## Calculus Of One Real Variable – By Pheng Kim Ving Chapter 13: Plane Curves – Section 13.1.2: Tangents And Sketching Of Parametric Curves

13.1.2
Tangents And Sketching Of Parametric Curves

 1. Smooth Curves

The parabola and the circle are smooth curves. The curve in Fig. 1.1 is smooth on R. The curve in Fig. 1.2 is smooth on any
interval not containing the origin (0, 0); it's not smooth on any interval containing the origin. A curve is said to be smooth if it
turns, well, smoothly, or continuously, without breaks or sharp points. So a curve is smooth if it has a tangent at each point of ### Fig. 1.1

Curve is smooth on R. ### Fig. 1.2

Curve isn't smooth on any interval containing origin (0, 0).

it and this tangent turns smoothly or continuously as the point of tangency moves along the curve. A tangent turning
continuously means that its slope changes continuously or is a continuous function of the position of the point of tangency. Thus

for the graph of a function y = F(x), the portion of it on an interval is smooth if the derivative dy/dx is continuous there. A curve is said to be smooth if it has a tangent at each point of it and this tangent turns smoothly or continuously as the point of tangency moves along the curve.   If the derivative dy/dx of a function y = F(x) is continuous on an interval, then the portion of its graph on that interval is smooth.   If dy/dx doesn't exist at some isolated points in an interval, the portion of the graph on that interval may or may not be smooth

 2. Tangents Of Parametric Curves

##### Slope Of Tangent

We're now going to investigate the Cartesian xy-equation of the tangent to the parametric curve x = f(t), y = g(t), using the
functions f and g, so that we don't have to find the corresponding Cartesian function y = F(x) or equation G(x, y) = 0.

To a given parameter value t1 there corresponds a unique point (x1, y1) = ( f(t1), g(t1)) on the curve. If the slope of the tangent
to the curve at the point (x1, y1) that corresponds to t1 is m then the Cartesian equation of the tangent is:

y = y1 + m(xx1).

We must specify that (x1, y1) is the one that corresponds to t1, because the same point (x1, y1) may correspond to more than 1
different value of t, as shown in Fig. 2.1, where (x1, y1) corresponds to 2 different t-values t1 and t2. We must specify which
t-value the point (x1, y1) corresponds to, because the tangent at (x1, y1) that corresponds to t1 may differ from the tangent at
(x1, y1) that corresponds to t2, as it does in Fig. 2.1.

Now back to the tangent. So our task is to find the slope of the tangent. Recall that the graph of the parametric curve
represents the x-y relationship, not the t-x or t-y relationship. Thus the slope of the tangent to it at the point (x, y)
corresponding to the parameter value t is dy/dx, not dx/dt or dy/dt. Utilizing the chain rule we get: so that:

 The slope of the tangent to the parametric curve x = f(t), y = g(t), at the point (x, y) corresponding to the parameter value t is:  ### Fig. 2.1

More than 1 different value of t may correspond to same
point (x1, y1) on curve. ### Fig. 2.2

y is a function of x on Nx that corresponds to Nt.

##### Conditions On f ' And g'

For the curve to have a tangent at (x1, y1) = ( f(t1), g(t1)) it must be smooth on an x-neighborhood Nx of x1 corresponding to a
t-neighborhood Nt of t1.

If f '(t) is continuous and never 0 on Nt, then by the intermediate-value theorem either f '(t) > 0 or f '(t) < 0 throughout Nt,
then x = f(t) is either increasing or decreasing on Nt, then the curve never reverses direction horizontally over Nx. Here Nx
corresponds to Nt. Then y is some function of x on Nx. Additionally, if g'(t) exists on Nt, then y is some differentiable function
of x on Nx. For the curve to be smooth on Nx, dy/dx must also be continuous on Nx, hence g'(t) must also, like f '(t), be
continuous on Nt. Therefore the piece of the curve is smooth and equations [2.1] and [2.2] are valid if both f '(t) and g'(t) are
continuous on a t-neighborhood of t1 and f '(t) is never 0 there, except possibly at isolated points. If f '(t1) = 0 then we must
consider the one-sided limits of g'(t)/ f '(t) at t1, as illustrated below.

