Exercises for Quadric Surfaces

For exercises 1 - 6, sketch and describe the cylindrical surface of the given equation.

1) [T] ( x^2+z^2=1)


The surface is a cylinder with the rulings parallel to the y-axis.

2) [T] ( x^2+y^2=9)

3) [T] ( z=cos(frac{π}{2}+x))


The surface is a cylinder with rulings parallel to the y-axis.

4) [T] ( z=e^x)

5) [T] ( z=9−y^2)


The surface is a cylinder with rulings parallel to the x-axis.

6) [T] ( z=ln(x))

For exercises 7 - 10, the graph of a quadric surface is given.

a. Specify the name of the quadric surface.

b. Determine the axis of symmetry of the quadric surface.


a. Cylinder; b. The (x)-axis



a. Hyperboloid of two sheets; b. The (x)-axis


For exercises 11 - 16, match the given quadric surface with its corresponding equation in standard form.

a. ( frac{x^2}{4}+frac{y^2}{9}−frac{z^2}{12}=1)

b. ( frac{x^2}{4}−frac{y^2}{9}−frac{z^2}{12}=1)

c. ( frac{x^2}{4}+frac{y^2}{9}+frac{z^2}{12}=1)

d. ( z^2=4x^2+3y^2)

e. ( z=4x^2−y^2)

f. ( 4x^2+y^2−z^2=0)

11) Hyperboloid of two sheets


12) Ellipsoid

13) Elliptic paraboloid


14) Hyperbolic paraboloid

15) Hyperboloid of one sheet


16) Elliptic cone

For exercises 17 - 28, rewrite the given equation of the quadric surface in standard form. Identify the surface.

17) ( −x^2+36y^2+36z^2=9)

( −frac{x^2}{9}+frac{y^2}{frac{1}{4}}+frac{z^2}{frac{1}{4}}=1,) hyperboloid of one sheet with the (x)-axis as its axis of symmetry

18) ( −4x^2+25y^2+z^2=100)

19) ( −3x^2+5y^2−z^2=10)

( −frac{x^2}{frac{10}{3}}+frac{y^2}{2}−frac{z^2}{10}=1,) hyperboloid of two sheets with the (y)-axis as its axis of symmetry

20) ( 3x^2−y^2−6z^2=18)

21) ( 5y=x^2−z^2)

( y=−frac{z^2}{5}+frac{x^2}{5},) hyperbolic paraboloid with the (y)-axis as its axis of symmetry

22) ( 8x^2−5y^2−10z=0)

23) ( x^2+5y^2+3z^2−15=0)

( frac{x^2}{15}+frac{y^2}{3}+frac{z^2}{5}=1,) ellipsoid

24) ( 63x^2+7y^2+9z^2−63=0)

25) ( x^2+5y^2−8z^2=0)

( frac{x^2}{40}+frac{y^2}{8}−frac{z^2}{5}=0,) elliptic cone with the (z)-axis as its axis of symmetry

26) ( 5x^2−4y^2+20z^2=0)

27) ( 6x=3y^2+2z^2)

( x=frac{y^2}{2}+frac{z^2}{3},) elliptic paraboloid with the (x)-axis as its axis of symmetry

28) ( 49y=x^2+7z^2)

For exercises 29 - 34, find the trace of the given quadric surface in the specified plane of coordinates and sketch it.

29) [T] ( x^2+z^2+4y=0,z=0)


Parabola ( y=−frac{x^2}{4},)

30) [T] ( x^2+z^2+4y=0,quad x=0)

31) [T] ( −4x^2+25y^2+z^2=100,quad x=0)


Ellipse ( frac{y^2}{4}+frac{z^2}{100}=1,)

32) [T] ( −4x^2+25y^2+z^2=100,quad y=0)

33) [T] ( x^2+frac{y^2}{4}+frac{z^2}{100}=1,quad x=0)


Ellipse ( frac{y^2}{4}+frac{z^2}{100}=1,)

34) [T] ( x^2−y−z^2=1,quad y=0)

35) Use the graph of the given quadric surface to answer the questions.

a. Specify the name of the quadric surface.

b. Which of the equations—( 16x^2+9y^2+36z^2=3600,9x^2+36y^2+16z^2=3600,) or ( 36x^2+9y^2+16z^2=3600) —corresponds to the graph?

c. Use b. to write the equation of the quadric surface in standard form.

a. Ellipsoid
b. The third equation
c. ( frac{x^2}{100}+frac{y^2}{400}+frac{z^2}{225}=1)

36) Use the graph of the given quadric surface to answer the questions.

a. Which of the equations—( 36z=9x^2+y^2,9x^2+4y^2=36z), or ( −36z=−81x^2+4y^2) —corresponds to the graph above?

c. to write the equation of the quadric surface in standard form.

