$\displaystyle x = t^{2} + 2t + 1$

$\displaystyle x = \left ( t + 1\right )^{2}$

$\displaystyle \pm \sqrt{x} = t + 1$

So 

$\displaystyle \sqrt{x} = t + 1$ or $\displaystyle -\sqrt{x} = t + 1$

We are restricting $\displaystyle t$ to values on $\displaystyle \left [ -1, 1\right ]$, so $\displaystyle t + 1$ is nonnegative; we choose 

$\displaystyle \sqrt{x} = t + 1$.

Also,

$\displaystyle y = t^{2} - 2t + 1$

$\displaystyle y= \left ( t - 1\right )^{2}$

$\displaystyle \sqrt{y} = \pm \left ( t - 1\right )$

So 

$\displaystyle \sqrt{y} = t- 1$ or $\displaystyle -\sqrt{y} = t- 1$

We are restricting $\displaystyle t$ to values on $\displaystyle \left [ -1, 1\right ]$, so $\displaystyle t - 1$ is nonpositive; we choose

$\displaystyle -\sqrt{y} = t- 1$

or equivalently,

$\displaystyle \sqrt{y} = -t+ 1$

to make $\displaystyle t - 1$ nonpositive.

Then,

$\displaystyle \sqrt{x}+ \sqrt{y} = (t + 1) + (-t + 1)$

and 

$\displaystyle \sqrt{x}+ \sqrt{y} = 2$

$\displaystyle \sqrt{y} = 2 - \sqrt{x}$

$\displaystyle y =\left ( 2 - \sqrt{x} \right )^2$

$\displaystyle y = 4 - 4\sqrt{x} + \left (\sqrt{x} \right )^{2}$

$\displaystyle y = x + 4 - 4\sqrt{x}$

$\displaystyle y = \sinh t = \frac{e ^{2t} -1}{2e ^{t}} = \frac{\left (e ^{ t} \right) ^{2} -1}{2e ^{t}}$

$\displaystyle e^{t} = 10 ^{\log e \cdot t} =\left ( 10 ^{ t} \right )^{\log e} = x ^{\log e}$, so

$\displaystyle y = \frac{\left ( x ^{\log e} \right) ^{2} -1}{2 x ^{\log e}}= \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}$

This makes the Cartesian equation

$\displaystyle y = \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}$.

Given $\displaystyle x=3+t$ and  $\displaystyle y=4t+7$, we can find $\displaystyle y$ in terms of $\displaystyle x$ by isolating $\displaystyle t$ in both equations:

$\displaystyle x=3+t\rightarrow t=x-3$

$\displaystyle y=4t+7\rightarrow t=\frac{y-7}{4}$

Since both of these transformations equal $\displaystyle t$, we can set them equal to each other:

$\displaystyle \frac{y-7}{4}=x-3$

$\displaystyle y-7=4(x-3)$

$\displaystyle y-7=4x-12$

$\displaystyle y=4x-5$

In order to find the arc length, we must use the arc length formula for parametric curves:

$\displaystyle L=\int_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}dt$.

Given  $\displaystyle x=t+10$ and $\displaystyle y=2t-9$, we can use using the Power Rule

 for all , to derive 

$\displaystyle \frac{dx}{dt}=(1)t^{1-1}+(0)10=1$ and 

$\displaystyle \frac{dy}{dt}=(1)2t^{1-1}-(0)9=2$.

Plugging these values and our boundary values for  into the arc length equation, we get:

$\displaystyle L=\int_{0}^{4}\sqrt{(1)^{2}+(2)^{2}}dt$

$\displaystyle L=\int_{0}^{4}(\sqrt{1+4})dt$

$\displaystyle L=\int_{0}^{4}(\sqrt{5})dt$

Now, using the Power Rule for Integrals

 for all ,

we can determine that:

$\displaystyle L=[t\sqrt{5}]_{0}^{4}\textrm{}dt$

$\displaystyle L=(4)\sqrt{5}-(0)\sqrt{5}$

$\displaystyle L=4\sqrt{5}$

In order to find the arc length, we must use the arc length formula for parametric curves:

$\displaystyle L=\int_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}dt$.

Given $\displaystyle x=4t-5$ and $\displaystyle y=6t+2$, we can use using the Power Rule

for all  , to derive

$\displaystyle \frac{dx}{dt}=(1)4t^{1-1}-(0)5=4$ and

$\displaystyle \frac{dy}{dt}=(1)6t^{1-1}+(0)2=6$ .

