PARAMETRIC EQUATIONS & POLAR COORDINATES So far, we have described plane curves by giving:  y as a function of x [y = f(x)] or x as a function of y [x = g(y)]  A relation between x and y that defines y implicitly as a function of x [f(x, y) = 0]

PARAMETRIC EQUATIONS & POLAR COORDINATES In this chapter, we discuss two new methods for describing curves.

Some curves—such as the cycloid—are best handled when both x and y are given in terms of a third variable t called a parameter [x = f(t), y = g(t)]. PARAMETRIC EQUATIONS x = at - a sin(t), y = a - a cos(t)

Other curves—such as the cardioid—have their most convenient description when we use a new coordinate system, called the polar coordinate system. POLAR COORDINATES edu/msrenault/ggb/po lar_grapher.html

INTRODUCTION Imagine that a particle moves along the curve C shown here.  It is impossible to describe C by an equation of the form y = f(x).  This is because C fails the Vertical Line Test.

However, the x- and y-coordinates of the particle are functions of time.  So, we can write x = f(t) and y = g(t). INTRODUCTION

Such a pair of equations is often a convenient way of describing a curve and gives rise to the following definition. INTRODUCTION

Suppose x and y are both given as functions of a third variable t (called a parameter) by the equations x = f(t) and y = g(t)  These are called parametric equations. PARAMETRIC EQUATIONS

Each value of t determines a point (x, y), which we can plot in a coordinate plane. As t varies, the point (x, y) = (f(t), g(t)) varies and traces out a curve C.  This is called a parametric curve. PARAMETRIC CURVE

The parameter t does not necessarily represent time.  In fact, we could use a letter other than t for the parameter. PARAMETER t

However, in many applications of parametric curves, t does denote time.  Thus, we can interpret (x, y) = (f(t), g(t)) as the position of a particle at time t.

Sketch and identify the curve defined by the parametric equations x = t 2 – 2t y = t + 1 Example 1 PARAMETRIC CURVES

Each value of t gives a point on the curve, as in the table.  For instance, if t = 0, then x = 0, y = 1.  So, the corresponding point is (0, 1). PARAMETRIC CURVES Example 1

Now, we plot the points (x, y) determined by several values of the parameter, and join them to produce a curve. PARAMETRIC CURVES Example 1

A particle whose position is given by the parametric equations moves along the curve in the direction of the arrows as t increases. PARAMETRIC CURVES Example 1

Notice that the consecutive points marked on the curve appear at equal time intervals, but not at equal distances.  That is because the particle slows down and then speeds up as t increases. PARAMETRIC CURVES Example 1

It appears that the curve traced out by the particle may be a parabola.  We can confirm this by eliminating the parameter t, as follows. PARAMETRIC CURVES Example 1

We obtain t = y – 1 from the equation y = t + 1. We then substitute it in the equation x = t 2 – 2t.  This gives: x = t 2 – 2t = (y – 1) 2 – 2(y – 1) = y 2 – 4y + 3  So, the curve represented by the given parametric equations is the parabola x = y 2 – 4y + 3 PARAMETRIC CURVES Example 1

This equation in x and y describes where the particle has been.  However, it doesn’t tell us when the particle was at a particular point. PARAMETRIC CURVES

The parametric equations have an advantage––they tell us when the particle was at a point. They also indicate the direction of the motion. ADVANTAGES

No restriction was placed on the parameter t in Example 1. So, we assumed t could be any real number.  Sometimes, however, we restrict t to lie in a finite interval. PARAMETRIC CURVES

For instance, the parametric curve x = t 2 – 2t y = t ≤ t ≤4 shown is a part of the parabola in Example 1.  It starts at the point (0, 1) and ends at the point (8, 5). PARAMETRIC CURVES

The arrowhead indicates the direction in which the curve is traced as t increases from 0 to 4. PARAMETRIC CURVES

In general, the curve with parametric equations x = f(t) y = g(t) a ≤ t ≤ b has initial point (f(a), g(a)) and terminal point (f(b), g(b)). INITIAL & TERMINAL POINTS

What curve is represented by the following parametric equations? x = cos t y = sin t 0 ≤ t ≤ 2π PARAMETRIC CURVES Example 2

If we plot points, it appears the curve is a circle.  We can confirm this by eliminating t. PARAMETRIC CURVES Example 2

Observe that: x 2 + y 2 = cos 2 t + sin 2 t = 1  Thus, the point (x, y) moves on the unit circle x 2 + y 2 = 1 PARAMETRIC CURVES Example 2

Notice that, in this example, the parameter t can be interpreted as the angle (in radians), as shown. PARAMETRIC CURVES Example 2

As t increases from 0 to 2π, the point (x, y) = (cos t, sin t) moves once around the circle in the counterclockwise direction starting from the point (1, 0). PARAMETRIC CURVES Example 2

