Assignment 5
This assignment concerns the topics covered in Calculus and Geometry of Curves
Only the Material of Problems I and II will be on the first Midterm
Problems
Mission to an Asteroid
This question has a rather involved story. Make sure to understand it fully before starting the problem, and draw yourself lots of pictures!
Background Story: One proposed method of preventing a future catastrophe like the impact that killed all non-avian dinosaurs 66 million years ago is asteroid redirection, where a spacecraft is launched at the dangerous asteroid and impacts it at a high velocity, slightly deflecting the asteroids trajectory and causing it to miss Earth. Such a redirect was attempted for the first time last year in 2022, when NASA’s dart mission impacted the asteroid Dimorphous, a 500ft rock in a relatively near-earth (but non-dangerous!) orbit. Here’s the rather overly dramatic “trailer video” from NASA about the mission:
The important science goals of the mission were to:
- (1) see how much the asteroid was deflected after the impact, and
- (2) understand better the physics of high-speed collisions, to be able to model re-direct possibilities more accurately in the future. Objective (1) was able to be accomplished by earth-based telescopes carefully plotting its trajectory in the night sky, but objective (2) required data from near the asteroid itself: ideally actual photographs of the impact moment. Unfortunately the DART spacecraft was unable to send such images back itself, as it impacted the asteroid at a speed of 4.1 miles per second and was instantly vaporised into plasma.
Here’s an actual video shot by the spacecraft on approach. Obviously, the video feed cuts out when the space ship hits the asterioid!
To get the crucial data back to earth, a small pair of italian-built cubesats were carried with DART, and released just days before the impact. These cube sats traveled on a slightly modified trajectory that just missed the asteroid, and so were able to take real-time photographs of the impact. One of these photographs is included below. You can clearly see the actual impact (bright spot with debris flying out into space) as well as the asteroid’s moon glowing bright in the glow of the molten material created. This image has been extremely crucial in understanding the actual physics of the impact.
The Problem Statement: In this problem, you are to imagine that you are a member of NASA’s DART team, and are tasked with the planning of these very important shots. Fix a system of coordinates \((x,y,z)\) (where \(x,y\) and \(z\) are measured in hundreds of miles) such that the asteroid is at the origin for all time. You propose a flight trajectory for the cubesat which departs (at \(t=0\)) three days before the impact, which takes the cubesat on the following parametric curve.
\[c(t)=\left(t-2,\sqrt{t}-\frac{t}{2},0\right)\]
The image below is a schematic of the flight path of the cubesat (red) and the impactor (orange).
The cubesat is oriented so that it remains tangent to its flightpath at all times.
Exercise 1
- At the time of the impact (t=3), how far is the cubesat from the asteroid?
- If the cubesat wants to take a picture of the impact, what angle must it rotate its camera by (where angle zero means the camera is pointed straight ahead, tangent to the flight path).
Ideas to compute this:
- Can you find the position of the cubesat at the time the asteroid is impacted by the main craft?
- Can you find the direction vector from the cube sat to the asteroid at this time?
- Can you find the direction the cubesat is pointed at this time?
- Can you find the angle between these?
Logo Design, Part I
Both this question and the next concerned wtih using the calculus of parametric curves to help with graphic design. Assume your graphic design firm is hired to produce a new logo for a company, and you decide on the following form:
This is made out of a square (the bounding box) and two curves, both of which are copies of the same parametric curve given below
\[\vec{c}(t)=\left(t^3-t,t^2\right)\]
Exercise 2 The curve defining your logo makes a sort of teardrop shaped loop in the middle. What is the inside angle of this teardrop?
Ideas to compute this:
- Can you find the location where the curve intersects itself?
- Can you find the two times when the curve is at that location?
- Can you find the tangent vectors to the curve there, and make sure they are pointing in the right direction for the angle you want (possibly, negate one of them: think about the direction the curve is traveling in)!
- Can you use the dot product as a tool to get your answer?
Logo Design, Part II
A lot of thought goes into the design of a logo, and geometric designs are often used to build simplicity and elegance into the final product. Below are some examples of logos, with the underlying geometry on display:
Sometimes the graphic design team goes a little too far into geometry, metaphor, and meaning behind a logo. If you’d like a good laugh check out the Pepsi Logo Redesign
Like Apple’s logo, at every point except the hard corner formed by the intersections (whose angle we computed in the last problem) we can approximate our logo by circles! Drawing a few of these gives the following geometric design (which we may submit to the company that hired us as part of the packet describing the geometry and meaning behind the logo).
Our goal here is to figure out how to actually compute these circles. We will focus on only a couple of them, drawn on the curve \(\vec{c}(t)\) itself below.
