how to draw 3d calculus

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In practice students taking multivariable calculus regularly have great difficulty visualising surfaces in 3 dimensions, despite the fact that we all live in iii dimensions. We'll now develop some technique to help us sketch surfaces in three dimensions¹.

Of form you lot could instead use some fancy graphing software, but office of the betoken is to build intuition. Not to mention that you can't use fancy graphing software on your test.

Nosotros all accept a fair fleck of feel drawing curves in two dimensions. Typically the intersection of a surface (in three dimensions) with a plane is a curve lying in the (2 dimensional) plane. Such an intersection is usually called a cross-department. In the special case that the plane is i of the coordinate planes, the intersection is sometimes called a trace. One tin often get a pretty practiced idea of what a surface looks like by sketching a bunch of cross-sections. Here are some examples.

Case 1.7.1 . \(4x^two+y^two-z^2=1\).

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

Solution

We'll commencement by fixing whatsoever 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 a horizontal cross-department. Any point \((ten,y,z)\) lying on that horizontal cross-section satisfies both

\begin{marshal*} &z=z_0\ \ \text{and}\ \ 4x^2+y^2-z^2=1\\ \iff &z=z_0\ \ \text{and}\ \ 4x^two+y^two=1+z_0^2 \end{align*}

Think of \(z_0\) as a constant. Then \(4x^two+y^two=one+z_0^2\) is a bend in the \(xy\)-plane. As \(1+z_0^ii\) is a constant, the bend is an ellipse. To make up one's mind its semi-axes², we observe that when \(y=0\text{,}\) we accept \(10=\pm\frac{one}{two}\sqrt{1+z_0^two}\) and when \(x=0\text{,}\) we have \(y=\pm\sqrt{ane+z_0^ii}\text{.}\) And so the curve is just an ellipse with \(x\) semi-centrality \(\frac{1}{two}\sqrt{1+z_0^2}\) and \(y\) semi-centrality \(\sqrt{1+z_0^2}\text{.}\) It's piece of cake to sketch.

The semi-axes of an ellipse are the line segments from the middle of the ellipse to the farthest points on the ellipse and to the nearest points on the ellipse. For a circle the lengths of all of these line segments are simply the radius.

Remember that this ellipse is the function of our surface that lies in the airplane \(z=z_0\text{.}\) Imagine that the sketch of the ellipse is on a single sheet of paper. Lift the canvass of paper upwards, move information technology around so that the \(x\)- and \(y\)-axes bespeak in the directions of the three dimensional \(x\)- and \(y\)-axes and place the sail of newspaper into the three dimensional sketch at height \(z_0\text{.}\) This gives a single horizontal ellipse in 3d, as in the figure below.

We can build upwards the full surface by stacking many of these horizontal ellipses — one for each possible acme \(z_0\text{.}\) And then we now depict a few of them every bit in the figure below. To reduce the amount of clutter in the sketch, nosotros take simply drawn the first octant (i.eastward. the function of three dimensions that has \(x\ge 0\text{,}\) \(y\ge 0\) and \(z\ge 0\)).

Here is why it is OK, in this instance, to simply sketch the first octant. Replacing \(x\) by \(-x\) in the equation \(4x^2+y^2-z^two=i\) does not change the equation. That means that a point \((x,y,z)\) is on the surface if and but if the bespeak \((-x,y,z)\) is on the surface. So the surface is invariant under reflection in the \(yz\)-plane. Similarly, the equation \(4x^ii+y^2-z^2=1\) does not change when \(y\) is replaced by \(-y\) or \(z\) is replaced past \(-z\text{.}\) Our surface is likewise invariant reflection in the \(xz\)- and \(yz\)-planes. One time we have the office in the outset octant, the remaining octants can exist gotten simply by reflecting well-nigh the coordinate planes.