For exercises 37 - 42, the equation of a quadric surface is given.

a. Use the method of completing the square to write the equation in standard form.

b. Identify the surface.

37) ( x^2+2z^2+6x−8z+1=0)

a. (frac{(x+3)^2}{16}+frac{(z−2)^2}{8}=1)
b. Cylinder centered at ( (−3,2)) with rulings parallel to the (y)-axis

38) ( 4x^2−y^2+z^2−8x+2y+2z+3=0)

39) ( x^2+4y^2−4z^2−6x−16y−16z+5=0)

a. (frac{(x−3)^2}{4}+(y−2)^2−(z+2)^2=1)
b. Hyperboloid of one sheet centered at ( (3,2,−2),) with the (z)-axis as its axis of symmetry

40) ( x^2+z^2−4y+4=0)

41) ( x^2+frac{y^2}{4}−frac{z^2}{3}+6x+9=0)

a. ((x+3)^2+frac{y^2}{4}−frac{z^2}{3}=0)
b. Elliptic cone centered at ( (−3,0,0),) with the (z)-axis as its axis of symmetry

42) ( x^2−y^2+z^2−12z+2x+37=0)

43) Write the standard form of the equation of the ellipsoid centered at the origin that passes through points ( A(2,0,0),B(0,0,1),) and ( C(12,sqrt{11},frac{1}{2}).)

( frac{x^2}{4}+frac{y^2}{16}+z^2=1)

44) Write the standard form of the equation of the ellipsoid centered at point ( P(1,1,0)) that passes through points ( A(6,1,0),B(4,2,0)) and ( C(1,2,1)).

45) Determine the intersection points of elliptic cone ( x^2−y^2−z^2=0) with the line of symmetric equations ( frac{x−1}{2}=frac{y+1}{3}=z.)

( (1,−1,0)) and ( (frac{13}{3},4,frac{5}{3}))

46) Determine the intersection points of parabolic hyperboloid ( z=3x^2−2y^2) with the line of parametric equations ( x=3t,y=2t,z=19t), where ( t∈R.)

47) Find the equation of the quadric surface with points ( P(x,y,z)) that are equidistant from point ( Q(0,−1,0)) and plane of equation ( y=1.) Identify the surface.

( x^2+z^2+4y=0,) elliptic paraboloid

48) Find the equation of the quadric surface with points ( P(x,y,z)) that are equidistant from point ( Q(0,2,0)) and plane of equation ( y=−2.) Identify the surface.

49) If the surface of a parabolic reflector is described by equation ( 400z=x^2+y^2,) find the focal point of the reflector.

( (0,0,100))

50) Consider the parabolic reflector described by equation ( z=20x^2+20y^2.) Find its focal point.

51) Show that quadric surface ( x^2+y^2+z^2+2xy+2xz+2yz+x+y+z=0) reduces to two parallel planes.

52) Show that quadric surface ( x^2+y^2+z^2−2xy−2xz+2yz−1=0) reduces to two parallel planes passing.

53) [T] The intersection between cylinder ( (x−1)^2+y^2=1) and sphere ( x^2+y^2+z^2=4) is called a Viviani curve.

a. Solve the system consisting of the equations of the surfaces to find the equation of the intersection curve. (Hint: Find ( x) and ( y) in terms of ( z).)

b. Use a computer algebra system (CAS) or CalcPlot3D to visualize the intersection curve on sphere ( x^2+y^2+z^2=4).


a. (x=2−frac{z^2}{2},y=±frac{z}{2}sqrt{4−z^2},) where ( z∈[−2,2];)


54) Hyperboloid of one sheet ( 25x^2+25y^2−z^2=25) and elliptic cone ( −25x^2+75y^2+z^2=0) are represented in the following figure along with their intersection curves. Identify the intersection curves and find their equations (Hint: Find y from the system consisting of the equations of the surfaces.)