Exercise 3
- The grey circle below is the circle which best approximates our logo at the very bottom of the teardrop shape. What is its radius?
- The two black circles are the two smallest circles that fit inside the teardrop shape. What’s an approximate value for their radii?
Ideas to compute this:
- Can you compute the curvature \(\kappa(t)\) of the logo’s curve as a function of \(t\)?
- For the bottom of the teardrop: what is the value of \(t\) for which \(\vec{c}(t)\) is exactly at the bottom (you should be able to find this by hand, looking at the equation for \(\vec{c}(t)\)). What is the curvature there?
- What is the relationship between curvature and the size of the best approximating circle?
- Can you put your function \(\kappa(t)\) into desmos, and use the graph to figure out what the radius of the smallest circles must be?
Solutions
Mission to an Asteroid
The cubesat is following the trajectory \(\vec{c}(t)=\left(t-2,\sqrt{t}-\tfrac{t}{2},0\right)\), so at the time of impact, \((t=3)\) the cubesat is at the position
\[\vec{c}(3)=\left(3-2,\sqrt{3}-\tfrac{3}{2}\right)=\left(1,\sqrt{3}-\tfrac{3}{2},0\right)\]
The distance from the origin is just the magnitude of this vector:
\[\begin{align*}\mathrm{dist}&=\sqrt{(1)^2+(\sqrt{3}-\tfrac{3}{2})^2+0^2}\\ &=\sqrt{1+3-3\sqrt{3}+\tfrac{9}{4}}\\ &=\sqrt{\frac{25}{4}-3\sqrt{3}}\\ &\approx 1.0265707 \end{align*}\]
To find the angle we need to rotate the camera, we need to first know two vectors:
- The vector tangent describing the direction the cubesat is pointed in
- The vector toAsteroid describing the direction from the cubesat to the asteroid.
We begin with the second of these. The asteroid is at the position \(O=(0,0)\) for all time. Thus, the direction vector from the cubesat to the origin is
\[\begin{align*}\mathrm{end}-\mathrm{start}&=O-\vec{c}(3)\\ &= (0,0,0)-\left(1,\sqrt{3}-\tfrac{3}{2},0\right)\\ &= \left(-1,\tfrac{3}{2}-\sqrt{3},0\right) \end{align*}\]
To find the direction the cubesat is facing, we need to find the tangent vector to its trajectort \(\vec{c}(t)\) at \(t=3\).
\[\vec{c}^\prime(t)=\left(1,\frac{1}{2\sqrt{t}}+\frac{1}{2},0\right)\] \[\vec{c}^\prime(3)=\left(1,\frac{1}{2\sqrt{3}}+\frac{1}{2}\right)\]
Now, we just need to find the angle between these two vectors. To do this, we can use the identity \[\cos\theta = \frac{u\cdot v}{\|u\|\|v\|}\]
Since the vector pointing at the asteroid is the negative of the position vector of the cubesat, its magnitude is the same as the distance to the asteroid, that we already found.
\[\|\mathrm{toAsteroid}\|\approx 1.0265707\]
The magnitude of the tangent vector is
\[\|\mathrm{tangent}\|=\sqrt{1+(\frac{1}{2\sqrt{3}}+\frac{1}{2})^2}\approx 1.273580962\]
Finally we need their dot product
\[\begin{align*}\mathrm{toAsteroid}\cdot \mathrm{tangent} &=\left(-1,\tfrac{3}{2}-\sqrt{3},0\right)\cdot \left(1,\frac{1}{2\sqrt{3}}+\frac{1}{2}\right)\\ &= (-1)(1)+\left(\frac{3}{2}-\sqrt{3}\right)\left(\frac{1}{2\sqrt{3}}+\frac{1}{2}\right)\\ &\approx -1.1830 \end{align*}\]
Putting these together we can compute the cosine of the angle:
\[\cos\theta \approx \frac{-1.1830}{(1.27358)(1.0265)}\approx -0.9053\]
This means that \(\theta\approx \arccos(-0.9053)\) which is 2.7029 radians, or 154.868 degress.
Logo Design I
The looped part of the new logo design is described by the parametric function \(\vec{c}(t)=\left(t^3-t,t^2\right)\). First, we try to understand what direction this curve is being traced out by the parameterization. When \(t=-2\) the \(x\)-coordinate of our function is \((-2)^3-(-2)=-8+2=-6\), whereas when \(t=2\) its \((2)^3-2=6\). Thus, as \(t\) increases we see the \(x\) coordinate going from negative to positive, so the curve is being traced from left to right.
