You'll have a chance to teach your examples to your peers
prior to the exam.
Conic Sections
Before we jump fully into $\mathbb{R}^3$,
we need some good examples of planar curves (curves in
$\mathbb{R}^2$) that we'll extend to object in 3D. These examples
are conic sections. We call them conic sections because you can
obtain each one by intersecting a cone and a plane (I'll show you
in class how to do this). Here's a definition.
Consider two identical, infinitely tall, right circular cones
placed vertex to vertex so that they share the same axis of
symmetry. A conic section is the intersection of this three
dimensional surface with any plane that does not pass through
the vertex where the two cones meet.
These intersections are called circles (when the plane is
perpendicular to the axis of symmetry), parabolas (when the plane
is parallel to one side of one cone), hyperbolas (when the plane
is parallel to the axis of symmetry), and ellipses (when the
plane does not meet any of the three previous criteria). The
definition above provides a geometric description of how to
obtain a conic section from cone. We'll not introduce an
alternate definition based on distances between points and lines,
or between points and points. Let's start with one you are
familiar with.
Consider the point $P=(a,b)$ and a positive number $r.$ A
circle circle with center $(a,b)$ and radius $r$ is the set of
all points $Q=(x,y)$ in the plane so that the segment $PQ$ has
length $r$.
Using the distance formula, this means that every circle
can be written in the form $(x-a)^2+(y-b)^2=r^2$.
The equation $4x^2+4y^2+6x-8y-1=0$ represents a circle (though
initially it does not look like it). Use the method of
completing the square to rewrite the equation in the form
$(x-a)^2 + (y-b)^2 = r^2$ (hence telling you the center and
radius). Then generalize your work to find the center and
radius of any circle written in the form $x^2+y^2+Dx+Ey+F=0$.
Parabolas
Before proceeding to parabolas, we need to
define the distance between a point and a line.
Let $P$ be a point and $L$ be a line. Define the distance
between $P$ and $L$ (written $d(P,L)$) to be the length of the
shortest line segment that has one end on $L$ and the other end
on $P$. Note: This segment will always be perpendicular to $L$.
Given a point $P$ (called the focus) and a line $L$ (called the
directrix) which does not pass through $P$, we define a
parabola as the set of all points $Q$ in the plane so that the
distance from $P$ to $Q$ equals the distance from $Q$ to $L$.
The vertex is the point on the parabola that is closest to the
directrix.
See page 658.
Consider the line $L:y=-p$, the point $P=(0,p)$, and
another point $Q=(x,y)$. Use the distance formula to show that
an equation of a parabola with directrix $L$ and focus $P$ is
$x^2=4py$. Then use your work to explain why an equation of a
parabola with directrix $x=-p$ and focus $(p,0)$ is $y^2=4px$.
Ask me about the reflective properties of parabolas in
class, if I have not told you already. They are used in satellite
dishes, long range telescopes, solar ovens, and more. The
following problem provides the basis to these reflective
properties and is optional. If you wish to present it, let me
know. I'll have you type it up prior to presenting in class.
Consider the parabola $x^2=4py$ with directrix $y=-p$ and focus
$(0,p)$. Let $Q=(a,b)$ be some point on the parabola. Let $T$
be the tangent line to $L$ at point $Q$. Show that the angle
between $PQ$ and $T$ is the same as the angle between the line
$x=a$ and $T$. This shows that a vertical ray coming down
towards the parabola will reflect of the wall of a parabola and
head straight towards the vertex.
The next two problems will help you use the basic equations
of a parabola, together with shifting and reflecting, to study
all parabolas whose axis of symmetry is parallel to either the
$x$ or $y$ axis.
See 11.6: 9-14
Once the directrix and focus are known, we can give an
equation of a parabola. For each of the following, give an
equation of the parabola with the stated directrix and focus.
Provide a sketch of each parabola.
- The focus is $(0,3)$ and the directrix is $y=-3$.
- The focus is $(0,3)$ and the directrix is $y=1$.
Give an equation of each parabola with the stated directrix and
focus. Provide a sketch of each parabola.
- The focus is $(2,-5)$ and the directrix is $y=3$.
- The focus is $(1,2)$ and the directrix is $x=3$.
See 11.6: 9-14
Each equation below represents a parabola. Find the
focus, directrix, and vertex of each parabola, and then provide
a rough sketch.