55) [T] Use a CAS or CalcPlot3D to create the intersection between cylinder ( 9x^2+4y^2=18) and ellipsoid ( 36x^2+16y^2+9z^2=144), and find the equations of the intersection curves.


two ellipses of equations ( frac{x^2}{2}+frac{y^2}{frac{9}{2}}=1) in planes ( z=±2sqrt{2})

56) [T] A spheroid is an ellipsoid with two equal semiaxes. For instance, the equation of a spheroid with the z-axis as its axis of symmetry is given by ( frac{x^2}{a^2}+frac{y^2}{a^2}+frac{z^2}{c^2}=1), where ( a) and ( c) are positive real numbers. The spheroid is called oblate if ( ca).

a. The eye cornea is approximated as a prolate spheroid with an axis that is the eye, where ( a=8.7mm) and ( c=9.6mm).Write the equation of the spheroid that models the cornea and sketch the surface.

b. Give two examples of objects with prolate spheroid shapes.

57) [T] In cartography, Earth is approximated by an oblate spheroid rather than a sphere. The radii at the equator and poles are approximately ( 3963)mi and ( 3950)mi, respectively.

a. Write the equation in standard form of the ellipsoid that represents the shape of Earth. Assume the center of Earth is at the origin and that the trace formed by plane ( z=0) corresponds to the equator.

b. Sketch the graph.

c. Find the equation of the intersection curve of the surface with plane ( z=1000) that is parallel to the xy-plane. The intersection curve is called a parallel.

d. Find the equation of the intersection curve of the surface with plane ( x+y=0) that passes through the z-axis. The intersection curve is called a meridian.


a. (frac{x^2}{3963^2}+frac{y^2}{3963^2}+frac{z^2}{3950^2}=1)


c. The intersection curve is the ellipse of equation ( frac{x^2}{3963^2}+frac{y^2}{3963^2}=frac{(2950)(4950)}{3950^2}), and the intersection is an ellipse.
d. The intersection curve is the ellipse of equation ( frac{2y^2}{3963^2}+frac{z^2}{3950^2}=1.)

58) [T] A set of buzzing stunt magnets (or “rattlesnake eggs”) includes two sparkling, polished, superstrong spheroid-shaped magnets well-known for children’s entertainment. Each magnet is ( 1.625) in. long and ( 0.5) in. wide at the middle. While tossing them into the air, they create a buzzing sound as they attract each other.

a. Write the equation of the prolate spheroid centered at the origin that describes the shape of one of the magnets.

b. Write the equations of the prolate spheroids that model the shape of the buzzing stunt magnets. Use a CAS or CalcPlot3D to create the graphs.

59) [T] A heart-shaped surface is given by equation ( (x^2+frac{9}{4}y^2+z^2−1)^3−x^2z^3−frac{9}{80}y^2z^3=0.)

a. Use a CAS or CalcPlot3D to graph the surface that models this shape.

b. Determine and sketch the trace of the heart-shaped surface on the xz-plane.



b. The intersection curve is ( (x^2+z^2−1)^3−x^2z^3=0.)

60) [T] The ring torus symmetric about the z-axis is a special type of surface in topology and its equation is given by ( (x^2+y^2+z^2+R^2−r^2)^2=4R^2(x^2+y^2)), where ( R>r>0). The numbers ( R) and ( r) are called are the major and minor radii, respectively, of the surface. The following figure shows a ring torus for which ( R=2) and ( r=1).

a. Write the equation of the ring torus with ( R=2) and ( r=1), and use a CAS or CalcPlot3D to graph the surface. Compare the graph with the figure given.

b. Determine the equation and sketch the trace of the ring torus from a. on the xy-plane.

c. Give two examples of objects with ring torus shapes.


Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at

Exercises and LaTeX edited by Paul Seeburger

NCERT Solutions for Class 10 Math’s – Quadratic Equations

NCERT Solutions For Class 10 Maths Chapter 4: The first step towards board exam preparation is solving NCERT textbook exercise questions. Thus, students searching for NCERT Solutions for Class 10 Maths Chapter 4 (Quadratic Equations) exercise questions can go through this article. The NCERT Solutions for Class 10 Maths Chapter 4 involves Exercise 4.1, Exercise 4.2, Exercise 4.3, Exercise 4.4 and Exercise 4.5. We have provided you with NCERT Class 10 Chapter 4 solutions in PDF so that you can refer to NCERT solutions offline as well. Moreover, Students can also download the NCERT solutions for Class 10 Maths chapter 4 in PDF format. Read on to get NCERT Class 10 Maths exercise solutions for Chapter 4 – Quadratic Equations.

List of Quadratic Equation Worksheets

Walk your students through this assortment of pdf worksheets! Acquaint them with finding the sum and product of the roots of a given quadratic equation. Equip them to utilize this sum and product to form the quadratic equation and determine the missing coefficients or constant in it.

This bunch of pdf exercises for high school students has some prolific practice in solving quadratic equations by factoring. Factor and solve for the real or complex roots of quadratic equations with integer, fractional, and radical coefficients.

Keep high school students au fait with the application of square root property in solving pure quadratic equations, with this assemblage of printable worksheets. Isolate the x 2 term on one side of the equation and the constant term on the other side, and solve for x by taking square roots.

Complete the square of the given quadratic equation and solve for the roots. Level up by working with equations involving radical, fractional, integer, and decimal coefficients.

Discern all the essential facts about a discriminant with this compilation of high school worksheets. Determine the discriminant by evaluating the expression b 2 - 4ac where a is the coefficient of x 2 , b the coefficient of x, and c the constant term in a quadratic equation.

Can you tell if the roots of a quadratic equation are equal or unequal without solving it? Take a quick jaunt into this collection of printable nature of roots handouts! Predict if the roots are equal or unequal and also if they are real or complex.

Be it finding the average or area or figuring out the slope or any other math calculation, formulas are important beyond doubt! Augment your ability to use the quadratic formula and find solutions to a quadratic equation with this set of practice resources!

Catch a glimpse of a variety of real-life instances where quadratic equations prove they have a significant role to play! Read each word problem carefully, form the equation with the given data, and solve for the unknown.

Exercises for Quadric Surfaces

1 Cone with sections

4 Hyperbolic paraboloid

5 Elliptic hyperboloid 1

6 Elliptic hyperboloid 2

7 Elliptic cylinder

8 Circular cylinder

9 Parabolic cylinder

10 Hyperbolic cylinder

11 Circular cone

12 Circular hyperboloid

13 Elliptic paraboloid

14 Ellipsoid with sections

A quadratic (or quadric) surface is a surface in three-space defined by an equation of degree two. Thus it is the three-dimensional analog of a conic section, which is a curve in two-space defined by an equation of degree two. The plane cross-sections of a quadratic surface must be conic sections, and the names for quadratic surfaces refer to the different types of plane cross-sections it possesses. The models in this group show the various types.

Surfaces of revolution are obtained by revolving a conic section about one of its axes. The surface of revolution obtained from an ellipse is called an ellipsoid, and that obtained from a parabola is called a paraboloid. A hyperbola gives rise to different surfaces of revolution, depending on whether it is revolved about the conjugate axis (which passes between the two branches of the hyperbola) or the transverse axis (which crosses the two branches). The first gives a hyperboloid of two sheets, the second a hyperboloid of one sheet. Surfaces of revolution have circular cross-sections perpendicular to the axis of revolution. More general surfaces have elliptic or hyperbolic cross-sections: thus one obtains elliptic and hyperbolic paraboloids, and elliptic hyperboloids of one or two sheets.

Degenerate quadratic surfaces occur when all the plane cross-sections through a given straight line are degenerate conics: either a pair of parallel straight lines, giving rise to a cylinder, or a pair of intersecting straight lines, giving rise to a cone.

Level Surfaces and Quadratic Surfaces

Interact on desktop, mobile and cloud with the free Wolfram Player or other Wolfram Language products.

For a function of three variables, , , and , the level surface of level is defined as the set of points in that are solutions of . A quadratic surface or quadric is a surface that is given by a second-order polynomial equation in the three variables , , and .