To find the spot where the loop crosses itself, we need to find two different \(t\) values that reach that same \((x,y)\) point on the curve. Because the \(y\) coordinate \(y=t^2\) is an even function it maps \(t=a\) and \(t=-a\) to the same point for any \(a\). But since \(x(t)=t^3-t\) is an odd function we have \(x(-a)=-x(a)\). And the only time \(x(a)\) can equal \(-x(a)\) is if they are both zero (thats the only number equal to its own negative)! Thus, we know \(x(a)=0\) so \(a^3=a\), or \(a=\pm 1\).
These are then our two time values, \(t=-1\) and \(t=1\), which both map to the point \(c(1)=(0,1)\). To find the angle of the teardrop at this point, we need to find the derivative of \(c(t)\) and both \(t=-1\) and \(t=1\).
This is just a calculation
\[\vec{c}^\prime(t)=\langle 3t^2-1, 2t\rangle\] \[\vec{c}^\prime(-1)=\langle 2,-2\rangle\hspace{1cm}\vec{c}^\prime(1)=\langle 2,2\rangle\]
These are the velocity vectors of the curve at the two points it crosses through \((0,1)\). To help us out, let’s draw them quick, remembering the direction the curve is being traced.
To find the angle of the teardrop, we need the two vectors tangent to the teardrop’s angle. From the above, we see that the tangent vector at \(t=-1\) is in the correct direction, but the one at \(t=1\) is in the wrong direction: we need to negate it! Thus, the two vectors we are interested in are \(\langle 2,-2\rangle\) and \(\langle -2,-2\rangle\). Again using the relationship of angles to the dot product,
\[\begin{align*}\cos\theta &= \frac{\langle 2,-2\rangle\cdot \langle -2,-2\rangle}{\|\langle 2,-2\rangle\| \| \langle -2,-2\rangle\|}\\ =\frac{-2+2}{\sqrt{8}\sqrt{8}}=0 \end{align*}\]
Thus, \(\theta=\pi/2\) or \(90\) degrees.
Logo Design II
The curvature of a curve is the reciprocal of the radius of the best approximating circle. Thus, the first step here is to find the curvature of our logo’s curve. For this, we can use the formula relating curvature to the curve’s first and second derivatives:
\[\kappa(t)=\frac{\|c^\prime(t)\times c^{\prime\prime}(t)\|}{\|c^\prime(t)\|^3}\]
We have computed the first derivative above, and differentiating again gives both velocity and acceleration. \[c^\prime(t)=\langle 3t^2-1, 2t\rangle\] \[c^{\prime\prime}(t)=\langle 6t, 2\rangle\]
We find the cross product of the two vectors we add in a zero for the \(z\) component, and use the determinant formula. \[\begin{align*} c^\prime(t)\times c^{\prime\prime}(t)&=\left|\begin{matrix} \ihat &\jhat &\khat\\ 3t^2-1 & 2t &0\\ 6t & 2 & 0 \end{matrix}\right|\\ &=\khat((3t^2-1)(2)-(2t)(6t))\\ &=-(6t^2+2)\khat \end{align*}\]
Then the magnitude of this,
\[\begin{align*} \|c^\prime\times c^{\prime\prime}\|&=\|\langle 0,0,-6t^2-2\rangle\|\\ &= 6t^2+2 \end{align*}\]
and also the magnitude of \(c^\prime\).
\[\begin{align*}\|c^\prime(t)\|&=\sqrt{(3t^2-1)^2+(2t)^2}\\ &=\sqrt{9t^4-2t^2+1} \end{align*}\]
Now we have all the components, and can assemble the curvature!
\[\kappa(t)=\frac{6t^2+2}{(9t^4-2t^2+1)^{3/2}}\]
Phew! Now we can finally answer the questions we are interested in. To answer the first, we need to find the curvature at the bottom of the loop. This is directly below where the durve intersects itself, which we learned in the previous part occurs at \(x=0\). Thus this point must also be at \(x=0\): the only other time value that gives this is \(t=0\), so this is where we need the curvature!
\[\kappa(0)=\frac{6(0)^2+2}{(9(0)^4-2(0)^2+1)^{3/2}}=2\]
Thus, the radius of the best approximating circle is \[R=\frac{1}{2}\]
For the second part, we are looking for the smallest approximating circles inside the teardrop, which means we want to find the maximum value of curvature between \(t=-1\) and \(t=1\). This function does not look fun to differentiate and set equal to zero, so we will take the problem’s suggestion and put it into desmos to look for the max:
Desmos shows that the maximum occurs around \(t=0.422\) with \(\kappa\approx 3.425\), so the radius of the smallest circles is
\[R=\frac{1}{\kappa}\approx 0.2919\]