How to Draw a 3d Math Plane

ane.7: Sketching Surfaces in 3d

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In practice students taking multivariable calculus regularly take nifty difficulty visualising surfaces in three dimensions, despite the fact that we all alive in iii dimensions. We'll now develop some technique to help us sketch surfaces in three dimensions^one.

Nosotros all have a off-white bit of experience drawing curves in 2 dimensions. Typically the intersection of a surface (in three dimensions) with a plane is a curve lying in the (two dimensional) airplane. Such an intersection is unremarkably called a cross-section. In the special example that the airplane is one of the coordinate planes, the intersection is sometimes called a trace. One can often become a pretty good idea of what a surface looks similar by sketching a bunch of cantankerous-sections. Here are some examples.

Example 1.seven.i. \(4x^2+y^2-z^ii=i\)

Sketch the surface that satisfies \(4x^two+y^2-z^ii=1\text{.}\)

Solution

We'll start by fixing any number \(z_0\) and sketching the office of the surface that lies in the horizontal plane \(z=z_0\text{.}\)

The intersection of our surface with that horizontal airplane is a horizontal cross-section. Any point \((x,y,z)\) lying on that horizontal cross-section satisfies both

\[\brainstorm{marshal*} &z=z_0\ \ \text{and}\ \ 4x^2+y^2-z^2=1\\ \iff &z=z_0\ \ \text{and}\ \ 4x^ii+y^2=i+z_0^2 \end{marshal*}\]

Think of \(z_0\) every bit a constant. Then \(4x^2+y^2=1+z_0^2\) is a curve in the \(xy\)-plane. Every bit \(i+z_0^ii\) is a constant, the curve is an ellipse. To determine its semi-axes², we discover that when \(y=0\text{,}\) we take \(10=\pm\frac{1}{ii}\sqrt{1+z_0^two}\) and when \(x=0\text{,}\) nosotros accept \(y=\pm\sqrt{1+z_0^2}\text{.}\) Then the curve is merely an ellipse with \(x\) semi-axis \(\frac{one}{two}\sqrt{1+z_0^ii}\) and \(y\) semi-axis \(\sqrt{1+z_0^2}\text{.}\) It's easy to sketch.

Remember that this ellipse is the role of our surface that lies in the airplane \(z=z_0\text{.}\) Imagine that the sketch of the ellipse is on a single canvas of newspaper. Lift the sail of paper up, move it around and then that the \(x\)- and \(y\)-axes point in the directions of the three dimensional \(x\)- and \(y\)-axes and place the canvass of paper into the three dimensional sketch at height \(z_0\text{.}\) This gives a single horizontal ellipse in 3d, every bit in the figure below.

We can build up the full surface past stacking many of these horizontal ellipses — one for each possible height \(z_0\text{.}\) So we now depict a few of them as in the figure below. To reduce the amount of clutter in the sketch, we have only drawn the kickoff octant (i.due east. the role of three dimensions that has \(ten\ge 0\text{,}\) \(y\ge 0\) and \(z\ge 0\)).

Here is why information technology is OK, in this example, to just sketch the first octant. Replacing \(x\) by \(-10\) in the equation \(4x^2+y^2-z^two=i\) does not modify the equation. That means that a signal \((x,y,z)\) is on the surface if and merely if the point \((-10,y,z)\) is on the surface. So the surface is invariant nether reflection in the \(yz\)-airplane. Similarly, the equation \(4x^2+y^2-z^2=1\) does non change when \(y\) is replaced by \(-y\) or \(z\) is replaced by \(-z\text{.}\) Our surface is also invariant reflection in the \(xz\)- and \(yz\)-planes. In one case we have the role in the get-go octant, the remaining octants can be gotten simply by reflecting about the coordinate planes.

We can get a more visually meaningful sketch by adding in some vertical cantankerous-sections. The \(x=0\) and \(y=0\) cross-sections (also called traces — they are the parts of our surface that are in the \(yz\)- and \(xz\)-planes, respectively) are

\[ x=0,\ y^2-z^2=1\qquad\text{and}\qquad y=0,\ 4x^2-z^2=1 \nonumber \]

