the derivatives and noting that $t_1=x_1$ and $t_2=x_2$, $$s=\int_{x_1}^{x_2}\sqrt{1+\left(\frac{\mathrm{d}f}{\mathrm{d}t}\right)^2}~\mathrm{d}t$$ Let $f'(x)=\frac{\mathrm{d}f}{\mathrm{d}x}$ and use the fact that $t=x$, $$s=\int_{x_1}^{x_2}\sqrt{1+f'(x)^2}~\mathrm{d}x$$