Chapter 10 Curves Given Parametrically

Introduction

So far we have had two different ways of giving a curve. The first was explicitly , i.e., in the form . For example, the equation gives us the upper semicircle of radius centered at the origin. The second way was implicitly, usually in the form . For example, the equation gives us the circle (both upper and lower semicircles) of radius centered at the origin.

Suppose we want to describe the points on the circumference of the circle of radius centered at the origin. A common way to do this is in terms of the angle which the line from the point to the origin makes with the positive x-axis (measured counterclockwise). That is, as goes from 0 to radians, we sweep once around the circle:

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= , = ,

The fact that the -coordinate is simply the cosine and the -coordinate is simply the sine is not surprising, since we defined the cosine and sine of an angle to be the - and -coordinates of the point that you get by going around the unit circle through the given angle.

If we want to shift the circle of radius so that its center is at the point , we can do this by adding to the x-formula and
k to the y-formula:

= ,
= ,

This is the circle which is given implicitly by the equation .

If instead of we had written, say, , then we would have described only half of the circle (the upper semicircle). If we had written , then we would have described going around the circle twice.

Let us suppose that we have put our ellipse on the axes in the following convenient way: its center is at the origin, its longer axis of symmetry --- of length --- is along the -axis, and its shorter axis of symmetry --- of length --- is along the -axis. Such an ellipse can be obtained by taking the unit circle and stretching it by a factor of in the x-direction and by a factor of in the y-direction. (If or is less than 1, the ``stretching'' is actually a contraction.) In terms of the parametric equations, we can accomplish this stretching simply by multiplying the x-equation for the unit circle by and multiplying the -equation for the unit circle by :

= , = , .

This is the parametric form for our ellipse --- the ellipse which can be given implicitly in the form

.

Note: A word of warning, though, about the parameter for an ellipse. It no longer has the meaning of the angle formed by the line from the point ( ) to the origin. For example, suppose a = 2, b = 1. Then the point on the ellipse corresponding to the parameter value = is the point = ,
= ; and it is easy to see that the line joining the
point ( ) to the origin does not make a 45^o angle with the x-axis.

There is a geometric way to interpret the variable as a suitable angle anyway. This is illustrated from the following picture of an ellipse with its auxiliary circle. The ellipse in space is oriented so that its projection on the x-y plane is a circle whose radius is , half the minor axis. It is easy to convince yourself that such a projection is possible - just rotate the plane of the ellipse on the minor axis of the ellipse minor axis until the projection of the rotated ellipse into the original plane is a circle. The angle, , spanned by the projections of half of the minor axis and the ray to the point P can be interpreted as the parameter that corresponds to P.

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It is very common to want to describe a curve in terms of the time parameter . In that case we sometimes call the curve the ``path'' or `` trajectory '' of our point (which we think of as a small object or particle). As goes from an initial time to a final time , the equations tell us where the particle is at that particular time .

Suppose a particle is traveling counterclockwise at 3 rev/sec around the circle of radius centered at the origin. It starts at the point ( ) at time . We want to give a parametric description of the motion in terms of . To do this, we find an expression in terms of for the angle through which the particle has rotated, and then we substitute this expression in the equations in Example 10.1. Because we shall be measuring
angles in radians, we first translate the angular velocity of rev/sec into units of rad/sec. Since , we obtain:
= rad/sec. Thus, starts at at time (because our point starts at ( ) on the positive x-axis), and it increases at the constant rate of . Hence, after seconds it has value . Finally, substituting this expression for in the equations in Example 10.1, we obtain:

= , = , ,

where we chose to take between and so that we would describe exactly one rotation of the particle around the circle.

Example 10.4: Motion at constant velocity between two points

= , = , .

We can check these formulas by noting that when they give us , (i.e., we're at the point ), and when they give us , (i.e., we're at the point ).

Again we are interested in an object falling under the influence of gravity (we neglect air resistance). In the past (see Example 6.9 ) our object was moving only in one dimension --- up and down. For example, if we drop a ball from a window meters high, its height at time is given by the formula = (here m/ ; the acceleration is negative, i.e., it is , because we are choosing up to be the positive direction).

Now suppose that, instead of dropping the ball from the window, we throw it horizontally out from the window at m/sec. The ball then moves in the -plane, where we have chosen the side of the building as the -axis and the line along the ground under the ball's path as the -axis. Its path is a parabola , as we shall soon see. Our task in this example is to write parametric equations for this motion.

To do this, we again consider the - and -directions separately. In the -direction, there is no force affecting the velocity, and so the m/sec horizontal component of velocity of the ball remains unchanged. That is, as in Example 10.4, in the -direction we are simply traveling at constant speed. Since we start at initial location , the formula for at time is:

.

In the vertical direction we have a falling body problem, and so can use the formula = . Here it is important to understand what means. is the initial velocity in the y- direction .

In this example is , since the ball's initial velocity is entirely horizontal. In other words, in the y-direction the problem is exactly as if we had simply dropped the ball from a height of meters. Thus, we have

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The velocity of an object moving in the plane is an example of a vector . A vector is something which has magnitude and direction . It can be visualized as an arrow of a certain length magnitude pointing in a certain direction. The magnitude of the velocity vector is called the speed of the object.

The same as Example 10.5, except that at time the ball is thrown at an angle of above the horizontal.

Here the initial velocity vector makes an angle of above the horizontal and has a magnitude of 10 m/sec. That is, it is partly horizontal and partly vertical. The first step in solving this problem is to resolve the initial velocity vector into its horizontal and vertical components . To resolve the velocity vector, we draw the velocity triangle having the velocity vector as its hypotenuse (see the figure).

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The same as Example 10.6, except that the ball is thrown at an angle of below the horizontal.

Here the only change is that the tip of the velocity triangle points downward, i.e., its vertical leg points in the negative direction. Thus, in the equations of motion the only thing that changes is the sign of in the falling body formula:

= , = .

Of course, the ball hits the ground much sooner (see following figure).

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Suppose we want to know the instant when the ball in Example 10.6 reaches the peak of its trajectory. At that instant the path has a horizontal tangent line, i.e., . Since , this slope is zero if and only if the numerator is . Thus, to find the instant when the ball reaches its highest point we set the vertical velocity equal to and solve for . This is the same principle that we saw in our earlier one-dimensional falling body problems: the ball is at the peak when its vertical velocity is zero. ( Warning : Be sure not to confuse this procedure, where we set equal to zero, with setting itself equal to , which we do to find the time when the ball hits the ground.) Since = ( ) = , we have , i.e., = 0.88 sec. If we also want to know the maximum height reached by the ball, we take this value of and substitute it in the formula for the height y, i.e.,

= m.

Suppose that we want to know the direction in which the ball in Example 10.6 is headed after 0.5 sec, and also at the instant when it hits the ground. We first take the time derivatives of and :

( ) = and ,.

Then we have:

= ,

Thus, when we have = . At that instant the ball's location is ( ) = ( ), so that the tangent line to the path at the point ( ) has slope . We can also determine the angle at which the ball is traveling at time . By this we mean the angle above the horizontal of the tangent line to the path. To do that we observe that the slope of a line is opposite over adjacent, i.e., the tangent of the angle that the line makes with the horizontal. So to determine the angle we use the inverse-tan on our calculator: = . At the same time we could also construct the velocity triangle, having horizontal leg and vertical leg , and compute the hypotenuse (which is the speed):

speed at time = m/sec.

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