We tin become a more than visually meaningful sketch past adding in some vertical cross-sections. The \(10=0\) 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

\begin{equation*} x=0,\ y^two-z^ii=1\qquad\text{and}\qquad y=0,\ 4x^2-z^two=1 \end{equation*}

These equations depict hyperbolae^three. If y'all don't remember how to sketch them, don't worry. Nosotros'll do it now. We'll first sketch them in 2d. Since

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

the sketches are

It's not simply a figure of speech!

Now we'll incorporate them into the 3d sketch. Once again imagine that each is a single sheet of paper. Pick each up and move it into the 3d sketch, carefully matching upwards the axes. The red (blueish) parts of the hyperbolas above go the scarlet (blue) parts of the 3d sketch below (assuming of class that yous are looking at this on a colour screen).

At present that we take a pretty good idea of what the surface looks like we tin clean upward 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 i canvass.

There are also hyperboloids of two sheets. For example, replacing the \(+1\) on the right hand side of \(x^2+y^2-z^2=1\) gives \(x^2+y^two-z^ii=-1\text{,}\) which is a hyperboloid of two sheets. We'll sketch information technology quickly in the adjacent instance.

Example one.7.2 . \(4x^2+y^2-z^2=-i\).

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

Solution

As in the last case, we'll start past fixing whatsoever 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 plane is

\begin{marshal*} &z=z_0\ \ \text{and}\ \ 4x^2+y^2=z_0^2-1 \end{marshal*}

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

• If \(|z_0| \lt 1\text{,}\) then \(z_0^ii-i \lt 0\) and there are no solutions to \(ten^2+y^2=z_0^2-1\text{.}\)
• If \(|z_0|=ane\) at that place is exactly one solution, namely \(10=y=0\text{.}\)
• If \(|z_0| \gt i\) so \(4x^2+y^ii=z_0^2-1\) is an ellipse with \(x\) semi-centrality \(\frac{1}{2}\sqrt{z_0^2-1}\) and \(y\) semi-centrality \(\sqrt{z_0^two-1}\text{.}\) These semi-axes are modest when \(|z_0|\) is close to \(1\) and abound as \(|z_0|\) increases.

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

Next nosotros add in the \(x=0\) and \(y=0\) cantankerous-sections (i.e. the parts of our surface that are in the \(yz\)- and \(xz\)-planes, respectively)

\brainstorm{equation*} x=0,\ z^2=1+y^2\qquad\text{and}\qquad y=0,\ z^two=1+4x^2 \end{equation*}

At present that nosotros accept a pretty practiced idea of what the surface looks like nosotros make clean upwardly and simplify the sketch.

Here is are ii figures created by graphing software.

This type of surface is called a hyperboloid of 2 sheets.

Case 1.7.3 . \(yz=ane\).

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

Solution

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

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

Next we'll move this 2d sketch into the \(yz\)-plane, i.e. the plane \(10=0\text{,}\) in 3d, except that we'll just draw in the office in the first octant.

The nosotros'll draw in \(10=x_0\) cross-sections for a couple of more values of \(x_0\)

and clean upward the sketch a scrap

Here are two figures created by graphing software.

Example 1.7.iv . \(xyz=four\).

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

Solution

Nosotros'll sketch this surface using much the aforementioned procedure equally we used in Examples ane.seven.1 and 1.7.2. We'll only sketch the part of the surface in the beginning 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 \(y,z \lt 0\text{,}\) \(x\ge0\)) are only reflections of the get-go octant office.

As usual, we start by fixing whatever 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

\begin{align*} &z=z_0\ \ \text{and}\ \ xy=\frac{4}{z_0} \terminate{marshal*}

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

Next we add some vertical cross-sections. Nosotros can't use \(x=0\) or \(y=0\) because any indicate on \(xyz=1\) must have all of \(10\text{,}\) \(y\text{,}\) \(z\) nonzero. And so we use

\begin{equation*} x=4,\ yz=1\qquad\text{and}\qquad y=4,\ xz=1 \terminate{equation*}

instead. They are again hyperbolae.

Finally, we clean up and simplify the sketch.

Here are two figures created by graphing software.