Plugging these values and our boundary values for into the arc length equation, we get:

$\displaystyle L=\int_{0}^{2}(\sqrt{(4)^{2}+(6)^{2}}dt$

$\displaystyle L=\int_{0}^{2}(\sqrt{16+36})dt$

$\displaystyle L=\int_{0}^{2}(\sqrt{52})dt$

$\displaystyle L=\int_{0}^{2}(\sqrt{4\times13})dt$

$\displaystyle L=\int_{0}^{2}(2\sqrt{13})dt$

Now, using the Power Rule for Integrals

for all ,

we can determine that:

$\displaystyle L=[2t\sqrt{13}]_{0}^{2}\textrm{}$

$\displaystyle L=2(2)\sqrt{13}-2(0)\sqrt{13}$

$\displaystyle L=4\sqrt{13}$

The length of a curve is found using the equation $\displaystyle L=\int_{\alpha }^{\beta }\sqrt{\left ( \frac{dx}{dt} \right )^{2}+\left ( \frac{dy}{dt} \right )^{2}}dt$

We use the product rule,

$\displaystyle \frac{d}{dt}(a*b)=a\frac{db}{dt}+b\frac{da}{dt}$, when $\displaystyle a$ and $\displaystyle b$ are functions of $\displaystyle t$,

the trigonometric rule,

$\displaystyle \frac{d}{dt}[cos(t)]=-sin(t)$ and  $\displaystyle \frac{d}{dt}[sin(t)]=cos(t)$

and exponential rule,

$\displaystyle \frac{d}{dt}[e^{t}]=e^{t}$ to find $\displaystyle \frac{dx}{dt}$ and $\displaystyle \frac{dy}{dt}$. 

In this case

$\displaystyle x=e^{t}cos(t)$,   $\displaystyle y=e^{t}sin(t)$

$\displaystyle \frac{dx}{dt}=\frac{d}{dt}e^{t}cos(t)=e^{t}\frac{d}{dt}cos(t)+cos(t)\frac{d}{dt}e^{t}$

$\displaystyle =e^{t}*(-sin(t))+cos(t)*e^{t}=e^{t}cos(t)-e^{t}sin(t)$

$\displaystyle \frac{dy}{dt}=\frac{d}{dt}e^{t}sin(t)=e^{t}\frac{d}{dt}sin(t)+sin(t)\frac{d}{dt}e^{t}$

$\displaystyle =e^{t}cos(t)+sin(t)*e^{t}=e^{t}cos(t)+e^{t}sin(t)$

$\displaystyle \alpha=0, \beta=2\pi$

The length of this curve is

$\displaystyle L=\int_{0}^{2\pi }\sqrt{\left (e^{t}cos(t)-e^{t}sin(t)\right )^{2}+\left (e^{t}cos(t)+e^{t}sin(t)\right )^{2}}dt$

Using the identity $\displaystyle (a\pm b)^{2}=a^{2}\pm2ab+b^{2}$

$\displaystyle L=\int_{0}^{2\pi }\sqrt{\binom{\left (e^{2t}cos^{2}(t)-2e^{t}cos(t)e^{t}sin(t)+e^{2t}sin^{2}(t)\right )}{\left +(e^{2t}cos^{2}(t)+2e^{t}cos(t)e^{t}sin(t)+e^{2t}sin^{2}(t)\right)}dt}$

$\displaystyle =\int_{0}^{2\pi }\sqrt{(2e^{2t}cos^{2}(t)+2e^{2t}sin^{2}(t))dt}$

$\displaystyle =\int_{0}^{2\pi }\sqrt{2e^{2t}(cos^{2}(t)+sin^{2}(t))dt}$

Using the identity $\displaystyle \sqrt{a*b}=\sqrt{a}\sqrt{b}$

$\displaystyle L=\int_{0}^{2\pi }\sqrt{2e^{2t}} \sqrt{cos^{2}(t)+sin^{2}(t)dt}$

Using the trigonometric identity $\displaystyle sin^{2}(at)+cos^{2}(at)=1$ where $\displaystyle a$ is a constant and $\displaystyle e^{2t}=(e^{t})^2$

$\displaystyle =\int_{0}^{2\pi }\sqrt{2}e^{t}dt}=\sqrt{2}\int_{0}^{2\pi }e^{t}dt}$

Using the exponential rule, $\displaystyle \int e^{t}dt}=e^{t}$

$\displaystyle \sqrt{2}\int_{0}^{2\pi }e^{t}dt}=\sqrt{2}e^{t}|_{0}^{2\pi}=\sqrt{2}\left ( e^{2 \pi}-e^{0}}\right )$