- $y=x^2$
- $(y-2)^2=4(x-1)$
Each equation below represents a parabola. Find the focus,
directrix, and vertex of each parabola, and then provide a
rough sketch.
- $y=-8x^2+3$
- $y=x^2-4x+5$
Ellipses
Given two points $F_1$ and $F_2$ (called foci) and a fixed
distance $d$, we define an ellipse as the set of all points $Q$
in the plane so that the sum of the distances $F_1Q$ and $F_2Q$
equals the fixed distance $d$. The center of the ellipse is the
midpoint of the segment $F_1F_2$. The two foci define a line.
Each of the two points on the ellipse that intersect this line
is called a vertex. The major axis is the segment between the
two vertexes. The minor axis is the largest segment
perpendicular to the major axis that fits inside the ellipse.
We can derive an equation of an ellipse in a manner very
similar to how we obtained an equation of a parabola. The
following problem will walk you through this. We will not have
time to present this problem in class. However, if you would like
to complete the problem and write up your solution on the wiki,
you can obtain presentation points for doing so. Let me know if
you are interested.
Consider the ellipse produced by the fixed distance $d$ and the
foci $F_1=(c,0)$ and $F_2=(-c,0)$. Let $(a,0)$ and $(-a,0)$ be
the vertexes of the ellipse.
- Show that $d=2a$ by considering the distances from $F_1$
and $F_2$ to the point $Q=(a,0)$.
- Let $Q=(0,b)$ be a point on the ellipse. Show that
$b^2+c^2=a^2$ by considering the distance between $Q$ and
each focus.
- Let $Q=(x,y)$. By considering the distances between $Q$
and the foci, show that an equation of the ellipse is
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$$
- Suppose the foci are along the $y$-axis (at $(0,\pm c)$)
and the fixed distance $d$ is now $d=2b$, with vertexes
$(0,\pm b)$. Let $(a,0)$ be a point on the $x$ axis that
intersect the ellipse. Show that we still have
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$$ but now we instead
have $a^2+c^2=b^2$.
You'll want to use the results of the previous problem to
complete the problems below. The key equation above is
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. The foci will be on the
$x$-axis if $a>b$, and will be on the $y$-axis if $b>a$.
The second part of the problem above shows that the distance from
the center of the ellipse to the vertex is equal to the
hypotenuse of a right triangle whose legs go from the center to a
focus, and from the center to an end point of the minor axis. The
next three problems will help you use the basic equations of an
ellipse, together with shifting and reflecting, to study all
ellipses whose major axis is parallel to either the $x$- or
$y$-axis.
See 11.6: 17-24
For each ellipse below, graph the ellipse and give the
coordinates of the foci and vertexes.
- $\ds 16x^2+25y^2=400$ [Hint: divide by 400.]
- $\ds \frac{(x-1)^2}{5}+\frac{(y-2)^2}{9}=1$
For the ellipse $x^2+2x+2y^2-8y=9$, sketch a graph and give the
coordinates of the foci and vertexes.
See 11.6: 25-26
Given an equation of each ellipse described below, and
provide a rough sketch.
- The foci are at $(2\pm 3,1)$ and vertices at $(2\pm 5,
1)$.
- The foci are at $(-1,3\pm 2)$ and vertices at $(-1, 3\pm
5)$.
Ask me about the reflective properties of an ellipse in
class, if I have not told you already. The following problem
provides the basis to these reflective properties and is
optional. If you wish to present it, let me know. I'll have you
type it up prior to presenting in class.
Consider the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with
foci $F_1=(c,0)$ and $F_2=(-c,0)$. Let $Q=(x,y)$ be some point
on the ellipse. Let $T$ be the tangent line to the ellipse at
point $Q$. Show that the angle between $F_1Q$ and $T$ is the
same as the angle between $F_2Q$ and $T$. This shows that a ray
from $F_1$ to $Q$ will reflect off the wall of the ellipse at
$Q$ and head straight towards the other focus $F_2$.
Hyperbolas
Given two points $F_1$ and $F_2$ (called foci) and a fixed
number $d$, we define a hyperbola as the set of all points $Q$
in the plane so that the difference of the distances $F_1Q$ and
$F_2Q$ equals the fixed number $d$ or $-d$. The center of the
hyperbola is the midpoint of the segment $F_1F_2$. The two foci
define a line. Each of the two points on the hyperbola that
intersect this line is called a vertex.
