Assignment 3
This assignment concerns the topics covered in Operations, Shapes and the beginnings of Parametrization.
1. The Area of a Parallelogram
We learned that the 2-dimensional cross product calculates the area of a parallelogram,
\[\langle a,b\rangle \times \langle c,d\rangle = \left|\begin{matrix}a&b\\ c&d\end{matrix}\right|=ad-bc\]
In this problem, your job is to verify this is actually the case. Below I’ve drawn a figure to help you out, for the paralleogram spanned by \(\langle a,b\rangle\) and \(\langle c,d\rangle\) surrounded by some other shapes that will be useful.
- Let the black vertex be the point \((0,0)\). Label all the vertices of the in the picture (the parallelogram/triangles/rectangles) using cartesian coordinates
- Use this labeling to find the side lengths of all the triangles and rectangles involved.
- Find the area of the parallelogram using the fact that you know how to find the areas of rectangles and triangles.
2. Using our Toolkit
The equation \(x^2+y^2+z^2+2x+6z=90\) defines a sphere in 3-dimensional space. A plane is said to be tangent to a sphere if it touches the sphere at exactly one point. Find the equations of the two planes parallel to \(3x+4y=5\) which are tangent to this sphere. Make sure to draw pictures along the way and also to explain your process carefully: what tools are you using and why?
A Hint: A line is tangent to a circle if it is orthogonal to the radius of the circle at its point of intersection. Similarly a plane is tangent to a sphere at a point \(P\) if the radius of the sphere from the sphere’s center to \(P\) is orthogonal to the plane.
3. Designing Parametric Curves
UPDATE: THIS PROBLEM IS NO LONGER ASSIGNED. Because of this remote weeek, we’ve pushed this material back.
Exercise 1 Create a parametric curve \(f(t)=(x(t),y(t))\) in the plane \(\mathbb{R}^2\) looks like the following (it doesn’t have to look exactly like this)
More precisely, find a parametric equation with the following properties:
- As \(t\) increases in \([0,\infty)\), the curve \(f(t)\) spirals around the origin at a uniform rate.
- At the beginning \(f(0)=(0,0)\) and then the spiral approaches a radius of 2 closer and closer, but never quite reaches it as \(t\to\infty\).
Hint: draw a graph of what you want the radius function to look like, then try to build such a function! There are infinitely many correct answers to this question, but explain your answer and how you came to it.
Solutions
Area of Parallelograms
Since the parallelogram is made using the vectors \(\langle a,b\rangle\) and \(\langle c,d \rangle\) we can label its edges with these direction vectors.
If we follow the hint and label the black vertex by the point \((0,0)\) these direction vectors let us name all the other points in the diagram:
Now that we have all the vertices labeled, we can find the side lengths of all the individual shapes involved. First, looking at the biggest rectangle we see its base has length \(a+c\) and its height is \(b+d\). Using all our labeled vertices we can similarly find the lengths of the rectangles and triangles within:
After labeling side lengths, we see that there are two triangles with sides \(a,b\), two triangles of side lengths \(c,d\) and two rectangles of side lengths \(b,c\). We can find the area of the parallelogram by subtracting the area of these shapes from the area of the big rectangle. Computing, we see the big rectangle has area
\[(a+c)(b+d)=ab+ad+bc+cd\]
From this we must subtract
\[2\left(\frac{1}{2}ab\right)+2\left(\frac{1}{2}cd\right)+2\left(bc\right)\] \[=ab+cd+2bc\]
The result is what we wanted:
\[ab+ad+bc+cd-(ab+cd+2bc)=ad+(ab-ab)+(cd-cd)+(bc-2bc)\] \[=ad-bc\]
Using our Toolkit
We are given an equation for a sphere, but it is hard to use as we know neither its center nor radius. So, we first complete the squares:
\[\begin{align*} x^2+y^2+z^2+2x+6z &= (x^2+2x+1-1)+y^2 + (z^2+6z+9-9)\\ &=(x+1)^2+y^2+(z+3)^2-10 \end{align*}\]
Thus, adding \(10\) to the other side of the original, we have \((x+1)^2+y^2+(z+3)^2=100\). This sphere is centered at \((-1,0,-3)\) and has radius \(\sqrt{100}=10\).
To find the points where our planes will be tangent, we need to think a bit about the plane \(x+y+z=0\). Its coefficient vector is \(\langle 3,4,0\rangle\), so this is orthogonal to the plane.
But, a plane is tangent to a sphere if the radius vector is orthogonal to the plane. This means the radius vector of the sphere from the center to the point of tangency is a scalar multiple of \(\langle 1,1,1\rangle\). But which scalar multiple? For it to pick out a point on the sphere, we know it must have length \(8\), as the sphere has radius 8. So we must find a vector of length \(10\) in the direction \(\langle 3,4,0\rangle\). But we know how to do this: find a unit vector in that direction, and multiply by 8! \[\|\langle 3,4,0\rangle\|=\sqrt{3^2+4^2+0^2}=\sqrt{9+16}=\sqrt{25}=5\] \[\textrm{Unit Vector}=\left\langle \frac{3}{5}, \frac{5}{5},0 \right\rangle\] \[\textrm{Radius Vector}= \left\langle \frac{30}{5}, \frac{40}{5}, 0\right\rangle=\langle 6,8,0\rangle\]
We can add this to the sphere’s center to get one of the points we are after:
\[P_1 = (-1,0,-3)+\langle 6,8,0\rangle = (5,8,-3)\]
We can get the other point by adding the negative of this vector:
\[P_2 = (-1,0,-3)-\langle 6,8,0\rangle= (-7,-8,-3)\]
In both cases we know the normal vector to the plane we want: its parallel to \(3x+4y=5\) so it has the same normal vector as this \(\langle 3,4,0\rangle\). Using this information its straightforward to find the equations of the planes:
\[\langle 3,4,0\rangle \cdot\langle x-5,y-8,z+3\rangle = 0\] \[\langle 3,4,0\rangle \cdot \langle x+7,y+8,z+3\rangle =0\]
Expanding these out and simplifying, we see the planes are
\[3x+4y=47\hspace{1cm}3x+4y=-53\]