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CLP-3 Multivariable Calculus

Joel Feldman, Andrew Rechnitzer, Elyse Yeager

Section 1.1 Points

Each point in two dimensions may be labeled by two coordinates 1 \((x,y)\) which specify the position of the point in some units with respect to some axes as in the figure below.
The set of all points in two dimensions is denoted 2  \(\bbbr^2\text{.}\) Observe that
  • the distance from the point \((x,y)\) to the \(x\)-axis is \(|y|\)
  • if \(y\gt 0\text{,}\) then \((x,y)\) is above the \(x\)-axis and if \(y\lt 0\text{,}\) then \((x,y)\) is below the \(x\)-axis
  • the distance from the point \((x,y)\) to the \(y\)-axis is \(|x|\)
  • if \(x\gt 0\text{,}\) then \((x,y)\) is to the right of the \(y\)-axis and if \(x\lt 0\text{,}\) then \((x,y)\) is to the left of the \(y\)-axis
  • the distance from the point \((x,y)\) to the origin \((0,0)\) is \(\sqrt{x^2+y^2}\)
Similarly, each point in three dimensions may be labeled by three coordinates \((x,y,z)\text{,}\) as in the two figures below.
The set of all points in three dimensions is denoted \(\bbbr^3\text{.}\) The plane that contains, for example, the \(x\)- and \(y\)-axes is called the \(xy\)-plane.
  • The \(xy\)-plane is the set of all points \((x,y,z)\) that satisfy \(z=0\text{.}\)
  • The \(xz\)-plane is the set of all points \((x,y,z)\) that satisfy \(y=0\text{.}\)
  • The \(yz\)-plane is the set of all points \((x,y,z)\) that satisfy \(x=0\text{.}\)
More generally,
  • The set of all points \((x,y,z)\) that obey \(z=c\) is a plane that is parallel to the \(xy\)-plane and is a distance \(|c|\) from it. If \(c \gt 0\text{,}\) the plane \(z=c\) is above the \(xy\)-plane. If \(c \lt 0\text{,}\) the plane \(z=c\) is below the \(xy\)-plane. We say that the plane \(z=c\) is a signed distance \(c\) from the \(xy\)-plane.
  • The set of all points \((x,y,z)\) that obey \(y=b\) is a plane that is parallel to the \(xz\)-plane and is a signed distance \(b\) from it.
  • The set of all points \((x,y,z)\) that obey \(x=a\) is a plane that is parallel to the \(yz\)-plane and is a signed distance \(a\) from it.
Observe that our 2d distances extend quite easily to 3d.
  • the distance from the point \((x,y,z)\) to the \(xy\)-plane is \(|z|\)
  • the distance from the point \((x,y,z)\) to the \(xz\)-plane is \(|y|\)
  • the distance from the point \((x,y,z)\) to the \(yz\)-plane is \(|x|\)
  • the distance from the point \((x,y,z)\) to the origin \((0,0,0)\) is \(\sqrt{x^2+y^2+z^2}\)
To see that the distance from the point \((x,y,z)\) to the origin \((0,0,0)\) is indeed \(\sqrt{x^2+y^2+z^2}\text{,}\)
  • apply Pythagoras to the right-angled triangle with vertices \((0,0,0)\text{,}\) \((x,0,0)\) and \((x,y,0)\) to see that the distance from \((0,0,0)\) to \((x,y,0)\) is \(\sqrt{x^2+y^2}\) and then
  • apply Pythagoras to the right-angled triangle with vertices \((0,0,0)\text{,}\) \((x,y,0)\) and \((x,y,z)\) to see that the distance from \((0,0,0)\) to \((x,y,z)\) is \(\sqrt{{\big(\sqrt{x^2+y^2}\big)}^2+z^2} =\sqrt{x^2+y^2+z^2}\text{.}\)
More generally, the distance from the point \((x,y,z)\) to the point \((x',y',z')\) is
\begin{equation*} \sqrt{(x-x')^2+(y-y')^2+(z-z')^2} \end{equation*}
Notice that this gives us the equation for a sphere quite directly. All the points on a sphere are equidistant from the centre of the sphere. So, for example, the equation of the sphere centered on \((1,2,3)\) with radius \(4\text{,}\) that is, the set of all points \((x,y,z)\) whose distance from \((1,2,3)\) is \(4\text{,}\) is
\begin{equation*} (x-1)^2+(y-2)^2+(z-3)^2=16 \end{equation*}
Here is an example in which we sketch a region in the \(xy\)-plane that is specified using inequalities.
In this example, we sketch the region
\begin{equation*} \Set{(x,y)}{ -12\le x^2-6x +y^2-4y \le -9,\ \ y\ge 1} \end{equation*}
in the \(xy\)-plane.
We do so in two steps. In the first step, we sketch the curves \(x^2-6x +y^2-4y=-12\text{,}\) \(x^2-6x +y^2-4y=-9\text{,}\) and \(y=1\text{.}\)