We can derive an equation of a hyperbola in a manner very
similar to how we obtained an equation of an ellipse. The
following problem will walk you through this. We will not have
time to present this problem in class.
Consider the hyperbola produced by the fixed number $d$ and the
foci $F_1=(c,0)$ and $F_2=(-c,0)$. Let $(a,0)$ and $(-a,0)$ be
the vertexes of the hyperbola.
- Show that $d=2a$ by considering the difference of the
distances from $F_1$ and $F_2$ to the vertex $(a,0)$.
- Let $Q=(x,y)$ be a point on the hyperbola. By considering
the difference of the distances between $Q$ and the foci,
show that an equation of the hyperbola is
$\frac{x^2}{a^2}-\frac{y^2}{c^2-a^2}=1,$ or if we let
$c^2-a^2=b^2$, then the equation is
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$$
- Suppose the foci are along the $y$-axis (at $(0,\pm c)$)
and the number $d$ is now $d=2b$, with vertexes $(0,\pm b)$.
Let $a^2=c^2-b^2$. Show that an equation of the hyperbola is
$$\frac{y^2}{b^2}-\frac{x^2}{a^2}=1.$$
You'll want to use the results of the previous problem to
complete the problems below.
See 11.6: 27-34
Consider the hyperbola
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$ Construct a box centered
at the origin with corners at $(a, \pm b)$ and $(-a,\pm b)$.
Draw lines through the diagonals of this box. Rewrite the
equation of the hyperbola by solving for $y$ and then factoring
to show that as $x$ gets large, the hyperbola gets really close
to the lines $y=\pm \frac{b}{a}x$. [Hint: rewrite so that you
obtain $y=\pm\frac{b}{a}x\sqrt{\text{something}}$]. These two
lines are often called oblique asymptotes. Now apply what you
have just done to sketch the hyperbola
$\frac{x^2}{25}-\frac{y^2}{9}=1$ and give the location of the
foci.
The next three problems will help you use the basic
equations of a hyperbola, together with shifting and reflecting,
to study all ellipses whose major axis is parallel to either the
$x$- or $y$-axis.
See 11.6: 27-34
For each hyperbola below, graph the hyperbola (include
the box and asymptotes) and give the coordinates of the foci
and vertexes.
- $\ds 16x^2-25y^2=400$ [Hint: divide by 400.]
- $\ds \frac{(x-1)^2}{5}-\frac{(y-2)^2}{9}=1$
For the hyperbola $x^2+2x-2y^2+8y=9$, sketch a graph (include
the box and asymptotes) and give the coordinates of the foci
and vertexes.
See 11.6: 35-38
Given an equation of each hyperbola described below, and
provide a rough sketch.
- The vertexes are at $(2\pm 3,1)$ and foci at $(2\pm 5,
1)$.
- The vertexes are at $(-1,3\pm 2)$ and foci at $(-1, 3\pm
5)$.
Ask me about the reflective properties of a hyperbola in
class, if I have not told you already. In particular, we can
discuss lasers and long range telescopes. The following problem
provides the basis to these reflective properties and is
optional. If you wish to present it, let me know. I'll have you
type it up prior to presenting in class.
Consider the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ with
foci $F_1=(c,0)$ and $F_2=(-c,0)$. Let $Q=(x,y)$ be a point on
the hyperbola. Let $T$ be the tangent line to the hyperbola at
point $Q$. Show that the angle between $F_1Q$ and $T$ is the
same as the angle between $F_2Q$ and $T$. This shows that if
you begin a ray from a point in the plane and head towards
$F_1$ (where the wall of the hyperbola lies between the start
point and $F_1$), then when the ray hits the wall at $Q$, it
reflects off the wall and heads straight towards the other
focus $F_2$.
Parametric Equations
In the vectors unit, we learned how to write the equation of a
line in $\mathbb{R}^2$ as $\vec r(t)=(u_1,u_2)t+(v_1,v_2)=\vec u t +
\vec v$. Since $\vec r(t)$ represented the $x$ and $y$ coordinates
of a point, we could also write this line as the two equations
$x=u_1t+v_1,\, y=u_2t+v_2$. We call these parametric
equations for the line.
Many times, the vector equation for a function
in $\mathbb{R}^2$ is simpler than the same curve expressed as $y=f(x)$. Let's explore that in an example.