Let , , and be nonzero constants. We plot level surfaces for quadratic functions in three variables, which give some well-known quadratic surfaces:

&Diamond gives ellipsoids when , this is a sphere centered at the origin of radius .

&Diamond or give elliptical cylinders with symmetry axes along the axis and axis, corresponding to and .

&Diamond gives elliptic paraboloids, opening up or down as or .

&Diamond and , with , give elliptic cones. For , the level surfaces are hyperboloids of one sheet.

&Diamond ( ) and ( ) give hyperboloids of two sheets.

Contributed by: Ana Moura Santos and João Pedro Pargana (Instituto Superior Técnico) (March 2011)
Open content licensed under CC BY-NC-SA

Exercises to Help Sciatica

Most exercises for sciatica are for the lower back. Check with your doctor before you try these exercises that you can do at home:

Knee-to-Chest Exercise

This simple stretch targets the lower buttock and upper thigh area.

  • Step 1: Lie on your back with your legs bent and your feet flat on the floor.
  • Step 2: Bring one knee to the chest while keeping the other foot on the floor.
  • Step 3: Keeping the lower back pressed to the floor, hold for up to 30 seconds.
  • Step 4: Repeat on the other side.

Try for 2 to 4 repetitions on each side. To make the exercise a little harder, keep one leg straight on the floor while raising the other to the chest. You can also bring both knees to the chest.

Standing Hamstring Stretch

Use care when doing this exercise. Hold on to something if necessary, and don't overstretch.

  • Step 1: Stand straight up and put one foot on a slightly higher surface, like a stair step.
  • Step 2: Straighten the leg on the step and point the toes up.
  • Step 3: Lean slightly forward while keeping the back straight.
  • Step 4: Hold for 20 to 30 seconds. Remember to breathe.
  • Step 5: Repeat with the other leg.

Try for 2 to 3 repetitions with each leg.

Pelvic Tilt Exercise

This is another deceptively simple exercise that is good for sciatica.

  • Step 1: Lie on your back with your legs bent and arms by your side.
  • Step 2: Tighten your stomach muscles, press your back into the floor, and rock the hips and pelvis slightly upward.
  • Step 3: Hold this position while imagining making your belly button touch your backbone. Don't forget to breathe.
  • Step 4: Release after a few seconds. Then repeat.

Try for 8 to 12 repetitions.

Glute Bridges

The glutes are a group of muscles in the buttocks. If they are tight, they can press on the sciatic nerve.

  • Step 1: Lie on your back on the floor with knees bent. Feet should be about shoulder-width apart. Relax your arms at your sides.
  • Step 2: Pushing through the heels, lift your hips until your body forms a straight line from knees to shoulders.
  • Step 3: Hold the position for a few seconds.
  • Step 4: Slowly lower the hips to the floor. Then repeat.

Good form is important for this exercise. Avoid arching or rounding the back. Try for 2 or 3 sets of 8 to 10 repetitions.

Lying Deep Gluteal Stretch

If you lack flexibility, you may need to modify this exercise slightly.

  • Step 1: Lie on your back with legs bent. Raise your right ankle, and rest it on your left knee.
  • Step 2: Using both hands, lace your fingers behind your left thigh and gently pull it toward you, keeping your head and back on the floor.
  • Step 3: Hold for 20 to 30 seconds.
  • Step 4: Repeat with the other leg.

You may need to elevate your head slightly with a book or firm cushion under it. If you can't reach your thigh easily, you can loop a towel around the thigh and use it to pull your thigh toward you. Do 2 to 3 repetitions with each leg.


  • Bright: Picard groups, definition and simple examples (P^2, quadric surface in P^3) notes (pdf)
  • van Luijk: Canonical divisor on hypersurface of degree d in P^n and on complete intersections very ample divisors notes (pdf)
  • Testa: Classification and the minimal model program the Hodge diamond notes (pdf)
  • Exercises 1 (pdf)


  • Bright: Riemann-Roch on surfaces, genus and effectivity of (-1)-curves, adjunction formula
  • van Luijk: Picard groups of del Pezzo surfaces notes (pdf)
  • Testa: Root Lattices and their automorphism groups Segre-Manin for del Pezzo surfaces I notes (pdf)
  • Exercises 2 (pdf)


