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:

$\displaystyle y = -3$

$\displaystyle x = 4 sin(t) - 2$

$\displaystyle y = 9cos(t) + 1$

Perform these operations:

$\displaystyle x + 2 = 4sin (t) \Rightarrow \frac{x+2}{4} = sin(t)$

$\displaystyle y - 1 = 9 cos(t) \Rightarrow \frac{y-1}{9} = cos(t)$

Now, we can use a Pythagorean trigonometric identity to transform the equation into a rectangular equation:

$\displaystyle \frac{(x+2)^2}{4^2} + \frac{(y-1)^2}{9^2} = sin^2t + cos^2 t = 1$

And this is the equation of an ellipse, centered at $\displaystyle (-2, 1)$ with horizontal axis $\displaystyle 4$ and vertical axis $\displaystyle 9$.

Given $\displaystyle x=-t+5$ and $\displaystyle y=9t+4$ let's solve both equations for $\displaystyle t$:

$\displaystyle x=-t+5\rightarrow t=5-x$

$\displaystyle y=9t+4\rightarrow t=\frac{y-4}{9}$

Since both equations equal $\displaystyle t$, let's set them equal to each other and solve for $\displaystyle y$:

$\displaystyle \frac{y-4}{9}=5-x$

$\displaystyle y-4=9(5-x)$

$\displaystyle y=9(5-x)+4$

$\displaystyle y=45-9x+4$

$\displaystyle y=49-9x$

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 $\displaystyle x=2t+7$ and $\displaystyle y=7t-2$, we can use using the Power Rule

for all ,

to derive 

$\displaystyle \frac{dx}{dt}=(1)2t^{1-1}+(0)7=2$ and $\displaystyle \frac{dy}{dt}=(1)7t^{1-1}-(0)2=7$.

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

$\displaystyle L=\int_{0}^{2}\sqrt{(2)^{2}+(7)^{2}}dt$

$\displaystyle L=\int_{0}^{2}(\sqrt{4+49})dt$

$\displaystyle L=\int_{0}^{2}(\sqrt{53})dt$

Now, using the Power Rule for Integrals $\displaystyle \int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $\displaystyle n\neq0$, we can determine that:

$\displaystyle L=\int_{0}^{2}(\sqrt{53})dt$

$\displaystyle L=[t\sqrt{53}]_{0}^{2}\textrm{}$

$\displaystyle L=2\sqrt{53}-0\sqrt{53}$

$\displaystyle L=2\sqrt{53}$

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 these derivatives we will need to use the power rule, chain rule, and rule of exponentials.

Power Rule: $\displaystyle x^n\ dx =nx^{n-1}$

Chain Rule: $\displaystyle f(g(x))\ dx=f'(g(x))\cdot g'(x)$

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

Applying these rules we get the following.

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

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

so we have 

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

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 rules and the rule for natural logs.

Trigonometric Rule for cosine: $\displaystyle cos(t)dt=-sin(t)$

Rule of Natural Log: $\displaystyle ln(u)dt=\frac{1}{u}\cdot \frac{du}{dt}$

Applying the above rules we get the following derivatives.

$\displaystyle \frac{dy}{dt}=\frac{1}{t}$

$\displaystyle \frac{dx}{dt}=-\sin t$

so we have 

$\displaystyle \frac{dy}{dx}=-\frac{1}{t\sin t}$.

Using the identity $\displaystyle \sec^2(t)-\tan^2(t)=1$, we can plug in the values $\displaystyle x$ for $\displaystyle \sec(t)$ and $\displaystyle y$ for $\displaystyle \tan(t)$ to obtain the equation $\displaystyle x^2-y^2=1$. This is the graph of a horizontal hyperbola with x-intercepts of $\displaystyle (-1,0)$ and $\displaystyle (1,0)$ with asympotes of $\displaystyle y=\pm{x}$. The picture is depicted below:

When $\displaystyle t=\frac{\pi}{4}$

we have that

$\displaystyle x=cos(\frac{\pi}{4})=\frac{\sqrt2}{2}$

$\displaystyle y=sin(\frac{\pi}{4})=\frac{\sqrt2}{2}$

This gives the coordinate

$\displaystyle (\frac{\sqrt2}{2},\frac{\sqrt2}{2})$

which is located in

$\displaystyle Quadrant\, I$

We substitute the given $\displaystyle t$ value into the parametric function.

$\displaystyle (3(-2), \,(-2)^3)$

$\displaystyle =(-6,-8)$

The resulting coordinate is in

$\displaystyle Quadrant\,III$