Given $\displaystyle x=7-9t$ and $\displaystyle y=3t^{2}$, let's solve both equations for :

$\displaystyle x=7-9t\rightarrow t=\frac{7-x}{9}$

$\displaystyle y=3t^{2}\rightarrow t=\sqrt{\frac{y}{3}}$

Since both equations equal , let's set them equal to each other and solve for :

$\displaystyle \sqrt{\frac{y}{3}}=\frac{7-x}{9}$

$\displaystyle \frac{y}{3}=\left(\frac{7-x}{9}\right)^{2}$

$\displaystyle y=3\left(\frac{7-x}{9}\right)^{2}$

Given $\displaystyle x=6t^{2}-14$ and $\displaystyle y=5t+1$, let's solve both equations for :

$\displaystyle x=6t^{2}-14\rightarrow t=\pm\sqrt{\frac{x+14}{6}}$

$\displaystyle y=5t+1\rightarrow t=\frac{y-1}{5}$

Since both equations equal , let's set them equal to each other and solve for :

$\displaystyle \frac{y-1}{5}=\pm\sqrt{\frac{x+14}{6}}$

$\displaystyle y-1=5\left(\pm\sqrt{\frac{x+14}{6}}\right)$

$\displaystyle y=5\left(\pm\sqrt{\frac{x+14}{6}}\right)+1$

Given $\displaystyle x=2t-7$ and $\displaystyle y=8-t^{2}$, let's solve both equations for :

$\displaystyle x=2t-7\rightarrow t=\frac{x+7}{2}$

$\displaystyle y=8-t^{2}\rightarrow t=\sqrt{8-y}$

Since both equations equal , let's set them equal to each other and solve for :

$\displaystyle \sqrt{8-y}=\frac{x+7}{2}$

$\displaystyle 8-y=\left(\frac{x+7}{2}\right)^2$

$\displaystyle y=8-\left(\frac{x+7}{2}\right)^2$

It is known that we can derive $\displaystyle \frac{dy}{dx}$ with the formula

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

So we just find $\displaystyle \frac{dy}{dt}, \frac{dx}{dt}$:

In order to find these derivatives we will need to use the power rule which states,

$\displaystyle x^n \ dx =nx^{n-1}$.

Applying the power rule we get the following.

$\displaystyle \frac{dy}{dt}=3t^2$

$\displaystyle \frac{dx}{dt}=2t+1$

so we have 

$\displaystyle \frac{dy}{dx}=\frac{3t^2}{2t+1}$.

It is known that we can derive $\displaystyle \frac{dy}{dx}$ with the formula

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

So we just find $\displaystyle \frac{dy}{dt}, \frac{dx}{dt}$:

To find the derivatives we will need to use trigonometric and exponential rules.

Trigonometric Rule for tangent: $\displaystyle tan(t)dt=sec^2(t)$

Rules of Exponentials: $\displaystyle e^u dx=e^u\cdot \frac{du}{dx}$

Thus, applying the above rules we get the following derivatives.

$\displaystyle \frac{dy}{dt}=e^t$

$\displaystyle \frac{dx}{dt}=\sec^2 t$

so we have 

$\displaystyle \frac{dy}{dx}=\frac{e^t}{\sec^2 t}$.

Finding the length of the curve requires simply applying the formula:

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

Where:

 $\displaystyle a=0, b =2$

Since we are also given $\displaystyle x$ and $\displaystyle y$, we can easily compute the derivatives of each:

$\displaystyle \frac{dx}{dt}=\frac{d}{dt}[5t+3]=5$

$\displaystyle \frac{dy}{dt}=\frac{d}{dt}[4t-5]=4$

Applying these into the above formula results in:

$\displaystyle L=\int_{0}^{2}\sqrt{(5)^2+(4)^2}\,dt=\int_{0}^{2}\sqrt{25+16}\,dt=\int_{0}^{2}\sqrt{41}\,dt$

$\displaystyle L=\sqrt{41}\int_{0}^{2}dt=\sqrt{41}\,(2-0)=2\sqrt{41}$

This is one of the answer choices.