  • By completing squares, we see that the equation \(x^2-6x +y^2-4y=-12\) is equivalent to \((x-3)^2 +(y-2)^2 =1\text{,}\) which is the circle of radius \(1\) centred on \((3,2)\text{.}\) It is sketched in the figure below.
  • By completing squares, we see that the equation \(x^2-6x +y^2-4y=-9\) is equivalent to \((x-3)^2 +(y-2)^2 =4\text{,}\) which is the circle of radius \(2\) centred on \((3,2)\text{.}\) It is sketched in the figure below.
  • The point \((x,y)\) obeys \(y=1\) if and only if it is a distance \(1\) vertically above the \(x\)-axis. So \(y=1\) is the line that is parallel to the \(x\)-axis and is one unit above it. This line is also sketched in the figure below.
In the second step we determine the impact that the inequalities have.
  • The inequality \(x^2-6x +y^2-4y\ge -12\) is equivalent to \((x-3)^2 +(y-2)^2 \ge 1\) and hence is equivalent to \(\sqrt{(x-3)^2 +(y-2)^2} \ge 1\text{.}\) So the point \((x,y)\) satisfies \(x^2-6x +y^2-4y\ge -12\) if and only if the distance from \((x,y)\) to \((3,2)\) is at least \(1\text{,}\) i.e. if and only if \((x,y)\) is outside (or on) the circle \((x-3)^2 +(y-2)^2 = 1\text{.}\)
  • The inequality \(x^2-6x +y^2-4y\le -9\) is equivalent to \((x-3)^2 +(y-2)^2 \le 4\) and hence is equivalent to \(\sqrt{(x-3)^2 +(y-2)^2} \le 2\text{.}\) So the point \((x,y)\) satisfies the inequality \(x^2-6x +y^2-4y\le -9\) if and only if the distance from \((x,y)\) to \((3,2)\) is at most \(2\text{,}\) i.e. if and only if \((x,y)\) is inside (or on) the circle \((x-3)^2 +(y-2)^2 = 4\text{.}\)
  • The point \((x,y)\) obeys \(y\ge 1\) if and only if \((x,y)\) is a vertical distance at least \(1\) above the \(x\)-axis, i.e. is above (or on) the line \(y=1\text{.}\)
  • So the region
    \begin{equation*} \Set{(x,y)}{ -12\le x^2-6x +y^2-4y \le -9,\ \ y\ge 1} \end{equation*}
    consists of all points \((x,y)\) that
    • are inside or on the circle \((x-3)^2 +(y-2)^2 = 4\) and
    • are also outside or on the circle \((x-3)^2 +(y-2)^2 = 1\) and
    • are also above or on the line \(y=1\text{.}\)
    It is the shaded region in the figure below.
Here are a couple of examples that involve spheres.
In this example, we are going to find the curve formed by the intersection of the \(xy\)-plane and the sphere of radius \(5\) centred on \((0,0,4)\text{.}\)
The point \((x,y,z)\) lies on the \(xy\)-plane if and only if \(z=0\text{,}\) and lies on the sphere of radius \(5\) centred on \((0,0,4)\) if and only if \(x^2+y^2+(z-4)^2=25\text{.}\) So the point \((x,y,z)\) lies on the curve of intersection if and only if both \(z=0\) and \(x^2+y^2+(z-4)^2=25\text{,}\) or equivalently
\begin{equation*} z=0,\quad x^2+y^2+(0-4)^2=25 \iff z=0,\quad x^2+y^2 = 9 \end{equation*}
This is the circle in the \(xy\)-plane that is centred on the origin and has radius \(3\text{.}\) Here is a sketch that show the parts of the sphere and the circle of intersection that are in the first octant. That is, that have \(x\ge 0\text{,}\) \(y\ge 0\) and \(z\ge 0\text{.}\)
In this example, we are going to find all points \((x,y,z)\) for which the distance from \((x,y,z)\) to \((9,-12,15)\) is twice the distance from \((x,y,z)\) to the origin \((0,0,0)\text{.}\)
The distance from \((x,y,z)\) to \((9,-12,15)\) is \(\sqrt{(x-9)^2+(y+12)^2+(z-15)^2}\text{.}\) The distance from \((x,y,z)\) to \((0,0,0)\) is \(\sqrt{x^2+y^2+z^2}\text{.}\) So we want to find all points \((x,y,z)\) for which
\begin{equation*} \sqrt{(x-9)^2+(y+12)^2+(z-15)^2} =2\sqrt{x^2+y^2+z^2} \end{equation*}
Squaring both sides of this equation gives
\begin{equation*} x^2-18x+81 +y^2+24y+144 +z^2-30z+225 = 4\big(x^2+y^2+z^2) \end{equation*}
Collecting up terms gives
\begin{alignat*}{1} 3x^2+18x +3y^2-24y +3z^2+30z \amp= 450\ \text{and, dividing by 3,} \\ x^2+6x +y^2-8y +z^2+10z \amp= 150\ \text{and, completing squares,}\\ x^2+6x +9 +y^2-8y +16 +z^2+10z +25 \amp= 200\ \text{or}\\ (x+3)^2+(y-4)^2+(z+5)^2 \amp=200 \end{alignat*}
This is the sphere of radius \(10\sqrt{2}\) centred on \((-3,4,-5)\text{.}\)

