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:

$\displaystyle \frac{dy}{dt}=5,\frac{dx}{dt}=\frac{1}{2\sqrt{t+5}}$

If we plug in t=4, into the above equations:

$\displaystyle \frac{dy}{dx}=\frac{5}{\frac{1}{2\sqrt{9}}}=\frac{5}{\frac{1}{2(3)}}=\frac{5}{\frac{1}{6}}=30$

This is one of the answer choices.

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.

To convert from parametric to rectangular form, we must eliminate the parameter by finding t in terms of x or y:

$\displaystyle t=x-5$

Now, plug t into the equation for y:

$\displaystyle y=\sec^2(x-5)$

Given $\displaystyle x=5t+12$ and $\displaystyle y=9-3t$,  let's solve both equations for :

$\displaystyle x=5t+12\rightarrow t=\frac{x-12}{5}$

$\displaystyle y=9-3t\rightarrow t=\frac{9-y}{3}$

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

$\displaystyle \frac{9-y}{3}=\frac{x-12}{5}$

$\displaystyle 9-y=3(\frac{x-12}{5})$

$\displaystyle y=9-\frac{3}{5}(x-12)$

To see why the parametric equations

$\displaystyle x(t)=4\cos t$, $\displaystyle y(t)=4\sin t$

describe the circle

$\displaystyle x^2+y^2=16$

we do the following manipulation:

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

which means

$\displaystyle \left(\frac{x(t)}{4} \right )^2+\left( \frac{y(t)}{4}\right )^2=1$

or 

$\displaystyle x(t)^2+y(t)^2=16$

which describes the circle of radius $\displaystyle 4$.

The reason that the parametric equations

$\displaystyle x(t)=3\sin t, y(t)=2\cos t$

don't describe the ellipse

$\displaystyle \frac{x^2}{4}+\frac{y^2}{9}=1$

is because if do some algebraic manipulation, we get

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

which describes a different ellipse than

$\displaystyle \frac{x^2}{4}+\frac{y^2}{9}=1$.

We can see that the parametric equations

$\displaystyle x(t)=\cos t, y(t)=\sin^2 t$

describe a section of the parabola

$\displaystyle y=1-x^2$

because if do some manipulations of the parametric equations, we get 

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

$\displaystyle \Rightarrow x(t)^2+y(t)=1$

So then we get

$\displaystyle y(t)=1-x(t)^2$

which describes part of the parabola 

$\displaystyle y=1-x^2$.

The parametric equations

$\displaystyle x(t)=2\cos t, y(t)=3\sin t$

describe an ellipse because we have

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

which means 

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

which is the equation for an ellipse.

Given $\displaystyle x=t^{2}-5$ and $\displaystyle y=3t+8$, et's solve both equations for :

$\displaystyle x=t^{2}-5 \rightarrow t=\pm\sqrt{x+5}$

$\displaystyle y=3t+8\rightarrow t=\frac{y-8}{3}$

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

$\displaystyle \frac{y-8}{3}=\pm\sqrt{x+5}$

$\displaystyle y-8=3(\pm\sqrt{x+5})$

$\displaystyle y=3(\pm\sqrt{x+5})+8$

Given $\displaystyle x=4+10t$ and $\displaystyle y=2-t$, let's solve both equations for :

$\displaystyle x=4+10t\rightarrow t=\frac{x-4}{10}$

$\displaystyle y=2-t\rightarrow t=2-y$

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

$\displaystyle 2-y=\frac{x-4}{10}$

$\displaystyle y=2-\frac{x-4}{10}$