Finding the length of the curve requires simply applying the formula:

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

Where:

 $\displaystyle a=0, b =1$

Since we are also given $\displaystyle x$ and $\displaystyle y$, we can easily compute the derivatives of each:

$\displaystyle \frac{dx}{dt}=\frac{d}{dt}[\frac{1}{4}t^3]=\frac{3}{4}t^2$

$\displaystyle \frac{dy}{dt}=\frac{d}{dt}[\frac{1}{10}t^2]=\frac{1}{5}t$

Applying these into the above formula results in:

$\displaystyle L=\int_{0}^{1}\sqrt{(\frac{3}{4}t^2)^2+(\frac{1}{5}t)^2}\,dt=\int_{0}^{1}\sqrt{\frac{9}{16}t^4+\frac{1}{25}t^2}\,dt$

Looking at this integral, it seems pretty intimidating at first, but we can do some algebraic manipulation to make it look like a simple u-substitution problem. First we notice that there is a $\displaystyle t^2$ term that is common within the two separate functions. If we factor it out then the integral becomes:

$\displaystyle \int_{0}^{1}\sqrt{t^2[\frac{9}{16}t^2+\frac{1}{25}]}\,dt=\int_{0}^{1}t\sqrt{\frac{9}{16}t^2+\frac{1}{25}}\,dt$

Now we can apply the rules of u-substitution:

$\displaystyle u=\frac{9}{16}t^2+\frac{1}{25},\,du=\frac{18}{16}t\,dt=\frac{9}{8}t\,dt$

Therefore:

$\displaystyle \frac{8}{9}du=t\,dt$

Now we must deal with the new bounds of our integral:

$\displaystyle u(0)=\frac{9}{16}(0)^2+\frac{1}{25}=\frac{1}{25}$

$\displaystyle u(1)=\frac{9}{16}(1)^2+\frac{1}{25}=\frac{241}{400}$

Our new integral becomes:

$\displaystyle \frac{8}{9}\int_{\frac{1}{25}}^{\frac{241}{400}}\sqrt{u}\,du=\frac{8}{9}[\frac{2}{3}u^{\frac{3}{2}}]_{\frac{1}{25}}^{\frac{241}{400}}=\frac{16}{27}[(\frac{241}{400})^\frac{3}{2}-(\frac{1}{25})^\frac{3}{2}]$

Plugging this result into a calculator results in:

$\displaystyle 0.2723945261$

Because the problem statement requires us to round this to three significant figures, the final result is:

$\displaystyle 0.272$

This is one of the answer choices.

Finding the length of the curve requires simply applying the formula:

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

Where:

 $\displaystyle a=0, b =\frac{\pi}{2}$

Since we are also given $\displaystyle x$ and $\displaystyle y$, we can easily compute the derivatives of each:

$\displaystyle \frac{dx}{dt}=\frac{d}{dt}[8\cos(3t)]=-8\sin(3t)\cdot3=-24\sin(3t)$

$\displaystyle \frac{dy}{dt}=\frac{d}{dt}[8\sin(3t)]=8\cos(3t)\cdot3=24\cos(3t)$

Applying these into the above formula results in:

$\displaystyle L=\int_{0}^{\frac{\pi}{2}}\sqrt{(-24\sin(3t))^2+(24\cos(3t))^2}\,dt=\int_{0}^{\frac{\pi}{2}}\sqrt{24^2\sin^2(3t)+24^2\cos^2(3t)}\,dt$We can factor out the common $\displaystyle 24^2$, and pull it outside of the square-root, and we will notice one of the most common trigonometric identities:

$\displaystyle \int_{0}^{\frac{\pi}{2}}\sqrt{24^2[\sin^2(3t)+\cos^2(3t)]}\,dt=\int_{0}^{\frac{\pi}{2}}24\sqrt{\sin^2(3t)+\cos^2(3t)}\,dt$

The term inside the square-root symbol can be simplified to $\displaystyle 1$.

$\displaystyle 24\int_{0}^{\frac{\pi}{2}}\sqrt{1}\,dt=24\int_{0}^{\frac{\pi}{2}}dt=24[t]_{0}^{\frac{\pi}{2}}=24[\frac{\pi}{2}-0]=\frac{24\pi}{2}=12\pi$

This is one of the answer choices.

Finding the length of the curve requires simply applying the formula:

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

Where:

 $\displaystyle a=0, b =\frac{\pi}{2}$

Since we are also given $\displaystyle x$ and $\displaystyle y$, we can easily compute the derivatives of each, using the Product Rule:

$\displaystyle \frac{dx}{dt}=\frac{d}{dt}[e^{3t}\cos(3t)]=-3e^{3t}\sin(3t)+3e^{3t}\cos(3t)=3e^{3t}[\cos(3t)-\sin(3t)]$

$\displaystyle \frac{dy}{dt}=\frac{d}{dt}[e^{3t}\sin(3t)]=3e^{3t}\cos(3t)+3e^{3t}\sin(3t)=3e^{3t}[\cos(3t)+\sin(3t)]$

