The formula for the length of a parametric curve in 3-dimensional space is \(\displaystyle l = \int_{t_0}^{t_1}\sqrt{(\frac{dx_1}{dt})^2+(\frac{dx_1}{dt})^2+(\frac{dx_3}{dt})^2} dt\)

Taking dervatives and substituting, we have 

\(\displaystyle l = \int_{0}^{5}\sqrt{(2e^{2t})^2+(-2e^{-2t})^2+(2\sqrt{2})^2} dt\)

\(\displaystyle l = \int_{0}^{5}\sqrt{4e^{4t}+4e^{-4t}+8} dt\)

\(\displaystyle l = 2\int_{0}^{5}\sqrt{e^{4t}+e^{-4t}+2} dt\). Factor a \(\displaystyle 4\) out of the square root.

\(\displaystyle l = 2\int_{0}^{5}\sqrt{e^{4t}+e^{-4t}+2e^{2t}e^{-2t}} dt\). "Uncancel" an \(\displaystyle e^{2t}, e^{-2t}\) next to the \(\displaystyle 2\). Now there is a perfect square inside the square root.

\(\displaystyle l = 2\int_{0}^{5}\sqrt{(e^{2t}+e^{-2t})^2} dt\). Factor

\(\displaystyle l = 2[\frac{1}{2}e^{2t}-\frac{1}{2}e^{-2t}]_{0}^{5}\). Take the square root, and integrate.

\(\displaystyle =2(\frac{e^{10}}{2}-\frac{e^{-10}}{2})-2({0})\)

\(\displaystyle =e^{10}-e^{-10}\)

To find the arc length of a function, we use the formula

\(\displaystyle \int_{t_{0}}^{t_{1}} \sqrt{(\frac{dx}{dt})^2 +(\frac{dy}{dt})^2 +(\frac{dz}{dt})^2} dt\).

Using \(\displaystyle t_0 =0, t_1 =\pi\) we have

\(\displaystyle =\int_0^\pi \sqrt{(-a\sin t)^2+(a\cos t)^2 +(a)^2} dt\)

\(\displaystyle =\int_0^\pi \sqrt{a^2\sin^2 t+a^2\cos^2 t +a^2} dt\)

\(\displaystyle =\int_0^\pi \sqrt{a^2(\sin^2 t+\cos^2 t) +a^2} dt\)

\(\displaystyle =\int_0^\pi \sqrt{a^2 +a^2} dt\)

\(\displaystyle =[\sqrt2at]_0^\pi\)

\(\displaystyle =\sqrt{2} a \pi\)

The formula for curvature of a Cartesian equation is \(\displaystyle \kappa(x)= \frac{|f''(x)|}{[1+(f'(x))^2]^{3/2}}\). (It's not the easiest to remember, but it's the most convenient form for Cartesian equations.)

We have \(\displaystyle f'(x) =20x, f''(x) = 20\), hence

\(\displaystyle \kappa(x)= \frac{|20|}{[1+(20x)^2]^{3/2}}\)

and \(\displaystyle \kappa(1)= \frac{|20|}{[1+(20(1))^2]^{3/2}} = \frac{20}{401^{3/2}}\).

To find the solution, we need to evaluate

\(\displaystyle L = \int_0^2 \left| \vec{r}(t)' \right| dt\).

First, we find 

\(\displaystyle \vec{r}(t)' = \left< \sqrt{10}t, t^2, 5\right>\), which leads to 

\(\displaystyle \left| \vec{r}(t)' \right| = \sqrt{(\sqrt{10}t)^2+(t^2)^2+(5)^2}\)

\(\displaystyle \left| \vec{r}(t)' \right| = \sqrt{t^4+10t^2+25}=\sqrt{(t^2+5)^2}=t^2+5\).

So we have a final expression to integrate for our answer

\(\displaystyle L= \int_0^2 \left| \vec{r}(t)' \right| dt = \int_0^2 \left(t^2+5 \right )dt= \frac{(2)^3}{3}+5(2)=\frac{38}{3}\)

The length of a curve r is given by:

\(\displaystyle L=\int_{t_1}^{t_2}|\mathbf{r'}(t)|dt\)

To solve:

\(\displaystyle \mathbf{r}'(t)=(2\cos t, \sqrt{5}, 2 \sin t)\)

\(\displaystyle |\mathbf{r'}(t)|=\sqrt{4\cos^2t+5+4\sin^2t}=\sqrt{4+5}=\sqrt{9}=3\)

\(\displaystyle L=\int^2_0 3 dt=3t|^2_0=6\)

To find the arc length of the curve function

\(\displaystyle c(t)\)

on the interval \(\displaystyle a\leq t\leq b\)

we follow the formula

\(\displaystyle \int_a^b \sqrt{(\frac{\mathrm{dx} }{\mathrm{d} t})^2+(\frac{\mathrm{dy} }{\mathrm{d} t})^2+(\frac{\mathrm{dz} }{\mathrm{d} t})^2}\,dt\)

