MATH1020 Calculus II	Maple Lab 3
Section 1-4	Due: Tuesday, September 28, 1999.

You should adhere to the lab policies when you submit your solution.

This lab explores parametric equations.

1.

 In this question, we investigate Bézier curves, which are used in the automotive industry to design the shape of cars. Bézier curves are also used by laser printers, which is the subject of this question. A cubic Bézier curve is determined by four control points, which we shall label P₀(x₀,y₀), P₁(x₁,y₁), P₂(x₂,y₂), and P₃(x₃,y₃). The curve is then given by the equations:

$$\begin{eqnarray*} x & := & x_0 (1-t)^3 + 3x_1t(1-t)^2 + 3x_2t^2(1-t) + x_3 t^3 \\ y & := & y_0 (1-t)^3 + 3y_1t(1-t)^2 + 3y_2t^2(1-t) + y_3 t^3, \end{eqnarray*}$$

where $0 \leq t \leq 1$. Notice that when t=0 we get (x,y)=(x₀,y₀), and when t=1 we have (x,y)=(x₃,y₃).

(a)

 Graph the Bézier curve with points P₀(4,1), P₁(28,48), P₂(50,42), and P₃(40,5). On the same figure, graph the line segments P₀P₁ and P₂P₃.

(b)

Prove that the tangent to the curve at P₀ passes through P₁, and that the tangent to the curve at P₃ passes through P₂.

(c)

Experiment with different choices of the control points to find a Bézier curve that gives a reasonable approximation to the letter C.

(d)

More complicated shapes can be obtained by piecing together two or more Bézier curves. Try to create an approximation to the letter S by using two Bézier curves.

2.

In this problem, you will estimate the area of the region enclosed by the loop of the curve corresponding to the parametric equations:

$$\begin{eqnarray*} u & := & 1 + \tan(t)(1-2 (\cos(t+1))^2) \\ v & := & 1 - \frac{2}{3} (\cos(t))^2. \\ \end{eqnarray*}$$

(a)

 Plot the curve for the interval [-Pi/3..Pi/3] with u on the horizontal axis and v on the vertical axis.

(b)

At what point does the curve cross itself? Estimate the u and v coordinates of the point of intersection by clicking the mouse on that point. Use the estimate of the value of v to find the values of t at the point of intersection. Call the smaller value t₀ and the larger value t₂. Check your answer by substituting t₀ and t₂ into the equation for u.

(c)

Find the value of t in the interval [-Pi/3..Pi/3] where the loop has a vertical tangent. Call this value t₁.

(d)

At what value of t in the interval [-Pi/3..Pi/3] does the loop have a horizontal tangent?

(e)

To determine the area of the loop in the interval [-Pi/3..Pi/3], we divide the curve into two distinct sections. Section #1 is from the intersection to the vertical tangent, and Section #2 is from the vertical tangent to the intersection. Use the following command to see these two sections:

plot $(\{[ u,v,t=-0.33*Pi..t1],[u,v,t=t1..0.33*Pi]\})$;

Include this plot in your worksheet. Label t₀, t₁, and t₂ by hand. Draw an arrow from t₀ to t₁ and another from t₁ to t₂.

(f)

 Calculate the area under each section of the curve and then calculate the total area A of the loop by adding the area under Section #1 and subtracting the area under Section #2. Use the following formula for A:

$$A := \int_{t_1}^{t_0} v (\frac{du}{dt}) dt \, - \, \int_{t_1}^{t_2} v (\frac{du}{dt}) dt.$$

Why do the limits on the first integral go from t₁ to t0?

(g)

Notice that in the equation for part (2f), we can reverse the limits of integration for the second integral and add this new integral to the other integral. By doing this we obtain an equation that allows us to calculate the area by integrating v*du from t₂ to t₀. Calculate this integral and verify that you obtain the same result as in part (2f).