Applying these into the above formula results in:

$\displaystyle L=\int_{0}^{\frac{\pi}{2}}\sqrt{(3e^{3t}[\cos(3t)-\sin(3t)])^2+(3e^{3t}[\cos(3t)+\sin(3t)])^2}\,dt$

Simplifying the above will require these two formulas:

$\displaystyle (a+b)^2=a^2+2ab+b^2 \rightarrow \,(1)$

$\displaystyle (a-b)^2=a^2-2ab+b^2\rightarrow\,(2)$

It may also be useful to know this formula:

$\displaystyle (x^a)^b=x^{ab}\rightarrow\,(3)$

We can factor out the common $\displaystyle (3e^{3t})^2$ to make the above expression easier to look at:

$\displaystyle L=\int_{0}^{\frac{\pi}{2}}\sqrt{(3e^{3t})^2([\cos(3t)-\sin(3t)]^2+[\cos(3t)+\sin(3t)])^2}\,dt$

We can take the $\displaystyle (3e^{3t})^2$ outside of the square-root by cancelling out the $\displaystyle 2$ representing the "square". Then we can apply formulas $\displaystyle (1)$ & $\displaystyle (2)$ to the trigonometric expressions:

$\displaystyle L=\int_{0}^{\frac{\pi}{2}}3e^{3t}\sqrt{[\cos^2(3t)-2\cos(3t)\sin(3t)+\sin^2(3t)]+[\cos^2(3t)+2\cos(3t)\sin(3t)+\sin^2(t)]}\,dt$

We can now simplify the terms inside the square-root to get:

$\displaystyle L=3e^{3t}\int_{0}^{\frac{\pi}{2}}\sqrt{[2\cos^2(3t)+2\sin^2(3t)]}\,dt$

If we factor out the common "2" above, we are left with the trigonometric identity, which simplifies to $\displaystyle 1$, since:

$\displaystyle \sin^2(t)+\cos^2(t)=1$

Therefore the integral now becomes:

$\displaystyle L=\int_{0}^{\frac{\pi}{2}}3e^{3t}\sqrt{2[\cos^2(3t)+\sin^2(3t)]}\,dt$

$\displaystyle L=\int_{0}^{\frac{\pi}{2}}3e^{3t}\sqrt{2(1)}\,dt=\int_{0}^{\frac{\pi}{2}}3\sqrt{2}e^{3t}\,dt$

This is a simple integral which can be solved using u-substitution. But first, we can factor out the constant term $\displaystyle 3\sqrt{2}$, outside of the integral:

$\displaystyle L=3\sqrt{2}\int_{0}^{\frac{\pi}{2}}e^{3t}\,dt$

We will make our substitutions:

$\displaystyle u=3t,\,du=3\,dt\rightarrow\frac{1}{3}du=dt$

We also need to change the bounds of the new integral:

$\displaystyle u(0)=3(0)=0$

$\displaystyle u(\frac{\pi}{2})=3(\frac{\pi}{2})=\frac{3\pi}{2}$

Our new integral becomes:

$\displaystyle L=3\sqrt{2}\int_{0}^{\frac{3\pi}{2}}\frac{1}{3}e^u\,du=\sqrt{2}[e^u]_{0}^{\frac{3\pi}{2}}=\sqrt{2}(e^{\frac{3\pi}{2}}-e^{0})=\sqrt{2}(e^{\frac{3\pi}{2}}-1)$

This is one of the answer choices.

At $\displaystyle t=2$, the graph passes through $\displaystyle (e^{-2}, 2^3 - 3\cdot 2^2 + 1) = (e^{-2}, -3)$

Now to find the slope, we will need both derivatives with respect to t, which are:

$\displaystyle \frac{dx}{dt} = -e^{-t}$

$\displaystyle \frac{dy}{dt} = 3t^2 - 6t$

So to obtain the slope, we just use:

 $\displaystyle \frac{\mathrm{dy} }{\mathrm{d} x} = \frac{dy/dt}{dx/dt}$,

and evaluate at $\displaystyle t=2$.

As it turns out, $\displaystyle \frac{\mathrm{dy} }{\mathrm{d} t}$ at $\displaystyle t=2$, and $\displaystyle \frac{\mathrm{dx} }{\mathrm{d} t} \neq 0$, so the slope will be 0 for this curve at the point $\displaystyle t=2$.

That means that $\displaystyle y = 0x +b = b$, and so solving at ordered pair $\displaystyle (e^{-2}, -3)$, the solution must be:

$\displaystyle y = -3$