More calculus with parametric curves
07 Theory - Distance, speed
Theory 3
Distance function
The distance function
returns the total distance traveled by the particle from a chosen starting time up to the (input) time : We need the dummy variable
so that the integration process does not conflict with in the upper bound.
Speed function
The speed of a moving particle is the rate of change of distance:
This formula can be explained in either of two ways:
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- Apply the Fundamental Theorem of Calculus to the integral formula for
. - Consider
for a small change : so the rate of change of arclength is , in other words .
08 Illustration
Example - Speed, distance, displacement
Speed, distance, displacement
The parametric curve
describes the position of a moving particle ( measuring seconds). (a) What is the speed function? Suppose the particle travels for
seconds starting at . (b) What is the total distance traveled? (c) What is the total displacement? Solution
(a)
Compute derivatives:
Now compute the speed:
(b)
Distance traveled by using speed.
Compute total distance traveled function:
Substitute
and . New bounds are and . Calculate: The distance traveled up to
is:
(c)
Displacement formula:
Now compute starting and ending points.
For starting point, insert
: For ending point, insert
: Insert
and : Link to original
09 Theory - Surface area of revolutions
Theory 4
Surface area of a surface of revolution: thin bands
Suppose a parametric curve
is revolved around the -axis or the -axis. The surface area is:
The radius
should be the distance to the axis: This formulas adds the areas of thin bands, but the bands are demarcated using parametric functions instead of input values of a graphed function.
The formula assumes that the curve is traversed one time as
Link to originalincreases from to .
10 Illustration
Example - Surface of revolution - parametric circle
Surface of revolution - parametric circle
By revolving the unit upper semicircle about the
-axis, we can compute the surface area of the unit sphere. Parametrization of the unit upper semicircle:
Therefore, the arc element:
Now for
we choose because we are revolving about the -axis.
Plugging all this into the integral formula and evaluating gives:
Notice: This method is a little easier than the method using the graph
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Example - Surface of revolution - parametric curve
Surface of revolution - parametric curve
Set up the integral which computes the surface area of the surface generated by revolving about the
-axis the curve for . Solution
For revolution about the
-axis, we set . Then compute
: Therefore the desired integral is:
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Polar curves
Videos
Review Videos
Videos, Organic Chemistry Tutor
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01 Theory - Polar points, polar curves
Theory 1
Polar coordinates are pairs of numbers
which identify points in the plane in terms of distance to origin and angle from -axis:
Converting
Polar coordinates have many redundancies: unlike Cartesian which are unique!
- For example:
- And therefore also
(negative can happen) - For example:
for every - For example:
for any Polar coordinates cannot be added: they are not vector components!
- For example
- Whereas Cartesian coordinates can be added:
The transition formulas
require careful choice of .
- The standard definition of
sometimes gives wrong
- This is because it uses the restricted domain
; the polar interpretation is: only points in Quadrant I and Quadrant IV (SAFE QUADRANTS) - Therefore: check signs of
and to see which quadrant, maybe need -correction!
- Quadrant I or IV: polar angle is
polar angle is
Equations (as well as points) can also be converted to polar.
For
, look for cancellation from . For
, try to keep inside of trig functions.
- For example:
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02 Illustration
Example - Converting to polar:
-correction Converting to polar: pi-correction
Compute the polar coordinates of
and of . Solution
For
we observe first that it lies in Quadrant II. Next compute:
This angle is in Quadrant IV. We add
to get the polar angle in Quadrant II: The radius is of course
since this point lies on the unit circle. Therefore polar coordinates are . For
we observe first that it lies in Quadrant IV. (No extra needed.) Next compute:
So the point in polar is
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Example - Shifted circle in polar
Shifted circle in polar
For example, let’s convert a shifted circle to polar. Say we have the Cartesian equation:
Then to find the polar we substitute
and and simplify: So this shifted circle is the polar graph of the polar function
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03 Theory - Polar limaçons
Theory 2
To draw the polar graph of some function, it can help to first draw the Cartesian graph of the function. (In other words, set
and , and draw the usual graph.) By tracing through the points on the Cartesian graph, one can visualize the trajectory of the polar graph. This Cartesian graph may be called a graphing tool for the polar graph.
A limaçon is the polar graph of
. The shape of a limaçon is determined by the value of
. Any limaçon can be rescaled to have this form:
: Limaçon satisfying : unit circle.
: Limaçon satisfying : ‘outer loop’ circle with ‘flat spot’, not quite a ‘dimple’:
: Limaçon satisfying : ‘cardioid’ ‘outer loop’ circle with ‘dimple’ that creates a cusp:
: Limaçon satisfying : ‘dimple’ pushes past cusp to create ‘inner loop’:
: Limaçon satisfying : ‘inner loop’ only, no outer loop exists:
: Limaçon satisfying : ‘inner loop’ and ‘outer loop’ and rotated :
Transitions between limaçon types,
:
Notice the transition points at
and : The flat spot occurs when
- Smaller
gives convex shape The cusp occurs when
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- Smaller
gives dimple (assuming ) - Larger
gives inner loop
04 Theory - Polar roses
Theory 3
Roses are polar graphs of this form:
The pattern of petals:
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(even): obtain petals
- These petals traversed once
(odd): obtain petals
- These petals traversed twice
- Either way: total-petal-traversals: always
05 Illustration
Example - Finding vertical tangents to a limaçon
Finding vertical tangents to a limaçon
Let us find the vertical tangents to the limaçon (the cardioid) given by
. Solution
(1) Convert to Cartesian parametric using
and :
(2) Compute
and :
(3) The vertical tangents occur when
. We must double check that at these points. Substitute
and observe quadratic: Solve:
Then find
:
(4) Compute the points. In polar coordinates:
In Cartesian coordinates:
At
: At
: At
:
(5) Correction:
is a cusp! The point
at is on the cardioid, but the curve is not smooth there, this is a cusp. Still, the left- and right-sided tangents exists and are equal, so in a certain sense we could say the curve has vertical tangent at
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