For the curve function in this problem we have

\(\displaystyle \frac{\mathrm{dx} }{\mathrm{d} t}=3\)

\(\displaystyle \frac{\mathrm{dy} }{\mathrm{d} t}=4cos(2t)\)

\(\displaystyle \frac{\mathrm{dz} }{\mathrm{d} t}=-4sin(2t)\)

and following the arc length formula we solve for the integral

\(\displaystyle \int_0^{10} \sqrt{(3)^2+(4cos(2t))^2+(-4sin(2t))^2}\,dt\)

\(\displaystyle =\int_0^{10} \sqrt{9+16sin^2(2t)+16cos^2(2t)}\,dt\)

\(\displaystyle =\int_0^{10} \sqrt{9+16(sin^2(2t)+cos^2(2t))}\,dt\)

\(\displaystyle =\int_0^{10} \sqrt{9+16}\,dt\)

\(\displaystyle =\int_0^{10} \sqrt{25}\,dt\)

\(\displaystyle =\int_0^{10} 5\,dt\)

\(\displaystyle =5t|_0^{10}\)

\(\displaystyle =5(10)-5(0)\)

Hence the arc length is \(\displaystyle 50\)

To find the arc length of the curve function

\(\displaystyle c(t)\)

on the interval \(\displaystyle a\leq t\leq b\)

we follow the formula

\(\displaystyle \int_a^b \sqrt{(\frac{\mathrm{dx} }{\mathrm{d} t})^2+(\frac{\mathrm{dy} }{\mathrm{d} t})^2+(\frac{\mathrm{dz} }{\mathrm{d} t})^2}\,dt\)

For the curve function in this problem we have

\(\displaystyle \frac{\mathrm{dx} }{\mathrm{d} t}=3\)

\(\displaystyle \frac{\mathrm{dy} }{\mathrm{d} t}=4\)

\(\displaystyle \frac{\mathrm{dz} }{\mathrm{d} t}=t^{\frac{1}{2}}\)

and following the arc length formula we solve for the integral

\(\displaystyle \int_0^{5} \sqrt{(3)^2+(4)^2+(t^{\frac{1}{2}})^2}\,dt\)

\(\displaystyle =\int_0^{10} \sqrt{t+25}\,dt\)

Using u-substitution, we have

\(\displaystyle u=t+25\) and \(\displaystyle du=dt\)

The integral then becomes

\(\displaystyle =\int_{25}^{30} \sqrt{u}\,du\)

\(\displaystyle =\frac{2}{3}u^{\frac{3}{2}}|_{25}^{30}\)

\(\displaystyle =\frac{2}{3}([30^{\frac{3}{2}}]-[25^{\frac{3}{2}}])\)

Hence the arc length is \(\displaystyle 26.2\)

To find the arc length of the curve function

 \(\displaystyle c(t)\)

on the interval 

 \(\displaystyle a\leq t \leq b\)

we follow the formula

 \(\displaystyle \int_{a}^{b}\sqrt{(\frac{\mathrm{dx} }{\mathrm{d} t})^2+(\frac{\mathrm{dy} }{\mathrm{d} t})^2+(\frac{\mathrm{dz} }{\mathrm{d} t})^2}\,dt\)

For the curve function in this problem we have

 \(\displaystyle \frac{\mathrm{d} x}{\mathrm{d} t}=t^{1/2}\)

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} t}=5\)

\(\displaystyle \frac{\mathrm{d} z}{\mathrm{d} t}=0\)

and following the arc length formula we solve for the integral

 \(\displaystyle \int_{0}^{11}\sqrt{(t^{1/2})^2+(5)^2+(0)^2}\,dt\)

\(\displaystyle \int_{0}^{11}\sqrt{t+25}\,dt\)

And using u-substitution, we set \(\displaystyle u=t+25\) and then solve the integral

\(\displaystyle \int_{25}^{36}\sqrt{u}\,du\)

\(\displaystyle =\frac{2}{3}u^{3/2}|_{25}^{36}\)

\(\displaystyle =\frac{2}{3}([36]^{3/2}-[25]^{3/2})\)

\(\displaystyle =\frac{2}{3}(216-125)\)

\(\displaystyle =\frac{2}{3}(91)\)

Which is approximately

\(\displaystyle 60.7\) units

Using the formula for curvature \(\displaystyle k=\frac{\left \| r'(t)\times r''(t) \right \|}{\left \| r'(t)\right \|^3}\). \(\displaystyle r'(t)=6ti+5j, r''(t)=6i\),  \(\displaystyle r'(t)\times r''(t) =-30k, \left \| r'(t)\times r''(t)\right \|=30\), and \(\displaystyle \left \| r'(t)\right \|\sqrt{36t^2+25}\). Plugging into the formula, we get \(\displaystyle k=\frac{30}{(36t^2+25)^{\frac{3}{2}}}\)