These equations describe hyperbolae³. If you don't recall how to sketch them, don't worry. We'll practise it now. We'll first sketch them in 2d. Since

\[\begin{alignat*}{2} y^2&=ane+z^2 & \quad\implies\quad &|y|\ge 1\\ &&&\text{ and }\quad y=\pm ane\text{ when }z=0\\ &&&\text{ and }\quad\text{for large } z,\ y\approx\pm z\\ 4x^2&=1+z^ii & \quad\implies\quad &|x|\ge \tfrac{1}{2}\\ &&&\text{ and }\quad x=\pm\tfrac{one}{two}\text{ when }z=0\\ &&&\text{ and }\quad\text{for large } z,\ x\approx\pm \tfrac{1}{2}z \cease{alignat*}\]

the sketches are

At present nosotros'll comprise them into the 3d sketch. Once over again imagine that each is a unmarried canvass of paper. Pick each up and move information technology into the 3d sketch, carefully matching up the axes. The red (blue) parts of the hyperbolas to a higher place become the cherry-red (blue) parts of the 3d sketch below (bold of course that you are looking at this on a colour screen).

Now that we accept a pretty adept idea of what the surface looks like nosotros can clean upwardly and simplify the sketch. Hither are a couple of possibilities.

Here are two figures created by graphing software.

This type of surface is called a hyperboloid of one canvass.

In that location are too hyperboloids of two sheets. For example, replacing the \(+ane\) on the correct hand side of \(10^2+y^ii-z^2=1\) gives \(ten^2+y^2-z^2=-1\text{,}\) which is a hyperboloid of 2 sheets. We'll sketch information technology quickly in the adjacent example.

Example 1.7.2. \(4x^2+y^two-z^2=-1\)

Sketch the surface that satisfies \(4x^2+y^2-z^2=-1\text{.}\)

Solution

Equally in the concluding instance, we'll start by fixing any number \(z_0\) and sketching the part of the surface that lies in the horizontal aeroplane \(z=z_0\text{.}\) The intersection of our surface with that horizontal plane is

\[\begin{align*} &z=z_0\ \ \text{and}\ \ 4x^2+y^2=z_0^two-i \end{align*}\]

Think of \(z_0\) every bit a constant.

• If \(|z_0| \lt ane\text{,}\) then \(z_0^2-one \lt 0\) and at that place are no solutions to \(x^two+y^2=z_0^2-1\text{.}\)
• If \(|z_0|=ane\) there is exactly one solution, namely \(x=y=0\text{.}\)
• If \(|z_0| \gt one\) and then \(4x^ii+y^2=z_0^2-1\) is an ellipse with \(x\) semi-axis \(\frac{ane}{two}\sqrt{z_0^two-i}\) and \(y\) semi-axis \(\sqrt{z_0^two-1}\text{.}\) These semi-axes are small when \(|z_0|\) is close to \(1\) and grow as \(|z_0|\) increases.

The first octant parts of a few of these horizontal cross-sections are fatigued in the figure below.

Side by side we add in the \(x=0\) and \(y=0\) cross-sections (i.e. the parts of our surface that are in the \(yz\)- and \(xz\)-planes, respectively)

\[ x=0,\ z^2=ane+y^two\qquad\text{and}\qquad y=0,\ z^2=1+4x^2 \nonumber \]

Now that we have a pretty good idea of what the surface looks like nosotros clean up and simplify the sketch.

Here is are 2 figures created by graphing software.

This type of surface is called a hyperboloid of two sheets.

Instance 1.vii.3. \(yz=one\)

Sketch the surface \(yz=1\text{.}\)

Solution

This surface has a special belongings that makes it relatively easy to sketch. There are no \(x\)'s in the equation \(yz=1\text{.}\) That means that if some \(y_0\) and \(z_0\) obey \(y_0z_0=i\text{,}\) then the point \((ten,y_0,z_0)\) lies on the surface \(yz=one\) for all values of \(10\text{.}\) Equally \(ten\) runs from \(-\infty\) to \(\infty\text{,}\) the bespeak \((ten,y_0,z_0)\) sweeps out a direct line parallel to the \(ten\)-axis. So the surface \(yz=1\) is a spousal relationship of lines parallel to the \(x\)-centrality. It is invariant nether translations parallel to the \(10\)-axis. To sketch \(yz=1\text{,}\) we simply need to sketch its intersection with the \(yz\)-plane and then translate the resulting curve parallel to the \(x\)-axis to sweep out the surface.

We'll kickoff with a sketch of the hyperbola \(yz=one\) in two dimensions.

Next we'll move this 2d sketch into the \(yz\)-aeroplane, i.due east. the plane \(10=0\text{,}\) in 3d, except that we'll only depict in the part in the start octant.

The we'll describe in \(x=x_0\) cross-sections for a couple of more than values of \(x_0\)

and clean up the sketch a bit

Here are 2 figures created by graphing software.

