Calculus Of One Real Variable – By Pheng Kim
Ving


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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 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 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 
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2. Tangents Of Parametric Curves 
We're now going to investigate the Cartesian xyequation 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 t_{1} there
corresponds a unique point (x_{1}, y_{1}) = ( f(t_{1}), g(t_{1})) on the curve. If the slope of the tangent
to the curve at the point (x_{1}, y_{1}) that corresponds to t_{1} is m then the Cartesian equation of the tangent is:
y = y_{1} + m(x – x_{1}).
We must specify that (x_{1}, y_{1}) is the one that corresponds to t_{1}, because the
same point (x_{1}, y_{1}) may correspond
to more than 1
different value of t, as shown
in Fig. 2.1, where (x_{1}, y_{1}) corresponds to 2 different tvalues
t_{1} and t_{2}. We must
specify which
tvalue the point (x_{1}, y_{1}) corresponds to, because the tangent at (x_{1}, y_{1}) that
corresponds to t_{1} may differ from the tangent at
(x_{1}, y_{1}) that
corresponds to t_{2}, 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 xy
relationship, not the tx or ty 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

Fig. 2.1 More than 1 different value of t
may correspond to same 
Fig. 2.2 y is a function of x on N_{x} that corresponds to N_{t}. 
For the curve to have a tangent at (x_{1}, y_{1}) = ( f(t_{1}), g(t_{1})) it must be smooth on an xneighborhood
N_{x} of x_{1} corresponding to a
tneighborhood N_{t}
of t_{1}.
If f '(t) is continuous and never 0 on N_{t}, then by the
intermediatevalue theorem either f '(t) > 0 or f '(t) < 0 throughout N_{t},
then x = f(t) is either increasing or decreasing on N_{t}, then the curve never
reverses direction horizontally over N_{x}.
Here N_{x}
corresponds to N_{t}.
Then y is some function of x on N_{x}.
Additionally, if g'(t) exists on N_{t},
then y is some differentiable function
of x on N_{x}.
For the curve to be smooth on N_{x},
dy/dx must also
be continuous on N_{x},
hence g'(t) must also, like f '(t), be
continuous on N_{t}.
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 tneighborhood of t_{1} and f '(t) is never 0 there, except possibly at isolated
points. If f '(t_{1}) = 0 then we
must
consider the onesided limits of g'(t)/ f '(t)
at t_{1}, as illustrated below._{}
We must specify that N_{x}
is the one that corresponds to N_{t},
because N_{x} may also be mapped to
by a different tinterval other
than N_{t}, as it is so in Fig.
2.2 by a tneighborhood to the right of N_{t}, as evidenced by the
existence of that upper portion of the
curve over N_{x}, that time in
increasing t and decreasing x.
We've seen that if f '(t) and g'(t) are continuous on a neighborhood of t_{1} and not both 0 at any point in it then the curve has a
tangent at the point x_{1} corresponding
to t_{1} and is smooth on the xneighborhood of x_{1} corresponding
to the tneighborhood of t_{1}.
Consider the tangent at the point (x_{1}, y_{1}) corresponding to a given parameter value t_{1}. Suppose f '(t) and g'(t)
are continuous

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 t_{1}. There can be more than 1 tangent at
(x_{1},
y_{1}) if (x_{1}, y_{1}) 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 '(t_{1}) and g'(t_{1}) are both 0.
If f '(t_{1}) and g'(t_{1})
are both 0 then the curve may or may not have a tangent at x_{1}. Cases 1 and
2d show that if f '(t_{1}) and g'(t_{1}) aren't both 0
then
the curve has a tangent at x_{1}.
Solution
EOS
Determine the Cartesian equation of the tangent to the curve
defined parametrically by x = s^{2}, y = cos
s, at the point where
s = 0.
Solution
When s = 0 we get x
= 0 and y = 1. We have:
EOS
Find the Cartesian equation of the tangent to the curve
given by the parametric equations x = t^{2} – 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
Find all points where the parametric curve x = s^{2} – 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 non0 there as it's never 0,
at s = 2 we have x
= 2^{2}
– 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
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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 d^{2}y/dx^{2} 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.
Utilize dy/dx = g'(t)/ f '(t). Clearly as t
is increasing in a tinterval,
the following behaviors take place on an xinterval
corresponding to that tinterval:
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.
Let's express d^{2}y/dx^{2} 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 xy relationship, thus its concavity is determined
by d^{2}y/dx^{2}. Employing the chain rule we have:
For the parametric equations x = f(t), y = g(t) we have:

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 “yIntercepts”
is plural. There may be more than 1 yintercept
if the curve isn't the graph of a function.
In this example it isn't and there are more than 1 yintercept.
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 yRow. For
each arc, first determine its direction, indicated by an arrow head. If its xinterval
contains increasing, respectively decreasing, values of x,
then horizontally it's directed rightward, respectively leftward. Or: If
its yinterval 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 yintervals. Its start and finish points are such
that it's directed from its start point to its
finish point. On any xinterval,
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 xinterval.
So be careful, eg the start point of the arc on the xinterval
[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.
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4. HigherOrder Derivatives 
We wish to express the higherorder derivatives of y with respect to x,
ie d^{2}y/dx^{2}, d^{3}y/dx^{3}, etc, in terms
of the parameter t.
The 2nd derivative d^{2}y/dx^{2} was already
done above. It appears here to be included in the
general formula for d^{n}y/dx^{n}.
Using
the chain rule (d^{k}y/dx^{k}
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 ordern

For the curve x = t^{2} – 3t – 2, y = t + 3, find d^{4}y/dx^{4} in terms of t.
Solution
EOS
To calculate the ordern 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 highorder 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 = t^{3}, y = t^{2} at the point
corresponding to t = 2.
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.
3. Find all points where the curve x = t^{2} + 2t – 1, y = e^{t} – t
+ 2 has a horizontal tangent. Determine the Cartesian equation of
each such tangent.
dx/dt = 2t + 2, dy/dt = e^{t} – 1,
dy/dt = 0 at t = 0, dx/dt is non0 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.
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.
6. If x = 2t + 4 and y = e^{3}^{t}:
a. Find d^{3}y/dx^{3} in terms of t.
b. Prove using mathematical induction
that d^{n}y/dx^{n} = (3/2)^{n}e^{3}^{t} for all positive
integer n.
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 + t^{2}), y = 2t/(1 + t^{2}), 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:
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