4. Parametric Equations and Polar Coordinates

Partition the interval $$ into $$ and evaluate $$ and plot points. The resulting curve is given below. The orientation is clockwise.

The initial point is $$ and final point is $$. It seems as though the curve is a parabola. To find a Cartesian equation, start with $$ and from the $$-equation we get $$ and thus $$.

Use the estimate $$. Evaluate the parametric equations along convenient $$ values:

The curve seems to be a portion of an ellipse. Recall that an ellipse centered at $$ with horizontal radius $$ and vertical radius $$ has Cartesian equation $$ Thus, we suspect that the Cartesian equation of the curve is $$ To eliminate the parameter start with $$ and then $$ Now $$ and thus $$ and thus $$

First determine a CCW orientation and then change the signs accordingly. The ellipse is:

The general equations of the parametrization of an ellipse in this case is $$ We need the interval to be $$. This gives a CCW orientation. To change the orientation, we can change the sign in front of the $$ term: $$

A parametrization is $$ with $$. We can find a right-to-left parametrization by changing all $$'s to $$ and changing the interval to $$. In this case, a right-to-left parametrization is then $$ with interval $$.

The slope of the line is $$ The equation of the line is then $$. A parametrization of the entire line is $$ and since we only want the line segment where $$ then the interval is $$.

We could solve for $$ in terms of $$: $$ However, in some cases we may only have a parametric equation for a curve and even if we had a Cartesian equation we may not be able to solve for $$ (could use implicit differentiation). Use instead a parametrization $$ We now need a way to find the slope of the tangent line in terms of the parametric equations. We do know that $$ near $$ and thus $$. Therefore, by the chain rule $$ and therefore $$ Because $$ this is sometimes written as $$ Hence, in this case $$ At what $$ value do we evaluate? It has to correspond to the $$ value where the parametrization passes through the point $$. The value $$ where the parametrization passes through the point $$ occurs when $$ From $$ either $$ or $$. In this case, need to take $$. Hence, we obtain $$ Therefore, equation of line is $$ which simplifies to $$

The curve is shown in Figure 4.7.

1. We compute $$ The curve passes through the point $$ when $$, so $$ but need $$. At $$ get $$ but at $$. Hence, $$ value is $$. Hence, $$ Equation of line is $$
2. Tangent line is horizontal when $$ which occurs at $$ values $$. The points are therefore $$ and $$.
3. Tangent line is vertical when $$ which occurs at $$. The point is $$.
4. To find a Cartesian equation notice that $$ and thus $$ is a Cartesian equation.

1. $$ which can be factored as $$ Now $$ represents only the origin. The other points on $$ therefore satisfy $$ which can be written as $$
2. Since $$ then $$ or $$.
3. Expanding gives $$ and then $$ or $$. Use $$ because when $$ get also $$.

1. $$ or $$ (vertical line)
2. $$ and completing square gives $$ (circle at $$ of radius $$)
3. Write as $$ and then $$ or $$ (line)
4. Get $$ and thus $$ or $$ and complete square to get $$.

Create a table of $$ coordinates by varying $$ at step-size $$:

This curve is called a cardioid. Since $$, we can obtain a parametrization for this curve as follows: $$ In general, if a curve is expressed in polar coordinates as $$ for $$ then a parametrization for the curve is: $$

Recall that $$ if given a parametrization $$ and $$. Here we have $$ We get $$ and evaluating at $$ we obtain $$. The point on the curve at $$ is $$ and thus the equation of the line is $$.

A complete revolution of the cardioid requires $$. Now $$ and then $$ Now $$ and thus $$

Compute $$

Apply the formula: $$

1. Evaluating $$ and $$ from $$ to $$ at step-size of $$, we obtain the following graph:

   

2. $$
3. $$
4. Recall that the Trapezoidal rule is for estimating an integral $$ Partitioning $$ into $$ equal subintervals, each subinterval has length $$. The Trapezoidal rule is then $$ where $$ are the points obtained after subdividing the interval $$. We wish to apply the Trapezoidal rule to $$ Hence, here $$, $$, $$, and the points to evaluate $$ are $$, $$, $$, $$, and $$. Hence, $$
5. Recall that given a parametrized curve $$ and $$, the slope of the line tangent to the curve at $$ is $$ The tangent line is vertical when $$. Here, $$ Then $$ In the interval $$, $$ when $$ and $$. Hence, the points are $$ and $$. This agrees with the sketch of the curve.

1. Evaluating $$ and $$ from $$ to $$ at step-size of $$, we obtain the curve shown in Figure 4.14.

   

2. Since $$, the arc length of the curve along for $$ is $$
3. Let $$. Applying Simpson's rule, we obtain that $$ and then the grid points are $$, $$, $$, $$, and $$. The symbolic form of Simpson's rule is then $$
4. The area enclosed by the curve is $$