What curve is represented by the given parametric equations? x = sin 2t y = cos 2t 0 ≤ t ≤ 2π Example 3 PARAMETRIC CURVES

Again, we have: x 2 + y 2 = sin 2 2t + cos 2 2t = 1  So, the parametric equations again represent the unit circle x 2 + y 2 = 1 PARAMETRIC CURVES Example 3

However, as t increases from 0 to 2π, the point (x, y) = (sin 2t, cos 2t) starts at (0, 1), moving twice around the circle in the clockwise direction. PARAMETRIC CURVES Example 3

Examples 2 and 3 show that different sets of parametric equations can represent the same curve. So, we distinguish between:  A curve, which is a set of points  A parametric curve, where the points are traced in a particular way PARAMETRIC CURVES

Find parametric equations for the circle with center (h, k) and radius r. Example 4 PARAMETRIC CURVES

We take the equations of the unit circle in Example 2 and multiply the expressions for x and y by r. We get: x = r cos t y = r sin t  You can verify these equations represent a circle with radius r and center the origin traced counterclockwise. Example 4 PARAMETRIC CURVES

Now, we shift h units in the x-direction and k units in the y-direction. Example 4 PARAMETRIC CURVES

Thus, we obtain the parametric equations of the circle with center (h, k) and radius r : x = h + r cos t y = k + r sin t 0 ≤ t ≤ 2π Example 4 PARAMETRIC CURVES

Sketch the curve with parametric equations x = sin t y = sin 2 t Example 5 PARAMETRIC CURVES

Observe that y = (sin t) 2 = x 2. Thus, the point (x, y) moves on the parabola y = x2. Example 5

However, note also that, as -1 ≤ sin t ≤ 1, we have -1 ≤ x ≤ 1.  So, the parametric equations represent only the part of the parabola for which -1 ≤ x ≤ 1. Example 5 PARAMETRIC CURVES

Since sin t is periodic, the point (x, y) = (sin t, sin 2 t) moves back and forth infinitely often along the parabola from (-1, 1) to (1, 1). Example 5 PARAMETRIC CURVES

Use a graphing device to graph the curve x = y 4 – 3y 2  If we let the parameter be t = y, we have the equations x = t 4 – 3t 2 y = t Example 6 GRAPHING DEVICES

 Using those parametric equations, we obtain this curve. Example 6 GRAPHING DEVICES

It would be possible to solve the given equation for y as four functions of x and graph them individually.  However, the parametric equations provide a much easier method. Example 6 GRAPHING DEVICES

In general, if we need to graph an equation of the form x = g(y) or y = f(x) we can use the parametric equations x = g(t) y = t or x = t y = f(t) GRAPHING DEVICES

For instance, these curves would be virtually impossible to produce by hand. COMPLEX CURVES

One of the most important uses of parametric curves is in computer-aided design (CAD). CAD

Special parametric curves called Bézier curves are used in manufacturing, especially in the automotive industry. They are also employed in specifying the shapes of letters and other symbols in laser printers. BÉZIER CURVES

CYCLOID The curve traced out by a point P on the circumference of a circle as the circle rolls along a straight line is called a cycloid. Example 7

Find parametric equations for the cycloid if:  The circle has radius r and rolls along the x-axis.  One position of P is the origin. Example 7 CYCLOIDS

We choose as parameter the angle of rotation θ of the circle (θ = 0 when P is at the origin). Suppose the circle has rotated through θ radians. Example 7 CYCLOIDS

As the circle has been in contact with the line, the distance it has rolled from the origin is: | OT | = arc PT = rθ  Thus, the center of the circle is C(rθ, r). Example 7 CYCLOIDS

Let the coordinates of P be (x, y). Then, from the figure, we see that:  x = |OT| – |PQ| = rθ – r sin θ = r(θ – sinθ)  y = |TC| – |QC| = r – r cos θ = r(1 – cos θ) Example 7 CYCLOIDS

Therefore, parametric equations of the cycloid are: x = r(θ – sin θ) y = r(1 – cos θ) θ R E. g. 7—Equation 1 CYCLOIDS

One arch of the cycloid comes from one rotation of the circle. So, it is described by 0 ≤ θ ≤ 2π. x = r(θ – sin θ) y = r(1 – cos θ) θ R Example 7 CYCLOIDS

It is possible to eliminate the parameter θ from Equations 1. However, the resulting Cartesian equation in x and y is:  Very complicated  Not as convenient to work with Example 7 PARAMETRIC VS. CARTESIAN

One of the first people to study the cycloid was Galileo.  He proposed that bridges be built in the shape.  He tried to find the area under one arch of a cycloid. CYCLOIDS