Exercises Exercises

Exercise Group.

Exercises — Stage 1
1.
Describe the set of all points \((x,y,z)\) in \(\bbbr^3\) that satisfy
  1. \(\displaystyle x^2 +y^2+z^2= 2x-4y+4\)
  2. \(\displaystyle x^2 +y^2+z^2 \lt 2x-4y+4\)
2.
Describe and sketch the set of all points \((x,y)\) in \(\bbbr^2\) that satisfy
  1. \(\displaystyle x=y\)
  2. \(\displaystyle x+y=1\)
  3. \(\displaystyle x^2+y^2=4\)
  4. \(\displaystyle x^2+y^2=2y\)
  5. \(\displaystyle x^2+y^2 \lt 2y\)
3.
Describe the set of all points \((x,y,z)\) in \(\bbbr^3\) that satisfy the following conditions. Sketch the part of the set that is in the first octant.
  1. \(\displaystyle z = x\)
  2. \(\displaystyle x + y + z = 1\)
  3. \(\displaystyle x^2 + y^2 + z^2 = 4\)
  4. \(x^2 + y^2 + z^2 = 4\text{,}\) \(z = 1\)
  5. \(\displaystyle x^2+y^2=4\)
  6. \(\displaystyle z = x^2 + y^2\)
4.
Let \(A\) be the point \((2,1,3)\text{.}\)
  1. Find the distance from \(A\) to the \(xy\)-plane.
  2. Find the distance from \(A\) to the \(xz\)-plane.
  3. Find the distance from \(A\) to the point \((x,0,0)\) on the \(x\)-axis.
  4. Find the point on the \(x\)-axis that is closest to \(A\text{.}\)
  5. What is the distance from \(A\) to the \(x\)-axis?

Exercise Group.

Exercises — Stage 2
5.
Consider any triangle. Pick a coordinate system so that one vertex is at the origin and a second vertex is on the positive \(x\)-axis. Call the coordinates of the second vertex \((a,0)\) and those of the third vertex \((b,c)\text{.}\) Find the circumscribing circle (the circle that goes through all three vertices).
6. (✳).
A certain surface consists of all points \(P=(x,y,z)\) such that the distance from \(P\) to the point \((0,0,1)\) is equal to the distance from \(P\) to the plane \(z+1=0\text{.}\) Find an equation for the surface, sketch and describe it verbally.
7.
Show that the set of all points \(P\) that are twice as far from \((3,-2,3)\) as from \((3/2,1,0)\) is a sphere. Find its centre and radius.

Exercise Group.

Exercises — Stage 3
8.
The pressure \(p(x,y)\) at the point \((x,y)\) is at least zero and is determined by the equation \(x^2-2px+y^2=3p^2\text{.}\) Sketch several isobars. An isobar is a curve with equation \(p(x,y)=c\) for some constant \(c\ge 0\text{.}\)
This is why the \(xy\)-plane is called “two dimensional” — the name of each point consists of two real numbers.
Not surprisingly, the \(2\) in \(\bbbr^2\) signifies that each point is labelled by two numbers and the \(\bbbr\) in \(\bbbr^2\) signifies that the numbers in question are real numbers. There are more advanced applications (for example in signal analysis and in quantum mechanics) where complex numbers are used. The space of all pairs \((z_1,z_2)\text{,}\) with \(z_1\) and \(z_2\) complex numbers is denoted \(\bbbc^2\text{.}\)
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