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

Where:

Since we are also given  and , we can easily compute the derivatives of each:

Applying these into the above formula results in:

We can factor out the common , and pull it outside of the square-root, and we will notice one of the most common trigonometric identities:

The term inside the square-root symbol can be simplified to .

This is one of the answer choices.

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

Where:

Since we are also given  and , we can easily compute the derivatives of each, using the Product Rule:

Applying these into the above formula results in:

Simplifying the above will require these two formulas:

It may also be useful to know this formula:

We can factor out the common  to make the above expression easier to look at:

We can take the  outside of the square-root by cancelling out the  representing the "square". Then we can apply formulas  &  to the trigonometric expressions:

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

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

Therefore the integral now becomes:

This is a simple integral which can be solved using u-substitution. But first, we can factor out the constant term , outside of the integral:

We will make our substitutions:

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

Our new integral becomes:

This is one of the answer choices.

At , the graph passes through 

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

So to obtain the slope, we just use:

 ,

and evaluate at .

As it turns out,  at , and , so the slope will be 0 for this curve at the point .

That means that , and so solving at ordered pair , the solution must be:

Perform these operations:

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

And this is the equation of an ellipse, centered at  with horizontal axis  and vertical axis .

Given  and  let's solve both equations for :

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

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

.

Given  and , we can use using the Power Rule

for all ,

to derive 

 and .

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

Now, using the Power Rule for Integrals  for all , we can determine that:

It is known that we can derive  with the formula

So we just find :

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

Power Rule: 

Chain Rule: 

Rules of Exponentials: 

Applying these rules we get the following.

so we have 

.

It is known that we can derive  with the formula

So we just find :

To find the derivatives we will need to use trigonometric rules and the rule for natural logs.

Trigonometric Rule for cosine: 

Rule of Natural Log: 

Applying the above rules we get the following derivatives.

so we have 

.

Using the identity , we can plug in the values for and  for to obtain the equation . This is the graph of a horizontal hyperbola with x-intercepts of and with asympotes of . The picture is depicted below:

When 

we have that

This gives the coordinate

which is located in