Subsection 1.7.i 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 role of two variables \(f(x,y)\text{.}\) Another adept mode to visualize the behaviour of a part \(f(ten,y)\) is to sketch what are called its level curves. By definition, a level curve of \(f(x,y)\) is a curve whose equation is \(f(x,y)=C\text{,}\) for some constant \(C\text{.}\) It is the set of points in the \(xy\)-plane where \(f\) takes the value \(C\text{.}\) Because it is a curve in 2d, it is usually easier to sketch than the graph of \(f\text{.}\) Here are a couple of examples.

Instance i.7.5 . \(f(x,y) = ten^2+4y^2-2x+2\).

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

Solution

Fix any real number \(C\text{.}\) Then, for the specified part \(f\text{,}\) the level curve \(f(x,y)=C\) is the set of points \((x,y)\) that obey

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

Now \((x-1)^ii + 4y^2\) is the sum of two squares, and so is always at to the lowest degree zero. And so if \(C-one \lt 0\text{,}\) i.e. if \(C \lt 1\text{,}\) in that location is no curve \(f(x,y)=C\text{.}\) If \(C-i=0\text{,}\) i.e. if \(C=ane\text{,}\) then \(f(x,y)=C-1=0\) if and only if both \((x-one)^2=0\) and \(4y^two=0\) and so the level curve consists of the single point \((one,0)\text{.}\) If \(C \gt ane\text{,}\) then \(f(x,y)=C\) become \((10-1)^two+4y^2=C-1 \gt 0\) which describes an ellipse centred on \((i,0)\text{.}\) It intersects the \(x\)-centrality when \(y=0\) and

\begin{equation*} (x-ane)^two = C-1 \iff x-1=\pm\sqrt{C-1} \iff x=1\pm \sqrt{C-1} \end{equation*}

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

\brainstorm{equation*} 4y^ii = C-ane \iff 2y=\pm\sqrt{C-1} \iff y=\pm\tfrac{one}{ii} \sqrt{C-one} \terminate{equation*}

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

It is often easier to develop an understanding of the behaviour of a function \(f(x,y)\) past looking at a sketch of its level curves, than information technology is by looking at a sketch of its graph. On the other mitt, yous can likewise apply a sketch of the level curves of \(f(x,y)\) as the first step in building a sketch of the graph \(z=f(x,y)\text{.}\) The side by side step would be to redraw, for each \(C\text{,}\) the level curve \(f(ten,y)=C\text{,}\) in the plane \(z=C\text{,}\) as we did in Example i.seven.one.

Example 1.vii.6 . \(e^{x+y+z}=1\).

The function \(f(x,y)\) is given implicitly by the equation \(eastward^{x+y+z}=1\text{.}\) Sketch the level curves of \(f\text{.}\)

Solution

This 1 is not equally nasty as it appears. That "\(f(ten,y)\) is given implicitly by the equation \(eastward^{x+y+z}=one\)" means that, for each \(ten,y\text{,}\) the solution \(z\) of \(due east^{x+y+z}=ane\) is \(f(x,y)\text{.}\) So, for the specified function \(f\) and any fixed real number \(C\text{,}\) the level curve \(f(ten,y)=C\) is the set of points \((ten,y)\) that obey

\begin{align*} eastward^{ten+y+C}=ane &\iff x+y+C = 0\qquad\text{(by taking the logarithm of both sides)}\\ &\iff x+y = -C \terminate{align*}

This is of course a direct line. It intersects the \(x\)-axis when \(y=0\) and \(ten=-C\) and it intersects the \(y\)-axis when \(x=0\) and \(y=-C\text{.}\) Hither is a sketch of some level curves.

We have simply seen that sketching the level curves of a function \(f(10,y)\) can aid u.s.a. understand the behaviour of \(f\text{.}\) We can generalise this to functions \(F(x,y,z)\) of three variables. A level surface of \(F(x,y,z)\) is a surface whose equation is of the grade \(F(x,y,z)=C\) for some constant \(C\text{.}\) It is the gear up of points \((x,y,z)\) at which \(F\) takes the value \(C\text{.}\)

Example i.7.7 . \(F(ten,y,z)=10^2+y^2+z^2\).