We must specify that Nx is the one that corresponds to Nt, because Nx may also be mapped to by a different t-interval other
than Nt, as it is so in Fig. 2.2 by a t-neighborhood to the right of Nt, as evidenced by the existence of that upper portion of the
curve over Nx, that time in increasing t and decreasing x. We've seen that if f '(t) and g'(t) are continuous on a neighborhood of t1 and not both 0 at any point in it then the curve has a
tangent at the point x1 corresponding to t1 and is smooth on the x-neighborhood of x1 corresponding to the t-neighborhood of t1.

 Consider the tangent at the point (x1, y1) corresponding to a given parameter value t1. Suppose f '(t) and g'(t) are continuous on a t-neighborhood of t1. Let m be the slope of the tangent. Then:  For part 2a, the reason for the equality is that dy/dx is continuous. For part 2b, that’s because by the uniqueness of the
tangent there can't be more than 1 tangent corresponding to the same single value t1. There can be more than 1 tangent at (x1,
y1) if (x1, y1) corresponds to more than 1 different value of t; see Fig. 2.1. For part 2c, the reason is similar. For part 2d, that's
because 2 overlapping vertical tangents constitute 1 vertical tangent.

Note that, unlike the case of dy/dx, cases 2a and 2b show that the fact that f '(t) and g'(t) are continuous doesn't always lead
to the curve being smooth and having a tangent. This situation may occur when f '(t1) and g'(t1) are both 0. If f '(t1) and g'(t1)
are both 0 then the curve may or may not have a tangent at x1. Cases 1 and 2d show that if f '(t1) and g'(t1) aren't both 0 then
the curve has a tangent at x1.

##### Example 2.1 Solution EOS

##### Example 2.2

Determine the Cartesian equation of the tangent to the curve defined parametrically by x = s2, y = cos s, at the point where
s = 0.

Solution
When s = 0 we get x = 0 and y = 1. We have: EOS

##### Example 2.3

Find the Cartesian equation of the tangent to the curve given by the parametric equations x = t2 – 6t + 1, y = t + 2, at the
point where t = 3.

Solution

If t = 3 then x = – 8 and y = 5. Now dx/dt = 2t – 6 = 2(t – 3) = 0 at t = 3, dy/dt = 1 for all t including t = 3. So the desired
tangent is vertical  with Cartesian equation x = – 8.

EOS

##### Example 2.4

Find all points where the parametric curve x = s2 – 4s – 2, y = s + 3, s in R, has a vertical tangent. Determine the Cartesian
equation of each vertical tangent.

Solution
dx/ds = 2s – 4 = 2(s – 2), dy/ds = 1,
dx/ds = 0 at s = 2, dy/ds is non-0 there as it's never 0,
at s = 2 we have x = 22 – 4(2) – 2 = – 6 and y = 2 + 3 = 5.
The vertical tangent is at (– 6, 5). Its Cartesian equation is of course x = – 6.
EOS

 3. Sketching Of Parametric Curves By Using Parametric Functions

In Section 13.1.1 Part 2 we sketched parametric curves by eliminating the parameter to obtain a Cartesian function y = F(x) or
equation G(x, y) = 0 and using this function or equation. Now we're interested in sketching the parametric curve x = f(t), y =
g(t) by using dy/dx and d2y/dx2 expressed in terms of the derivatives of the parametric functions f and g, so that we don't
have to find the corresponding Cartesian function or equation.

##### Tangent ##### Increasing/Decreasing

Utilize dy/dx = g'(t)/ f '(t). Clearly as t is increasing in a t-interval, the following behaviors take place on an x-interval
corresponding to that t-interval:

If dy/dx > 0, then y as a function of x is increasing.

Or: If f '(t) > 0 and g'(t) > 0 then x is increasing and y is increasing (curve goes to up right), then y as a function of x is
increasing.

Or: If f '(t) < 0 and g'(t) < 0 then x is decreasing and y is decreasing (curve goes to down left), then y as a function of x is
increasing.

If dy/dx < 0, then y as a function of x is decreasing.

Or: If f '(t) > 0 and g'(t) < 0 then x is increasing and y is decreasing (curve goes to down right), then y as a function of x is
decreasing.

Or: If f '(t) < 0 and g'(t) > 0 then x is decreasing and y is increasing (curve goes to up left), then y as a function of x is
decreasing.

##### Concavity/Second Derivative

Let's express d2y/dx2 in terms of the derivatives of f and g and thus in terms of the variable t. Again keep in mind that the
curve represents the x-y relationship, thus its concavity is determined by d2y/dx2. Employing the chain rule we have: For the parametric equations x = f(t), y = g(t) we have: ##### In Section 13.1.1 Example 2.6 we sketched the curve defined by the parametric equations x = t3 – 3t, y = t2, t in [–2, 2], using the ”lo-tech” table of values and some properties of the curve. Now sketch it by using derivatives.