  • Testa: Segre-Manin for del Pezzo surfaces II
  • van Luijk: Growth of rational points
  • Bright: Explicitly computed examples of Brauer-Manin obstruction through quadratic reciprocity notes (pdf)


  • Testa: Central simple algebras and Brauer groups notes (pdf)
  • Bright: Reformulate easy examples the enlightening way notes (pdf)
  • van Luijk: Galois cohomology I
  • Exercises 3 (pdf)


  • van Luijk Galois Cohomology II
  • Bright and Testa: How to find the algebra from Galois cohomology? Magma Program

Prerequisites. Some familiarity with basic algebraic and arithmetic geometry is assumed, at the level of the first two chapters of Silverman's book "Arithmetic of Elliptic Curves".

Preliminary Reading. Participants may find it helpful to work through the first few chapters of "Chapters on Algebraic Surfaces" by Miles Reid (available online). Preliminary reading on group cohomology will also be useful, a good reference is chapter 2 of Milne's notes on Class Field Theory.

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Particpants from universities belonging to the GTEM network might be able to obtain travel and accommodation expenses from GTEM please contact your local scientist-in-charge.

Exercises for Quadric Surfaces

Optimizing scale. The default mode of gradient descent is to choose the scale that minimizes the energy in that direction of motion. This involves a line search along the line of motion for the minimum energy. Evolver does the search by evaluating the energy for several scales, raising or lowering the scale until it has three scales for which the middle energy is lower than for the outer two. Then it uses quadratic intepolation to get the final scale. The steps of one iteration are listed in detail here.

Scale limit. It is possible for the aforementioned line search to go too far in certain circumstances, so there is an upper bound set for the scale, the default limit being 1.0. This can be changed by using the "scale_limit value" phrase in the top of the datafile, or by setting the scale_limit variable at run time, or in response to the prompt produced by the "m" command.

Reasonable scales. Trouble in evolving is usually signaled by a small scale, which means there is some obstacle to evolution. Of course, that means you have to know what a reasonable scale is, and that depends on the type of energy you are using and how refined your surface si. In normal evolution, the size of the scale is set by the development of small-scale roughness in the surface. Combined with a little dimensional analysis, that leads to the conclusion that the scale should vary as L 2-q , where L is the typical edge length and the units of energy are length q . The dimensional analysis goes like this: Let D be the perturbation of one vertex away from an equilibrium surface. In general, energy is quadratic around an equibrium, so

So scale is on the order of L 2-q . Some examples: Dimensional Dependence of Scale
EnergyEnergy dimension ScaleExample file
Area of soapfilm L 2 L 0 quad.fe
Length of string L 1 L 1 flower.fe
Squared curvature of string L -1 L 3 elastic8.fe
Squared mean curvature of soapfilm L 0 L 2 sqcube.fe
In particular, the scale for area evolution is independent of refinement, but for most other energies the scale decreases with refinement.

Another common influence on the scale for area evolution is the surface tension. Doing a liquid solder simulation in a system of units where the surface tension of facets is assigned a value 470, say, means that all calculated gradients are multiplied by 470, so the scale decreases by a factor of 470 to get the same geometric motion. Thus you should set scale_limit to be the inverse of the surface tension.

Fixed scale. It can be useful to iterate with a fixed scale. For example, if you make a change to the volume of a body and want that adjustment to take effect without the complication of simultaneously trying to minimize energy, then iterate once with a zero scale. For example, if you run cube.fe and do you will see pure volume adjustment. The 'm' command toggles back and forth between optimizing and fixed scale, except when you follow it by a number it sets a fixed scale.

Another use for fixed scale is simulating grain growth. Here the motion of grain boundaries is defined to have a velocity proportional to their mean curvature, so you want to keep the scale small enough so the linear approximation to the motion is reasonably good.

Convergence. It is impossible to tell, in general, when gradient descent is close to convergence. Surfaces can be arbitrarily far from the minimum, and move toward it arbitrarily slowly. As an example, run the capillary3.fe datafile. This is a slightly pinched tube with a soapfilm across it. The minimum energy comes when the film is exactly at the middle of the neck. Try this evolution: Not at the neck yet. Try more g steps. See how many it takes you to converge to the minimum energy of 3.13262861328124 to, say, 8 decimal places.