Later, this curve arose in connection with the brachistochrone problem—proposed by the Swiss mathematician John Bernoulli in 1696:  Find the curve along which a particle will slide in the shortest time (under the influence of gravity) from a point A to a lower point B not directly beneath A. BRACHISTOCHRONE PROBLEM

Bernoulli showed that, among all possible curves that join A to B, the particle will take the least time sliding from A to B if the curve is part of an inverted arch of a cycloid. BRACHISTOCHRONE PROBLEM

The Dutch physicist Huygens had already shown that the cycloid is also the solution to the tautochrone problem:  No matter where a particle is placed on an inverted cycloid, it takes the same time to slide to the bottom. TAUTOCHRONE PROBLEM

Huygens mounted cycloidal-shaped metal 'cheeks' next to the pivot in his 1673 clock, that constrained the suspension cord and forced the pendulum to follow a cycloid arc.  Then, the pendulum takes the same time to make a complete oscillation— whether it swings through a wide or a small arc.  This was a great improvement over existing mechanical clocks; their best accuracy was increased from around 15 minutes a day to around 15 seconds a day. CYCLOIDS & PENDULUMS

PARAMETRIC EQUATIONS & POLAR COORDINATES We have seen how to represent curves by parametric equations. Now, we apply the methods of calculus to these parametric curves.

TANGENTS we have seeb that some curves defined by parametric equations x = f(t) and y = g(t) can also be expressed—by eliminating the parameter—in the form y = F(x). If we substitute x = f(t) and y = g(t) in the equation y = F(x), we get: g(t) = F(f(t)) So, if g, F, and f are differentiable, the Chain Rule gives: g’(t) = F’(f(t))f’(t) = F’(x)f’(t)

If f’(t) ≠ 0, we can solve for F’(x): TANGENTS Equation 1

The slope of the tangent to the curve y = F(x) at (x, F(x)) is F’(x). Thus, Equation 1 enables us to find tangents to parametric curves without having to eliminate the parameter. TANGENTS

Using Leibniz notation, we can rewrite Equation 1 in an easily remembered form: TANGENTS Equation 2

If we think of a parametric curve as being traced out by a moving particle, then  dy/dt and dx/dt are the vertical and horizontal velocities of the particle.  Formula 2 says that the slope of the tangent is the ratio of these velocities. TANGENTS

From Equation 2, we can see that the curve has:  A horizontal tangent when dy/dt = 0 (provided dx/dt ≠ 0).  A vertical tangent when dx/dt = 0 (provided dy/dt ≠ 0). TANGENTS

This information is useful for sketching parametric curves.

It is also useful to consider d 2 y/dx 2. This can be found by replacing y by dy/dx in Equation 2: TANGENTS

A curve C is defined by the parametric equations x = t 2, y = t 3 – 3t. a.Show that C has two tangents at the point (3, 0) and find their equations. b.Find the points on C where the tangent is horizontal or vertical. c.Determine where the curve is concave upward or downward. d.Sketch the curve. TANGENTS Example 1

Notice that y = t 3 – 3t = t(t 2 – 3) = 0 when t = 0 or t = ±.  Thus, the point (3, 0) on C arises from two values of the parameter: t = and t = –  This indicates that C crosses itself at (3, 0). TANGENTS Example 1 a

Since the slope of the tangent when t = ± is: dy/dx = ± 6/(2 ) = ± TANGENTS Example 1 a

So, the equations of the tangents at (3, 0) are: TANGENTS Example 1 a

C has a horizontal tangent when dy/dx = 0, that is, when dy/dt = 0 and dx/dt ≠ 0.  Since dy/dt = 3t 2 - 3, this happens when t 2 = 1, that is, t = ±1.  The corresponding points on C are (1, -2) and (1, 2). TANGENTS Example 1 b

C has a vertical tangent when dx/dt = 2t = 0, that is, t = 0.  Note that dy/dt ≠ 0 there.  The corresponding point on C is (0, 0). TANGENTS Example 1 b

To determine concavity, we calculate the second derivative: x = t 2,  The curve is concave upward when t > 0.  It is concave downward when t < 0. TANGENTS Example 1 c

Using the information from (b) and (c), we sketch C. TANGENTS Example 1 d

a.Find the tangent to the cycloid x = r(θ – sin θ), y = r(1 – cos θ ) at the point where θ = π/3. a.At what points is the tangent horizontal? When is it vertical? TANGENTS Example 2

The slope of the tangent line is: TANGENTS Example 2 a

When θ = π/3, we have and TANGENTS Example 2 a

Hence, the slope of the tangent is. Its equation is: TANGENTS Example 2 a

The tangent is sketched here. TANGENTS Example 2 a

The tangent is horizontal when dy/dx = 0.  This occurs when sin θ = 0 and 1 – cos θ ≠ 0, that is, θ = (2n – 1)π, n an integer.  The corresponding point on the cycloid is ((2n – 1)πr, 2r). x = r(θ – sin θ), y = r(1 – cos θ ) Example 2 b TANGENTS