Example i.vii.4. \(xyz=iv\)

Sketch the surface \(xyz=4\text{.}\)

Solution

We'll sketch this surface using much the aforementioned procedure as nosotros used in Examples one.7.1 and 1.7.2. We'll just sketch the role of the surface in the starting time octant. The remaining parts (in the octants with \(ten,y \lt 0\text{,}\) \(z\ge 0\text{,}\) with \(x,z \lt 0\text{,}\) \(y\ge 0\) and with \(y,z \lt 0\text{,}\) \(ten\ge0\)) are just reflections of the first octant part.

As usual, nosotros get-go by fixing any number \(z_0\) and sketching the part of the surface that lies in the horizontal plane \(z=z_0\text{.}\) The intersection of our surface with that horizontal plane is the hyperbola

\[\brainstorm{align*} &z=z_0\ \ \text{and}\ \ xy=\frac{4}{z_0} \end{align*}\]

Annotation that \(x\rightarrow\infty\) equally \(y\rightarrow 0\) and that \(y\rightarrow\infty\) every bit \(x\rightarrow 0\text{.}\) So the hyperbola has both the \(x\)-axis and the \(y\)-axis as asymptotes, when drawn in the \(xy\)-plane. The first octant parts of a few of these horizontal cross-sections (namely, \(z_0=4\text{,}\) \(z_0=ii\) and \(z_0=\frac{1}{2}\)) are drawn in the figure below.

Next we add some vertical cross-sections. Nosotros tin't use \(x=0\) or \(y=0\) because any point on \(xyz=i\) must have all of \(x\text{,}\) \(y\text{,}\) \(z\) nonzero. And so we apply

\[ x=4,\ yz=ane\qquad\text{and}\qquad y=4,\ xz=1 \nonumber \]

instead. They are again hyperbolae.

Finally, nosotros make clean up and simplify the sketch.

Here are two figures created by graphing software.

Level Curves and Surfaces

Often the reason you are interested in a surface in 3d is that it is the graph \(z=f(x,y)\) of a function of ii variables \(f(ten,y)\text{.}\) Another skillful manner to visualize the behaviour of a office \(f(x,y)\) is to sketch what are called its level curves. By definition, a level bend of \(f(10,y)\) is a curve whose equation is \(f(x,y)=C\text{,}\) for some constant \(C\text{.}\) It is the set up of points in the \(xy\)-plane where \(f\) takes the value \(C\text{.}\) Considering it is a curve in 2nd, it is commonly easier to sketch than the graph of \(f\text{.}\) Hither are a couple of examples.

Example 1.7.5. \(f(x,y) = x^2+4y^2-2x+2\)

Sketch the level curves of \(f(ten,y) = x^2+4y^2-2x+2\text{.}\)

Solution

Fix whatsoever real number \(C\text{.}\) And then, for the specified function \(f\text{,}\) the level curve \(f(ten,y)=C\) is the ready of points \((x,y)\) that obey

\[\begin{align*} x^2+4y^two-2x+2=C &\iff x^2-2x+1 + 4y^2 +1 =C\\ &\iff (x-1)^2 + 4y^ii = C-one \finish{align*}\]

At present \((x-1)^2 + 4y^2\) is the sum of two squares, and then is ever at least naught. And so if \(C-1 \lt 0\text{,}\) i.eastward. if \(C \lt 1\text{,}\) in that location is no curve \(f(x,y)=C\text{.}\) If \(C-1=0\text{,}\) i.e. if \(C=1\text{,}\) and then \(f(x,y)=C-ane=0\) if and merely if both \((x-1)^2=0\) and \(4y^two=0\) and and then the level bend consists of the unmarried point \((i,0)\text{.}\) If \(C \gt 1\text{,}\) so \(f(x,y)=C\) become \((x-1)^2+4y^2=C-1 \gt 0\) which describes an ellipse centred on \((ane,0)\text{.}\) It intersects the \(x\)-axis when \(y=0\) and

\[ (x-ane)^2 = C-ane \iff 10-1=\pm\sqrt{C-1} \iff x=1\pm \sqrt{C-1} \nonumber \]

and it intersects the line \(x=1\) (i.east. the vertical line through the middle) when

\[ 4y^2 = C-one \iff 2y=\pm\sqrt{C-1} \iff y=\pm\tfrac{1}{2} \sqrt{C-1} \nonumber \]