Using the exponential rule, $\displaystyle e^{0}=1$, gives us the final solution

$\displaystyle L=\sqrt{2}\left ( e^{2 \pi}-1}\right )$

The formula for dy/dx for parametric equations is given as:

$\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

From the problem statement:

$\displaystyle \frac{dy}{dt}=3\sec^2(t),\frac{dx}{dt}=5\sec(t)\tan(t)$

If we plug these into the above equation we end up with:

$\displaystyle \frac{dy}{dx}=\frac{3\sec^2(t)}{5\sec(t)\tan(t)}=\frac{3\sec(t)}{5\tan(t)}=\frac{3\frac{1}{cos(t)}}{5\frac{sin(t)}{cos(t)}}$

$\displaystyle \frac{dy}{dx}=\frac{3}{5}\frac{1}{\cos(t)}\frac{\cos(t)}{\sin(t)}=\frac{3}{5}\frac{1}{\sin(t)}$

If we plug in our given value for t, we end up with:

$\displaystyle \frac{3}{5}\frac{1}{\sin(\frac{\pi}{6})}=\frac{3}{5}\frac{1}{\frac{1}{2}}=\frac{3}{5}*2=\frac{6}{5}$

This is one of the answer choices.

Between $\displaystyle 0$ and $\displaystyle \frac{\pi }{2}$, the radius approaches $\displaystyle 1$ from $\displaystyle 0$.

From $\displaystyle \frac{\pi }{2}$ to $\displaystyle \pi$ the radius goes from $\displaystyle 1$ to $\displaystyle 0$.

Between $\displaystyle \pi$ and $\displaystyle \frac{3\pi }{2}$, the curve is redrawn in the opposite quadrant, the first quadrant as the radius approaches $\displaystyle -1$.

From $\displaystyle \frac{3\pi }{2}$ and $\displaystyle 2\pi$, the curve is redrawn in the second quadrant as the radius approaches $\displaystyle 0$ from $\displaystyle -1$.   

Because this function has a period of $\displaystyle \pi$, the amplitude of the graph $\displaystyle y=sin(2x)$  appear at a reference angle of $\displaystyle \frac{\pi }{4}$ (angles halfway between the angles of the axes).  

Between $\displaystyle 0$ and $\displaystyle \frac{\pi }{4}$ the radius approaches 1 from 0.

Between $\displaystyle \frac{\pi }{4}$ and $\displaystyle \frac{\pi }{2}$, the radius approaches 0 from 1.

From $\displaystyle \frac{\pi }{2}$ to $\displaystyle \frac{3\pi }{4}$ the radius approaches -1 from 0 and is drawn in the opposite quadrant, the fourth quadrant because it has a negative radius.

Between $\displaystyle \frac{3\pi }{4}$ and $\displaystyle \pi$, the radius approaches 0 from -1, and is also drawn in the fourth quadrant.

From $\displaystyle \pi$ and $\displaystyle \frac{5\pi }{4}$, the radius approaches 1 from 0. Between $\displaystyle \frac{5\pi }{4}$ and $\displaystyle \frac{3\pi }{2}$, the radius approaches 0 from 1.

Then between $\displaystyle \frac{3\pi }{2}$ and $\displaystyle \frac{7\pi }{4}$ the radius approaches -1 from 0. Because it is a negative radius, it is drawn in the opposite quadrant, the second quadrant. Likewise, as the radius approaches 0 from -1. Between $\displaystyle \frac{7\pi }{4}$ and $\displaystyle 2\pi$, the curve is drawn in the second quadrant.                  

Taking the graph of $\displaystyle y=cos(2x)$, we only want the areas in the positive first quadrant because the radius is squared and cannot be negative.

This leaves us with the areas from $\displaystyle 0$ to $\displaystyle \frac{\pi }{4}$, $\displaystyle \frac{3\pi }{4}$ to $\displaystyle \frac{5\pi }{4}$, and $\displaystyle \frac{7\pi }{4}$ to $\displaystyle 2\pi$.

Then, when we take the square root of the radius, we get both a positive and negative answer with a maximum and minimum radius of $\displaystyle \pm 1$.

To draw the graph, the radius is 1 at $\displaystyle 0$ and traces to 0 at $\displaystyle \frac{\pi }{4}$. As well, the negative part of the radius starts at -1 and traces to zero in the opposite quadrant, the third quadrant.

From $\displaystyle \frac{3\pi }{4}$ to $\displaystyle \pi$, the curves are traced from 0 to 1 and 0 to -1 in the fourth quadrant. Following this pattern, the graph is redrawn again from the areas included in $\displaystyle \pi$ to $\displaystyle 2\pi$.