3D graphs Manual

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x

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Example 1. A circular cone

At the right is the graph of the function

𝑧 = 𝑔(𝑥) = 16 − 4𝑥 (0 ≤ 𝑥 ≤ 4)

Put a scale on the axes.

Calculate g(2) and illustrate this on the diagram: g(2) = 8

Now we are going to introduce a new horizontal axis that we will

call y, so that we have a 3-dimensional coordinate system (x, y, z).

Then we rotate the graph of g about the z-axis through 360°. What

we get looks like a wizard’s hat. Technically, it’s called a cone,

more precisely a “circular cone” because its base is a circle.

This cone can be regarded as the graph of a function

𝑧 = 𝑓(𝑥, 𝑦)

defined at all points (x, y) in the circular base. At any point (x, y)

in the base circle, z is the height of the cone above that point.

Your job is to find a formula for 𝑓(𝑥, 𝑦).

Calculate

𝑓(2,0) = 8

𝑓(0,2) = 8

𝑓(2,2) = 16 − 4√8 = 16 − 8√2 This is often a critical breakthrough case. Students will struggle

before they see the significance of the distance of the point from

the origin. A critical aha! is obtained from drawing the circle

through which that point rotates. See graph below right.

𝑓(1,3) = 16 − 4√10

𝑓(𝑥, 𝑦) = 16 − 4√𝑥2 + 𝑦2

Hands on! Some students have trouble visualizing the cone. I find

it helps a lot to have a forked twig holding the straight part vertical

and rotating the branch. Another idea is to get a piece of fairly

stiff see-through plastic and make a cone. To allow the students to

“see” the dependence of z on x and y, an effective visual is to run a

thin vertical rod (e.g. a piece of spaghetti) down a small hole in the

cone, onto the point (x, y).

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Example 2. A circular paraboloid

Start with a pair of z-x axes and draw the parabola 𝑧 = 16 − 𝑥2.

Now introduce a new horizontal y-axis so that we have a 3-D

coordinate system. Rotate the parabola about the z-axis through

360°. What we get is called a circular paraboloid and it is the

graph of a function 𝑧 = 𝑓(𝑥, 𝑦). Your job is to find a formula for

this function, that is, to find the height of the paraboloid at any

point in the circular base.

What is f(2, 0)? That is, how high is

the paraboloid at x = 2, y = 0?

𝑓(2,0) = 16 − 4 = 12

Also calculate

𝑓(0,2) = 16 − 4 = 12

𝑓(2,2) = 16 − 8 = 8

𝑓(1,3) = 16 − 10 = 6

𝑓(𝑥, 𝑦) = 16 − (𝑥2 + 𝑦2)

The following exercises will use contour diagrams. An effective introduction to these is found in

the Khan Academy video: https://www.khanacademy.org/math/multivariable-calculus/thinking-

about-multivariable-function/visualizing-scalar-valued-functions/v/contour-plots

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𝑧 = 16 − (𝑥2 + 𝑦2)

In the diagram above, the contours plots for the parabola are drawn at intervals of z = 1. They

are circles and each circle corresponds to a fixed value of z (a fixed height). Label a few of them

with their z-value. Discuss the significance of the fact that the outer contours are more closely

spaced than the inner ones.

Now draw on the contour diagram the line

𝑦 = 4 − 2𝑥

Starting at (0, 4) walk up the hill following the path

(as if the contour diagram was the map of your

route) and then down the other side until you get

back to the z = 0 circle.

The contour lines should give you a good sense of

the nature of your walk. Figure out when you are

ascending and when you are descending. Mark the

point on your path when you attain your maximum

height. Draw a rough graph of your elevation z

against x.

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𝑧 = 16 − (𝑥2 + 𝑦2)

Now we will use the tools of algebra to do what we did above graphically. The line y = 4 – 2x is

drawn on the contour diagram. Your walk on the hill starts at (0, 4) and the 2-dimensional

“map” of your route is the diagram above.

Put the equation of the line and the equation of the surface together to get an equation for z in

terms of x along the line. Use this equation to draw a more accurate graph of z in terms of x and

identity precisely the point of maximum elevation.

Solution

The equation 𝑧 = 16 − (𝑥2 + 𝑦2) gives us z in terms of x and y.