- What does the graph of $\vec r(t)=(\cos t, \sin
t)$, where $0\leq t\leq 2\pi$, look like? [Hint: try plotting a
bunch of points for different $t$ values.]
- What does the graph $\vec r(t)=(3\cos t, 3\sin t)$, $0\leq t\leq 2\pi$ look like? [Hint: try plotting a bunch of points]
- If $R$ is a constant, what does the graph $\vec r(t)=(R\cos t,
R\sin t)$, $0\leq t\leq 2\pi$ look like? What does $R$ tell you
about the graph? [Hint: try plotting a bunch of points]
- Since each vector $\vec r(t)$ gives $x$ and $y$ values for a point,
we could write the first part above as $x=\cos t$, $y=\sin t$,
which we call parametric equations for this graph.
Let's use this to find an equation for the curve in the first
part. First, find an equation relating $x$ and $y$ ($t$ should
not be in this equation). [Hint: use
a trig identity.]
- Solve the equation you just found for $y$. If you graphed this,
it should give you the same curve as you found in the first part.
- Which is simpler, the $y=f(x)$ equation from the last part, or the
$\vec r(t)$ vector equation from the first part?
- Write a vector equation $\vec r(t)=(?,?)$ so that the graph is the part
of the curve $y=2x-3$ for $-3\leq x \leq 5$. [Hint: if $\vec
r(t)=(t,?)$, what should the $y$ component of $\vec r(t)$ be?]
- Write a vector equation $\vec r(t)=(?,?)$ so that the graph is the curve
$y=x^2$ for $-\infty\leq x\leq \infty$.
- Write a vector equation $\vec r(t)=(?,?)$ so that the graph is the
curve $y=\pm \sqrt{x}$ for $0\leq x\leq 9$. [Hint: you might find
it helpful to draw the curve first.]
Different vector equations can describe the same curve in a graph. It's
like walking to school and back on the same path. The graph is
the path---just a drawing of where we walked. The vector
equation, however, tells us exactly where we are at every instant,
so the vector equation for walking to school is different than the
vector equation for walking home, even if the path that we walk on
is the same in each case.
See 11.1: 1-18. This is the same for all
the problems below.
By plotting points, construct separate graphs of the
parametric curves given below (for a bunch of $t$ values, find $x$
and $y$ and plot the point). Place an arrow on your
graph to show the direction of motion, and label the start and end
of the curve.
- $\vec r(t)=(\cos t, \sin t)$, for $0\leq t\leq 2\pi$.
- $\vec r(t)=(\cos t, \sin t)$ for $0\leq t\leq 4\pi$.
- $\vec r(t)=(\sin t, \cos t)$, for $0\leq t\leq 2\pi$.
- $\vec r(t)=(\sin(2t), \cos(2t))$ for $0\leq t\leq
2\pi$.
- $\vec r(t)=(\cos t, \sin t, t)$, for $0\leq t\leq
4\pi$. [Hint: this will be a curve in 3D, so you'll find $x,y,z$
coordinates for each $t$ value.]
Here's how to check the answer to the first part of the previous
problem using Sage (without the labels and arrow).
Plot the path traced out by the parametric curve $x=1+2\cos t,
y=3+5\sin t$. Then use the trig identity $\cos^2t+\sin^2t=1$ to
give a Cartesian equation of the curve (an equation that only
involves $x$ and $y$). What are the foci of the resulting
object (it's a conic section).
What we did in the previous chapter should help here.
Find parametric equations $x=?,y=?,z=?$ for a line that passes through
the points $(0,1,2)$ and $(3,-2,4)$.
Plot the path traced out by the parametric curve $\vec r(t)=
(t^2+1, 2t-3).$ Give a Cartesian equation of the curve
(eliminate the parameter $t$), and then find the focus of the
resulting curve.
Consider the parametric curve given by $x=\tan t, y=\sec t$.
Plot the curve for $-\pi/2<t<\pi/2$. Give a Cartesian
equation of the curve (a trig identity will help). Then find
the foci of the resulting conic section. [Hint: this problem
will probably be easier to draw if you first find the Cartesian
equation, and then plot the curve.]
Derivatives and Tangent
lines
See Larson 12.2 and 12.3
The derivative from calculus 1 measures how much the output of a
function changes if the input changes a little bit. Likewise, the
derivative of a vector-valued function $\vec r(t)$ measures how much
the output $\vec r(t)$ changes if the input $t$ changes a little
bit. This means that if $\vec r(t)$
represents the position of some object at time $t$, then $d\vec
r/dt$ represents how fast the position is changing, or the
velocity.
