# Curves

This unit covers the following ideas. In preparation for the quiz and exam, make sure you have a lesson plan containing examples that explain and illustrate the following concepts.
1. Recognize vector equations as parametric equations.
2. Find derivatives and tangent lines for parametric equations. Explain how to find velocity, speed, and acceleration from parametric equations.
3. Use integrals to find the lengths of parametric curves.
1. Be able to describe, graph, give equations of, and find foci for conic sections (parabolas, ellipses, hyperbolas).
2. Model motion in the plane using parametric equations. In particular, describe conic sections using parametric equations.
3. Find derivatives and tangent lines for parametric equations. Explain how to find velocity, speed, and acceleration from parametric equations.
4. Use integrals to find the lengths of parametric curves.

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.
1. The focus is $(0,3)$ and the directrix is $y=-3$.
2. 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.
1. The focus is $(2,-5)$ and the directrix is $y=3$.
2. 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.
1. $y=x^2$
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.
1. $y=-8x^2+3$
2. $y=x^2-4x+5$

# 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.
1. Show that $d=2a$ by considering the distances from $F_1$ and $F_2$ to the point $Q=(a,0)$.
2. 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.
3. Let $Q=(x,y)$. By considering the distances between $Q$ and the foci, show that an equation of the ellipse is $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$$
4. 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.
1. $\ds 16x^2+25y^2=400$ [Hint: divide by 400.]
2. $\ds \frac{(x-1)^2}{5}+\frac{(y-2)^2}{9}=1$
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.
1. The foci are at $(2\pm 3,1)$ and vertices at $(2\pm 5, 1)$.
2. 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.
1. Show that $d=2a$ by considering the difference of the distances from $F_1$ and $F_2$ to the vertex $(a,0)$.
2. 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 $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$$
3. 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.
1. $\ds 16x^2-25y^2=400$ [Hint: divide by 400.]
2. $\ds \frac{(x-1)^2}{5}-\frac{(y-2)^2}{9}=1$
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.
1. The vertexes are at $(2\pm 3,1)$ and foci at $(2\pm 5, 1)$.
2. 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.
1. 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.]
2. 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]
3. 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]
4. 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.]
5. 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.
6. Which is simpler, the $y=f(x)$ equation from the last part, or the $\vec r(t)$ vector equation from the first part?
Let's practice finding the vector equations for some other graphs.
1. 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?]
2. Write a vector equation $\vec r(t)=(?,?)$ so that the graph is the curve $y=x^2$ for $-\infty\leq x\leq \infty$.
3. 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.
1. $\vec r(t)=(\cos t, \sin t)$, for $0\leq t\leq 2\pi$.
2. $\vec r(t)=(\cos t, \sin t)$ for $0\leq t\leq 4\pi$.
3. $\vec r(t)=(\sin t, \cos t)$, for $0\leq t\leq 2\pi$.
4. $\vec r(t)=(\sin(2t), \cos(2t))$ for $0\leq t\leq 2\pi$.
5. $\vec r(t)=(\cos t, \sin t, t)$, 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 $t=3$.
1. 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.)
2. Draw the vectors $\vec r(3)$ and $\vec r(3.1)$.
3. Draw the vector $\vec r(3.1)-\vec r(3)$ between the ends of $\vec r(3)$ and $\vec r(3.1)$.
4. 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 $\vec r(t)=( 3\cos t, 3\sin t )$.
1. Graph the curve $\vec r$, and compute $\frac{d\vec r}{dt}$ and $\frac{d^2\vec r}{dt^2}$.
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 $t=\pi/4$.
3. 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 $\vec r(t) = (2t+3, 4(2t-1)^2)$.
1. Construct a graph of $\vec r$ for $0\leq t\leq 2$.
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 $t=1$?
3. Find a vector equation of the tangent line to $\vec r$ at $t=1$. Include this on your graph.
4. 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 $\vec r(t) = (t^2,t^3)$.
1. Compute $dy/dx$ and $\frac{d^2y}{dx^2}$ at $t=2$ using the theorem above.
2. 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 $a\leq t\leq b$.
1. Show that the object's speed $|d\vec r/dt|$ at any time $t$ is $\ds\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}$.
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?
3. 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.
1. The parabola $\vec p(t) = (t,t^2)$ for $t\in[0,3]$.
2. The ellipse $\vec e(t) = (4\cos t,5\sin t)$ for $t\in[0,2\pi]$.
3. 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).