So, when \(C \gt 1\text{,}\) \(f(ten,y)=C\) is the ellipse centred on \((1,0)\) with \(10\) semi-axis \(\sqrt{C-1} \) and \(y\) semi-axis \(\tfrac{1}{two}\sqrt{C-1}\text{.}\) Here is a sketch of some representative level curves of \(f(ten,y) = 10^ii+4y^two-2x+two\text{.}\)

Information technology is often easier to develop an understanding of the behaviour of a part \(f(x,y)\) by looking at a sketch of its level curves, than it is by looking at a sketch of its graph. On the other mitt, you tin can also use a sketch of the level curves of \(f(x,y)\) as the showtime pace in building a sketch of the graph \(z=f(x,y)\text{.}\) The next step would be to redraw, for each \(C\text{,}\) the level bend \(f(x,y)=C\text{,}\) in the plane \(z=C\text{,}\) every bit we did in Instance 1.seven.1.

Example 1.7.6. \(e^{x+y+z}=ane\)

The role \(f(x,y)\) is given implicitly by the equation \(due east^{x+y+z}=one\text{.}\) Sketch the level curves of \(f\text{.}\)

Solution

This 1 is non every bit nasty every bit it appears. That "\(f(ten,y)\) is given implicitly past the equation \(e^{x+y+z}=1\)" ways that, for each \(ten,y\text{,}\) the solution \(z\) of \(due east^{10+y+z}=one\) is \(f(x,y)\text{.}\) So, for the specified function \(f\) and any stock-still real number \(C\text{,}\) the level curve \(f(ten,y)=C\) is the prepare of points \((x,y)\) that obey

\[\brainstorm{marshal*} eastward^{x+y+C}=1 &\iff x+y+C = 0\qquad\text{(by taking the logarithm of both sides)}\\ &\iff x+y = -C \end{align*}\]

This is of course a directly line. Information technology intersects the \(x\)-axis when \(y=0\) and \(10=-C\) and it intersects the \(y\)-centrality when \(10=0\) and \(y=-C\text{.}\) Here is a sketch of some level curves.

We have only seen that sketching the level curves of a office \(f(x,y)\) can help united states sympathise the behaviour of \(f\text{.}\) We tin generalise this to functions \(F(10,y,z)\) of iii variables. A level surface of \(F(x,y,z)\) is a surface whose equation is of the course \(F(ten,y,z)=C\) for some constant \(C\text{.}\) It is the set of points \((ten,y,z)\) at which \(F\) takes the value \(C\text{.}\)

Example i.7.7. \(F(x,y,z)=x^2+y^2+z^two\)

Permit \(F(ten,y,z)=x^2+y^2+z^2\text{.}\) If \(C \gt 0\text{,}\) and so the level surface \(F(x,y,z)=C\) is the sphere of radius \(\sqrt{C}\) centred on the origin. Here is a sketch of the parts of the level surfaces \(F=one\) (radius \(1\)), \(F=4\) (radius \(ii\)) and \(F=9\) (radius \(3\)) that are in the outset octant.

Case one.7.8. \(F(x,y,z)=x^2+z^2\)

Let \(F(x,y,z)=x^two+z^2\) and \(C \gt 0\text{.}\) Consider the level surface \(x^2+z^ii=C\text{.}\) The variable \(y\) does not announced in this equation. And then for any fixed \(y_0\text{,}\) the intersection of the our surface \(ten^two+z^ii=C\) with the plane \(y=y_0\) is the circumvolve of radius \(\sqrt{C}\) centred on \(ten=z=0\text{.}\) Here is a sketch of the first quadrant part of one such circumvolve.

The full surface is the horizontal stack of all of those circles with \(y_0\) running over \(\mathbb{R}\text{.}\) It is the cylinder of radius \(\sqrt{C}\) centred on the \(y\)-axis. Hither is a sketch of the parts of the level surfaces \(F=i\) (radius \(1\)), \(F=iv\) (radius \(two\)) and \(F=9\) (radius \(3\)) that are in the first octant.

Example 1.7.ix. \(F(x,y,z)=e^{x+y+z}\)

Let \(F(x,y,z)=e^{10+y+z}\) and \(C \gt 0\text{.}\) Consider the level surface \(e^{x+y+z}=C\text{,}\) or equivalently, \(ten+y+z=\ln C\text{.}\) It is the plane that contains the intercepts \((\ln C,0,0)\text{,}\) \((0,\ln C,0)\) and \((0,0,\ln C)\text{.}\) Hither is a sketch of the parts of the level surfaces

• \(F=east\) (intercepts \((1,0,0)\text{,}\) \((0,1,0)\text{,}\) \((0,0,1)\)),
• \(F=e^two\) (intercepts \((2,0,0)\text{,}\) \((0,2,0)\text{,}\) \((0,0,2)\)) and
• \(F=due east^3\) (intercepts \((3,0,0)\text{,}\) \((0,3,0)\text{,}\) \((0,0,3)\))

that are in the first octant.