In calculus 1, the derivative of a function $f(t)$ is defined as
$$\frac{df}{dt} =\ds\lim_{h\to0}\frac{f(t+h)-f(t)}{h}.$$ We extend
this definition to vectors by using the same definition, but we just
use vector functions instead of calculus 1 functions.
If $\vec r(t)$ is a vector equation of a
curve, then we define the derivative to be $$\frac{d\vec
r}{dt}=\ds\lim_{h\to 0}\frac{\vec r(t+h)-\vec r(t)}{h}.$$
Let $\vec r(t)=(t^2-t-6, 2t-3)$ be a function representing the $(x,y)$ position of
an object at time $t$. In this problem, we'll calculate and
draw an approximation to $d\vec r/dt$ at $t=3$.
- Draw the parametric curve $\vec r(t)$ for $2\leq t\leq
4$.Check your plot of the curve with Sage. (Click
the word "Sage" to see how to do it.)
- Draw the vectors $\vec r(3)$ and $\vec r(3.1)$.
- Draw the vector $\vec r(3.1)-\vec r(3)$ between the ends of
$\vec r(3)$ and $\vec r(3.1)$.
- Draw the vector $\frac {\vec r(3.1)-\vec r(3)}{0.1}$,
starting at the head of $\vec r(3)$. This vector approximates
$d\vec r/dt$ ($h=0.1$ in this case).
The next problem figures out an easy way to calculate the exact derivative $d\vec
r/dt$.
See page 728.
See the first part of Larson 12.2
Show that if $\vec r(t) = (f(t),g(t))$, then the
derivative is just $\frac{d\vec r}{dt} = (f'(t),g'(t))$.
[Hint: The
definition above says that $\frac{d\vec r}{dt}=\ds\lim_{h\to
0}\frac{\vec r(t+h)-\vec r(t)}{h}$. We were told $\vec r(t) =
(f(t),g(t))$, so use this in the derivative definition. Then
try to modify the equation to obtain $\frac{d\vec r}{dt} =
(f'(t),g'(t))$.]
The previous problem shows you can take the derivative of a
vector valued function by just differentiating each component
separately. For practice, calculate the exact derivative at $t=3$ of the
problem above where you approximated the derivative. Was the
approximation close? The next problem shows you that velocity and
acceleration are still connected to the first and second
derivatives.
See 13.1:5-8 and 13.1:19-20
Consider the parametric curve given by $\vec r(t)=(
3\cos t, 3\sin t )$.
- Graph the curve $\vec r$, and compute $\frac{d\vec
r}{dt}$ and $\frac{d^2\vec r}{dt^2}$.
- On your graph, draw the vectors $\frac{d\vec
r}{dt}\left(\frac{\pi}{4}\right)$ and $\frac{d^2\vec
r}{dt^2}\left(\frac{\pi}{4}\right)$ with their tail placed on
the curve at $\vec r\left(\frac{\pi}{4}\right)$. These
vectors represent the velocity and acceleration vectors at time $t=\pi/4$.
- Give a vector equation of the tangent line to this curve
at $t=\frac{\pi}{4}$. (You know a point and a direction
vector.)
If an object moves
along a path $\vec r(t)$, we can find the velocity and
acceleration by just computing the first and second
derivatives. The velocity is $\frac{d\vec r}{dt}$, and the
acceleration is $\frac{d^2\vec r}{dt^2}$. Speed is a scalar,
not a vector. The speed of an object is just the length of the
velocity vector.
Consider the curve $\vec r(t) = (2t+3, 4(2t-1)^2)$.
- Construct a graph of $\vec r$ for $0\leq t\leq 2$.
- If this curve represented the path of a horse running
through a pasture, find the velocity of the horse at any time
$t$, and then specifically at $t=1$. What is the horse's
speed at $t=1$?
- Find a vector equation of the tangent line to $\vec r$ at
$t=1$. Include this on your graph.
- Show that the slope of the line is $$\ds
\frac{dy}{dx}\big|_{x=5} = \frac{ (dy/dt)\big|_{t=1} }{
(dx/dt)\big|_{t=1} }.$$ [How can you turn the direction
vector, which involves $(dx/dt)$ and $(dy/dt)$ into a slope
$(dy/dx)$?]