But here we need an equation for z at each x. So what are we to

take for y? Well the path we are following gives us y in terms of x.

So we plug that equation

𝑦 = 4 − 2𝑥

into the equation

𝑧 = 16 − (𝑥2 + 𝑦2)

of the surface. We get:

𝑧 = 16 − (𝑥2 + (4 − 2𝑥)2)

Simplify:

𝑧 = 16 − (𝑥2 + 16 − 16𝑥 + 4𝑥2)

= 16𝑥 − 5𝑥2

= 𝑥(16 − 5𝑥)

This is a parabola with zeros at x = 0 and x = 16/5 = 3.2.

The max height is at its vertex which is halfway between the roots

at x = 8/5 = 1.6.

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Example 3. Another walk on the circular paraboloid

𝑧 = 16 − (𝑥2 + 𝑦2)

(a) This time you are going to start at the bottom of the hill at (4, 0, 0) and walk up the hill along

a semi-circular path to the top at (0, 0, 16). On the contour diagram above draw the top half of

the circle centred at (2, 0) of radius 2. This will be the route you will take.

Using the information contained in the contour diagram, draw a rough graph of your height z

against x. Pay attention to the spacing of the contour lines!

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𝑧 = 16 − (𝑥2 + 𝑦2)

(b) You start at the point (4, 0, 0) and walk up the hill following the semi-circle with centre (2,0)

and radius 2. Find an equation for your elevation z against x and draw its graph.

The circle has equation

(𝑥 − 2)2 + 𝑦2 = 4

The surface has equation

𝑧 = 16 − (𝑥2 + 𝑦2)

Solve the circle equation for y

𝑦2 = 4 − (𝑥 − 2)2

and put that into the equation of the surface:

𝑧 = 16 − (𝑥2 + 4 − (𝑥 − 2)2)

= 16 − (𝑥2 + 4 − (𝑥2 − 4𝑥 + 4))

= 16 − (𝑥2 + 4 − 𝑥2 + 4𝑥 − 4))

= 16 − 4𝑥

The graph is a straight line 𝑧 = 16 − 4𝑥. That’s a surprise!

3D graphs Manual

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𝑧 = 16 − (𝑥2 + 𝑦2)

𝑧 = 16 − 4𝑥

(c) When we walk up the hill following the semicircular path, the graph of z against x is a

straight line. That’s a real surprise. In fact it’s a bit hard to decide what it means in “real”

terms––graphically and physically.

Graphically: How can you interpret that straight line in terms of the geometry of the contour

diagram? Find a way to illustrate the straight-line behavior on the contour diagram

Physically: How can you interpret that straight line in terms of your walk up the hill? Straight

lines have a constant slope––in what sense is the slope of your path up the hill a constant?

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It’s not clear to me what the students will do with this. But here are some ideas.

Graphically: Interpret that straight line in terms of the geometry of the contour diagram.

Well what it says is that as x increases, z will increase at a constant rate. So equal changes in x

produce equal increases in elevation. In particular going from one contour line down to the next,

will always correspond to the same increase in x. Okay––on the interval 0 ≤ x ≤ 4, there are there

are 16 z-intervals of size 1 (0 ≤ z ≤ 16), so each unit change in z will correspond to an x-change

of 4/16 = 1/4.

We can illustrate that on the contour diagram. There is one vertical line at each contour and

these lines are spaced at equal x-intervals of 1/4.

𝑧 = 16 − 4𝑥

Physically: How can you interpret that straight line in terms of your walk on the hill? Straight

lines have a constant slope––in what sense is the slope of your path up the hill a constant?

Hmm. Slope is rate of change of height, but that word “rate” is tricky. The constant slope of the

above line is rate of change of elevation with respect to x. It might be easier to think in terms of

time. Suppose you took 4 minutes to make the walk and varied your speed around the semi-

circle so that you ascended 1 unit in every 15 seconds (¼ minute). Then your x-coordinate is

increasing at a constant rate 1 unit/min and your elevation is increasing at a constant rate 4

units/min. How would your speed vary?––you’d be going fast at the beginning and the end and

more slowly in the middle. [Nice question: find a formula for your speed as a function of time.]