Solution In the table of curve behavior in Fig. 3.1, HT = horizontal tangent, VT = vertical tangent, and IP = inflection point. The curve is
sketched in Fig. 3.2. ### Fig. 3.1

Table Of Curve Behavior. ### Fig. 3.2

Curve For Example 3.1.

EOS

Remark that the word “y-Intercepts” is plural. There may be more than 1 y-intercept if the curve isn't the graph of a function.
In this example it isn't and there are more than 1 y-intercept. In the table of curve behavior in Fig. 3.1, observe that the values
of t are appropriately in increasing order, but those of x in this case aren't in any order and repeat themselves, which signifies
that the curve isn't the graph of a function.

The Directed Arcs In The y-Row. For each arc, first determine its direction, indicated by an arrow head. If its x-interval
contains increasing, respectively decreasing, values of x, then horizontally it's directed rightward, respectively leftward. Or: If
its y-interval contains increasing, respectively decreasing, values of y, then vertically it's directed upward, respectively
downward. Then determine and label its start and finish points. The start and finish points of the interval [a, b] are of course a
and b respectively, whether a < b or a > b. The start, respectively finish, point of an arc has as coordinates the start,
respectively finish, points of its x- and y-intervals. Its start and finish points are such that it's directed from its start point to its
finish point. On any x-interval, horizontally the arc, no matter how it's directed, always goes in the direction of the arrow head
from the start point to the finish point of the x-interval. So be careful, eg the start point of the arc on the x-interval [2, 0] is (2,
1), not (0, 0).

Recall that the double vertical bar in a row means that the quantity in that row isn't defined at the value of the variable above
the double bar.

 4. Higher-Order Derivatives

We wish to express the higher-order derivatives of y with respect to x, ie d2y/dx2, d3y/dx3, etc, in terms of the parameter t.
The 2nd derivative d2y/dx2 was already done above. It appears here to be included in the general formula for dny/dxn. Using
the chain rule (dky/dxk is a function of t and t is a function of x) we have: For the parametric equations x = f(t), y = g(t), where f and g have derivatives of sufficiently high order, the order-n derivative of y with respect to x in terms of t is: ##### Example 4.1

For the curve x = t2 – 3t – 2, y = t + 3, find d4y/dx4 in terms of t.

Solution EOS

To calculate the order-n derivative we first have to calculate the derivatives of all orders from 1 to n – 1. It's a straightforward
but often tedious matter to calculate a high-order derivative expressed in terms of the parameter.

# Problems & Solutions

1. Find the Cartesian equations of the tangent and normal lines to the parametric curve x = t3, y = t2 at the point
corresponding to t = 2.

##### Solution

At t = 2 we have x = 8 and y = 4. The slope of the tangent line is:   Note: The cycloid was introduced in Section 13.1.1 Example 3.4 – The Cycloid.

##### Solution  3. Find all points where the curve x = t2 + 2t – 1, y = ett + 2 has a horizontal tangent. Determine the Cartesian equation of
each such tangent.

##### Solution

dx/dt = 2t + 2, dy/dt = et – 1,
dy/dt = 0 at t = 0, dx/dt is non-0 there,
at t = 0 we have x = –1 and y = 3.

The horizontal tangent is at (–1, 3). Its Cartesian equation is of course y = 3. 4. Find all points where the curve x = cos 2s, y = sin s, s in R, has a vertical tangent. Determine the Cartesian equation of
each such tangent.

##### Solution The vertical tangent is at (1, 0). Its Cartesian equation is of course x = 1. 5. Sketch the parametric curve: making use of the 1st 2 derivatives.

##### Solution    6. If x = 2t + 4 and y = e3t:
a. Find d3y/dx3 in terms of t.
b. Prove using mathematical induction that dny/dxn = (3/2)ne3t for all positive integer n.

##### Solution So it's also true for n + 1. Thus it's true for all positive integer n. 7. Consider the parametric curve x = 1/(1 + t2), y = 2t/(1 + t2), t in R.
a. Find the maximum distance from the origin to a point on the curve.
b. Determine the points of maximum distance from the origin.

Solution

a.
The distance s from the origin to an arbitrary point (x, y) on the curve corresponding to the parameter value t is:  