The previous problem introduced the following key theorem.
Its proof is just the chain rule.
If $\vec r(t) = (x(t),y(t))$ is a parametric curve, then the
slope $dy/dx$ of the curve can be found using the formula
$$\ds\frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx} =
\frac{dy/dt}{dx/dt}.$$ The second derivative is then
$\ds\frac{d^2y}{dx^2} = \frac{d(y'(x))}{dx} =
\frac{d(dy/dx)}{dx}=\frac{d(dy/dx)/dt}{dx/dt}$.
An easy way to remember this theorem is to find
$\frac{dy}{dx}$, just find the derivative of $y$ with respect
to $t$, and then divide by $dx/dt$. This will allow you to
connect derivatives of vector valued functions to slopes and
derivatives back in first semester calculus.
See 11.2:1-14
See Larson 10.2, Example 2
Consider the parametric curve given by $\vec r(t) =
(t^2,t^3)$.
- Compute $dy/dx$ and $\frac{d^2y}{dx^2}$ at $t=2$ using the
theorem above.
- Eliminate the parameter $t$ (get a Cartesian equation
for the curve). Then find $y'$ and $y'$ at $t=2$
using first semester calculus.
Here are some Sage examples letting you practice finding $\frac
{dy}{dx}$. Click "Evaluate" and do several examples to make sure you
understand the concept. When typing in expressions, make sure to
use a * to signify multiplication.
Here are some Sage examples letting you practice finding $\frac
{d^2y}{dx^2}$. Click "Evaluate" and do several examples to make sure you
understand the concept. When typing in expressions, make sure to
use a * to signify multiplication.
Arc Length
See the last part of Larson 10.3
If an object moves at a
constant speed, then the distance travelled is $$\text{distance}
= \text{speed}\times\text{time}.$$ This requires that the speed
be constant. What if the speed is not constant? Over a really
small time interval $dt$, the speed is almost constant, so we can
still use the idea above. The following problem will help you
develop the key formula for arc length.
Suppose an object moves along the path given by $\vec
r(t)=(x(t),y(t))$ for $a\leq t\leq b$.
- Show that the object's speed $|d\vec r/dt|$ at any time $t$ is
$\ds\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}$.
- If you move over a really small time interval, say of
length $dt$, then the speed is almost constant. If you move
at constant speed
$\ds\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}$
for a time length $dt$, what's the distance $ds$ you have
traveled?
- Explain why the length of the path given by $\vec r(t)$
for $a\leq t\leq b$ is
This is the arc length formula. Ask me in class for an
alternate way to derive this formula.
$$s=\int ds=\int_a^b \left|\frac{d\vec r}{dt}\right|
dt=\int_a^b
\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt.$$
-
See page 639.
Draw a small curve. Pick two points close
together. Construct a straight line segment between
them (call this $ds$). Then draw a right triangle
that shows the change in $x$ and change in $y$
(written $dx$ and $dy$). Use the Pythagorean theorem
to show that
$ds=\sqrt{dx^2+dy^2}=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt$.
- If the curve is in space (so $\vec
r(t)=(x(t),y(t),z(t))$ is the path), then what is the
arc length of the curve?
See 11.2: 25-30
See
Sage
for a plot of the curve
Find the length of the curve $\ds \vec r(t) =
\left(t^3,\frac{3t^2}{2}\right)$ for $t\in[1,3]$. The notation
$t\in[1,3]$ means $1\leq t\leq 3$. Be prepared to show us your
integration steps in class (you'll need a $u$-substitution).
For each curve below, set up an integral formula which would
give the length, and sketch the curve. Do not worry about
integrating them.
The reason I don't want you to actually compute the integrals
is that they will get ugly really fast. Try doing one in
Wolfram Alpha and see what the computer gives.
- The parabola $\vec p(t) = (t,t^2)$ for $t\in[0,3]$.
- The ellipse $\vec e(t) = (4\cos t,5\sin t)$ for
$t\in[0,2\pi]$.
- The hyperbola $\vec h(t) = (\tan t,\sec t)$ for
$t\in[-\pi/ 4,\pi/4]$.
To actually compute the integrals above and find the
lengths, we would use a numerical technique to approximate the
integral (something akin to adding up the areas of lots and lots
of rectangles as you did in first semester calculus).