When θ = 2nπ, both dx/dθ and dy/dθ are 0. It appears from the graph that there are vertical tangents at those points. TANGENTS Example 2 b

We can verify this by using l’Hospital’s Rule as follows: TANGENTS Example 2 b

A similar computation shows that dy/dx → -∞ as θ → 2nπ –.  So, indeed, there are vertical tangents when θ = 2nπ, that is, when x = 2nπr. TANGENTS Example 2 b

AREAS We know that the area under a curve y = F(x) from a to b is where F(x) ≥ 0.

Suppose the curve is traced out once by the parametric equations x = f(t) and y = g(t), α ≤ t ≤ β.  Then, we can calculate an area formula by using the Substitution Rule for Definite Integrals. AREAS

Find the area under one arch of the cycloid x = r(θ – sin θ) y = r(1 – cos θ) AREAS Example 3

One arch of the cycloid is given by 0 ≤ θ ≤ 2π. Using the Substitution Rule with x = r(θ – sin θ) y = r(1 – cos θ) and dx = r(1 – cos θ) dθ, we have the following result. AREAS Example 3

AREAS Example 3

AREAS The result of Example 3 says that the area under one arch of the cycloid is three times the area of the rolling circle that generates the cycloid.  Galileo guessed this result.  However, it was first proved by the French mathematician Roberval and the Italian mathematician Torricelli. Example 3

ARC LENGTH We already know how to find the length L of a curve C given in the form y = F(x), a ≤ x ≤ b

ARC LENGTH Formula 3 in Section 8.1 says that, if F’ is continuous, then Equation 3

Suppose that C can also be described by the parametric equations x = f(t) and y = g(t), α ≤ t ≤ β, where dx/dt = f’(t) > 0.  This means that C is traversed once, from left to right, as t increases from α to β and f(α) = a and f(β) = b. ARC LENGTH

Putting Formula 2 into Formula 3 and using the Substitution Rule, we obtain: ARC LENGTH

Since dx/dt > 0, we have: ARC LENGTH Formula 4

Even if C can’t be expressed in the form y = f(x), Formula 4 is still valid. However, we obtain it by polygonal approximations. ARC LENGTH

Thus, ARC LENGTH Equation 5

The sum in Equation 5 resembles a Riemann sum for the function. However, is not exactly a Riemann sum because t i * ≠ t i ** in general. ARC LENGTH

Nevertheless, if f’ and g’ are continuous, it can be shown that the limit in Equation 5 is the same as if t i * and t i ** were equal, namely, ARC LENGTH

Thus, using Leibniz notation, we have the following result—which has the same form as Formula 4. ARC LENGTH

Let a curve C is described by the parametric equations x = f(t), y = g(t), α ≤ t ≤ β, where:  f’ and g’ are continuous on [α, β].  C is traversed exactly once as t increases from α to β. Then, the length of C is: THEOREM Theorem 6

Suppose we use the representation of the unit circle : x = cos t y = sin t 0 ≤ t ≤ 2π  Then, dx/dt = –sin t and dy/dt = cos t ARC LENGTH Example 4

 So, as expected, Theorem 6 gives: ARC LENGTH Example 4

On the other hand, suppose we use the representation : x = sin 2t y = cos 2t 0 ≤ t ≤ 2π  Then, dx/dt = 2cos 2t and dy/dt = – 2sin 2t ARC LENGTH Example 4

 Then, the integral in Theorem 6 gives: ARC LENGTH Example 4

Notice that the integral gives twice the arc length of the circle ?  This is because, as t increases from 0 to 2π, the point (sin 2t, cos 2t) traverses the circle twice. ARC LENGTH Example 4

In general, when finding the length of a curve C from a parametric representation, we have to be careful to ensure that C is traversed only once as t increases from α to β. ARC LENGTH Example 4

SURFACE AREA In the same way as for arc length, we can obtain a formula for surface area.

Let the curve given by the parametric equations x = f(t), y = g(t), α ≤ t ≤ β, be rotated about the x-axis, where:  f’, g’ are continuous.  g(t) ≥ 0. Then, the area of the resulting surface is given by: SURFACE AREA Formula 7

The general symbolic formulas S = ∫ 2πy ds and S = ∫ 2πx ds. However, for parametric curves, we use: SURFACE AREA

Show that the surface area of a sphere of radius r is 4πr 2.  The sphere is obtained by rotating the semicircle x = r cos t y = r sin t 0 ≤ t ≤ π about the x-axis. SURFACE AREA Example 6

 So, from Formula 7, we get: SURFACE AREA Example 6