Exercises

Phase 1

i. ✳

Friction match the following equations and expressions with the corresponding pictures. Cartesian coordinates are \((10, y, z)\text{,}\) cylindrical coordinates are \((r, \theta, z)\text{,}\) and spherical coordinates are \((\rho, \theta, \vec{a}rphi)\text{.}\)

\[\begin{alignat*}{7} &\text{(a)}\quad& \vec{a}rphi&=\pi/3 & &\text{(b)}\quad& r&=two\cos\theta & &\text{(c)}\quad& x^2+y^ii&=z^ii+i\\ &\text{(d)}& y&=ten^2+z^2\qquad & &\text{(e)}& \rho&=two\cos\vec{a}rphi\qquad & &\text{(f)}& z&=x^iv+y^4-4xy & \end{alignat*}\]

two

In each of (a) and (b) below, you are provided with a sketch of the first quadrant parts of a few level curves of some office \(f(ten,y)\text{.}\) Sketch the first octant part of the respective graph \(z=f(x,y)\text{.}\)

iii

Sketch a few level curves for the function \(f(x,y)\) whose graph \(z=f(x,y)\) is sketched below.

Phase 2

four

Sketch some of the level curves of

1. \(\displaystyle f(ten,y)=x^2+2y^2\)
2. \(\displaystyle f(10,y)=xy\)
3. \(\displaystyle f(ten,y)=xe^{-y}\)

5. ✳

Sketch the level curves of \(f(ten,y)=\frac{2y}{x^ii+y^2}\text{.}\)

6. ✳

Describe a "contour map" of \(f(x, y) = e^{-x^2 +4y^two}\), showing all types of level curves that occur.

7. ✳

A surface is given implicitly by

\[ x^ii + y^2 - z^2 + 2z = 0 \nonumber \]

1. Sketch several level curves \(z = \)abiding.
2. Depict a crude sketch of the surface.

8. ✳

Sketch the hyperboloid \(z^ii=4x^2+y^2-i\text{.}\)

9

Describe the level surfaces of

1. \(\displaystyle f(x,y,z)=x^2+y^2+z^ii\)
2. \(\displaystyle f(x,y,z)=ten+2y+3z\)
3. \(\displaystyle f(x,y,z)=x^ii+y^2\)

10

Sketch the graphs of

1. \(\displaystyle f(10,y)=\sin x\qquad 0\le x\le two\pi,\ 0\le y\le 1\)
2. \(\displaystyle f(x,y)=\sqrt{ten^ii+y^2}\)
3. \(\displaystyle f(10,y)=|x|+|y|\)

11

Sketch and draw the following surfaces.

1. \(\displaystyle 4x^2+y^2=16\)
2. \(\displaystyle x+y+2z=4\)
3. \(\displaystyle \frac{y^2}{ix}+\frac{z^2}{4}=1+\frac{ten^two}{16}\)
4. \(\displaystyle y^ii=x^2+z^2\)
5. \(\displaystyle \frac{x^two}{9}+\frac{y^2}{12}+\frac{z^2}{9}=i\)
6. \(x^2+y^2+z^ii+4x-past+9z-b=0\) where \(b\) is a constant.
7. \(\displaystyle \frac{x}{iv}=\frac{y^2}{4}+\frac{z^ii}{9}\)
8. \(\displaystyle z=10^2\)

Stage three

12

The surface below has circular level curves, centred along the \(z\)-axis. The lines given are the intersection of the surface with the right half of the \(yz\)-aeroplane. Give an equation for the surface.

1. Of course you could instead use some fancy graphing software, just role of the betoken is to build intuition. Non to mention that y'all tin't utilise fancy graphing software on your exam.
2. The semi-axes of an ellipse are the line segments from the middle of the ellipse to the uttermost points on the ellipse and to the nearest points on the ellipse. For a circle the lengths of all of these line segments are just the radius.
3. It'south non just a figure of spoken communication!

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Source: https://math.libretexts.org/Bookshelves/Calculus/CLP-3_Multivariable_Calculus_%28Feldman_Rechnitzer_and_Yeager%29/01:_Vectors_and_Geometry_in_Two_and_Three_Dimensions/1.07:_Sketching_Surfaces_in_3d

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