Let \(F(10,y,z)=x^ii+y^2+z^2\text{.}\) If \(C \gt 0\text{,}\) then 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 \(F=1\) (radius \(1\)), \(F=4\) (radius \(ii\)) and \(F=9\) (radius \(3\)) that are in the get-go octant.

Instance one.vii.8 . \(F(ten,y,z)=x^two+z^2\).

Permit \(F(ten,y,z)=x^2+z^ii\) and \(C \gt 0\text{.}\) Consider the level surface \(10^2+z^2=C\text{.}\) The variable \(y\) does not appear in this equation. So for any fixed \(y_0\text{,}\) the intersection of the our surface \(ten^ii+z^2=C\) with the plane \(y=y_0\) is the circumvolve of radius \(\sqrt{C}\) centred on \(x=z=0\text{.}\) Here is a sketch of the showtime quadrant role of ane such circle.

The full surface is the horizontal stack of all of those circles with \(y_0\) running over \(\bbbr\text{.}\) It is the cylinder of radius \(\sqrt{C}\) centred on the \(y\)-axis. Here is a sketch of the parts of the level surfaces \(F=1\) (radius \(ane\)), \(F=4\) (radius \(ii\)) and \(F=ix\) (radius \(3\)) that are in the outset octant.

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

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

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

that are in the first octant.

Exercises ane.vii.2 Exercises

Exercises — Stage 1

one. ✳.

Lucifer the following equations and expressions with the corresponding pictures. Cartesian coordinates are \((ten, y, z)\text{,}\) cylindrical coordinates are \((r, \theta, z)\text{,}\) and spherical coordinates are \((\rho, \theta, \varphi)\text{.}\)

(A)

(B)

(C)

(D)

(Eastward)

(F)

\brainstorm{alignat*}{7} &\text{(a)}\quad& \varphi&=\pi/iii & &\text{(b)}\quad& r&=ii\cos\theta & &\text{(c)}\quad& 10^two+y^2&=z^2+1\\ &\text{(d)}& y&=ten^two+z^two\qquad & &\text{(e)}& \rho&=2\cos\varphi\qquad & &\text{(f)}& z&=x^iv+y^4-4xy & \stop{alignat*}

2.

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 function \(f(x,y)\text{.}\) Sketch the first octant part of the respective graph \(z=f(ten,y)\text{.}\)

(a)

(b)

3.

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

Exercises — Stage 2

4.

Sketch some of the level curves of

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

5. ✳.

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

6. ✳.

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

seven. ✳.

A surface is given implicitly by

\begin{equation*} x^ii + y^two - z^2 + 2z = 0 \cease{equation*}

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

eight. ✳.

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

9.

Describe the level surfaces of

1. \(\displaystyle f(x,y,z)=10^2+y^two+z^2\)
2. \(\displaystyle f(x,y,z)=x+2y+3z\)
3. \(\displaystyle f(10,y,z)=ten^two+y^2\)

10.

Sketch the graphs of

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

11.

Sketch and draw the following surfaces.

1. \(\displaystyle 4x^2+y^ii=sixteen\)
2. \(\displaystyle x+y+2z=4\)
3. \(\displaystyle \frac{y^ii}{9}+\frac{z^2}{4}=one+\frac{x^ii}{16}\)
4. \(\displaystyle y^2=x^two+z^2\)
5. \(\displaystyle \frac{x^2}{9}+\frac{y^2}{12}+\frac{z^2}{nine}=1\)
6. \(x^ii+y^2+z^2+4x-by+9z-b=0\) where \(b\) is a abiding.
7. \(\displaystyle \frac{ten}{four}=\frac{y^two}{4}+\frac{z^2}{9}\)
8. \(\displaystyle z=x^2\)

Exercises — Stage iii

12.

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

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Source: https://personal.math.ubc.ca/~CLP/CLP3/clp_3_mc/sec_sketching.html

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