3D graphs Manual

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peter.taylor@queensu.ca

x

y

rcos

rsinr

(d) Thinking of time I cannot resist one final question perhaps accessible only for grade 12.

Suppose you travel the semicircle at constant speed, again taking 4 minutes for the trip so that (in

terms of angle) you walk at 45º/minute. Find a formula for your height z against time t, and plot

its graph.

How do we go around the circle at constant speed? All

students have seen the angular parameterization of the

circle illustrated at the right

x = rcosθ

y = rsinθ

The point is that as θ changes at a constant rate, the point

moves around the circle at constant speed. This is a non-

trivial result and relies on the symmetry of the circle. For

example it is not true if the circle is replaced by an ellipse.

Okay––the semi-circle has radius r = 2, but its centre is not

the origin, rather it is (2, 0). So

x = 2 + 2cosθ

y = 2sinθ

Finally when θ = π we want t = 4, so 𝜃 =𝜋

4𝑡 and we have:

x = 2 + 2cos𝜋

4𝑡

y = 2sin𝜋

4𝑡

To get z in terms of t we put these into the equation of the

surface.

𝑧 = 16 − (𝑥2 + 𝑦2)

= 16 − [(2 + 2cos𝜋

4𝑡)

2

+ (2cos𝜋

4𝑡)

2

]

= 16 − (4 + 8cos𝜋

4𝑡 + 4cos2

𝜋

4𝑡 + 4sin2

𝜋

4𝑡)

= 16 − (4 + 8cos𝜋

4𝑡 + 4)

= 8 − 8cos𝜋

4𝑡

= 8 (1 − cos𝜋

4𝑡)

This tells us that we should turn the cos graph upside-

down, lift it 1 unit and then rescale both axes. Since you

are walking at constant horizontal speed, the slope of the z-t

graph will represent the slope of the hill in the direction of

your path. In particular we learn that the slope is

symmetric about t = 2 and attains its maximum at t = 2, the

halfway point.

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Example 4. (Homework?) A hyperbolic paraboloid

𝒛 = 𝒙𝟐 − 𝒚𝟐

(a) Factor the expression 𝑥2 − 𝑦2 and hence

identify and mark two lines on the contour diagram

at which z = 0.

𝑧 = 𝑥2 − 𝑦2 = (𝑥 − 𝑦)(𝑥 + 𝑦) = 0

Hence the zero contour is the pair of lines x = y

and x = –y. These are a pair of perpendicular

horizontal lines that are contained in the surface.

Can you “find them” in the diagram of the surface

above?

(b) The contour lines are spaced at intervals of z =

1. Mark their z-values on the contour diagram for

all even integers z.

(c) On the contour diagram, draw the path y = 2 – x.

Starting at (–2, 4) walk on the hill along that path.

Using the information from the contour diagram,

make a rough sketch on the pair of axes at the right

of the graph of your elevation z against x.

Which is the x-axis and which is

the y-axis? Take care with this.

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𝒛 = 𝒙𝟐 − 𝒚𝟐

The line y = 2 – x is drawn on the contour diagram above.

Your walk on the hill starts at (–2, 4) and the projection onto

the x-y plane of your path is the line y = 2 – x. Use algebra to

find an equation for your elevation z at any x and draw its

graph. This will provide a check for your graph on the

previous page.

The equation 𝑧 = 𝑥2 − 𝑦2 gives us z in terms of x and y.

But here we need an equation for z at each x.

Well the path we are following gives us y in terms of x.

So we plug that equation

𝑦 = 2 − 𝑥

into the equation

𝑧 = 𝑥2 − 𝑦2

of the surface. We get:

𝑧 = 𝑥2 − (2 − 𝑥)2

Simplify:

𝑧 = 𝑥2 − (4 − 4𝑥 + 𝑥2)

= 𝑥2 − 4 + 4𝑥 − 𝑥2)

= −4 + 4𝑥

We get a straight line 𝑧 = 4𝑥 − 4. We are walking on the surface but we are walking in a

straight line––like walking up a ladder! That’s a surprise––it gives us another straight line that

is contained in the surface. In fact it turns out that the surface is made up entirely of straight

lines! Read more by googling “ruled surfaces.”