Another problem can be saddle points. If you start with a symmetric surface, then under 'g' iteration the surface should stay symmetric. If the surface has a symmetric saddle point, then the iteration could approach it and look like it was converging to a minimum.

Conjugate gradient

In practice, the conjugate gradient method remembers a cumulative "history vector", which it combines with the ordinary gradient to figure out the conjugate gradient direction. The upshot is that conjugate gradient can converge much faster than ordinary gradient descent. In the ideal case of a quadratic energy function in N variables and perfect numerical accuracy, conjugate gradient will reach the exact minimum within N steps.

Conjugate gradient can be toggled with the U command, or with the conj_grad toggle. It should always be used with optimizing scale.

To see the dramatic improvement conjugate gradient can produce, run capillary3.fe again, with this evolution: Conjugate gradient can get in trouble, particularly when you use it too early when the surface is making large adjustments. To see an example, run capillary3.fe with this evolution: and take a close look at the film in the center.

In sum, one should run a few steps of ordinary gradient descent at the start of evolving a surface or after making big changes, but the bulk of iteration should be done in conjugate gradient mode.

Notes: The conjugate gradient method is designed for quadratic energy functions. As long as the energy function is nearly quadratic, as it should be near an energy minimum, conjugate gradient works well. Otherwise, it may misbehave, either by taking too big steps or by getting stalled. Both effects are due to the history vector being misleading. To prevent too big steps, one should iterate without conjugate gradient for a few steps whenever significant changes are made to the surface (refining, changing a constraint, etc.). On the other hand, if it looks like conjugate gradient is converging, it may have simply become confused by its own history. See the catenoid example for a case in point. A danger signal is the scale factor going to zero. If you are suspicious, toggle conjugate gradient off and on ("U 2" does nicely) to erase the history vector and start over.

Exercises for Quadric Surfaces

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Franklin Bradley, AK

The first thing I'll do here is multiply through on the left-hand side, and then I'll move the 4 over from the right-hand side to the left-hand side:

Since there are no factors of (1)(&ndash4) = &ndash4 that add up to &ndash2 , then this quadratic does not factor. (In other words, there is no possible way that the faux-factoring solution of " x = 4, x &ndash 2 = 4 " could be even slightly correct.)

So factoring won't work, but I can use the Quadratic Formula in this case, I'll be plugging in the values a = 1, b = &ndash2, and c = &ndash4 :

x = &ndash1.24, x = 3.24 , rounded to two decimal places.

For reference, here's what the graph of the associated quadratic, y = x 2 &ndash 2x &ndash 4 , looks like:

As you can see, the solutions from the Quadratic Formula match up with the x -intercepts. The locations where the graph crosses the x -axis give the values that solve the original equation.

There is another connection between the solutions from the Quadratic Formula and the graph of the parabola: you can tell how many x -intercepts you're going to have from the value inside the square root. The argument (that is, the contents) of the square root, being the expression b 2 &ndash 4ac , is called the "discriminant" because, by using its value, you can "discriminate" between (that is, be able to tell the difference between) the various solution types.

In this case, the value of the discriminant b 2 &ndash 4ac was 20 in particular, the value was not zero and was not negative. Because the value was not negative, the equation was going to have at least one (real-valued) solution because the value was not zero, the two solutions were going to be distinct (that is, they were going to be different from each other).

Solve 9x 2 + 12x + 4 = 0 . Leave your answer in exact form.

Using a = 9, b = 12, and c = 4 , the Quadratic Formula gives me:

In the first example on this page, I had gotten two solutions because the value of the discriminant (that is, the value inside the square root) was non-zero and positive. As a result, the "plus-minus" part of the Formula gave me two distinct values one for the "plus" part of the numerator and another for the "minus" part. In this case, though, the square root reduced to zero, so the plus-minus didn't count for anything.

This sort of solution, where you get only one value because "plus or minus zero" didn't change anything, is called a "repeated" root, because x is equal to , but it's equal to this value kind-of twice: and .

You can see this repetition better if you factor the quadratic (and, because the solutions were nice neat fractions, the quadratic must factor): 9x 2 + 12x + 4 = (3x + 2)(3x + 2) = 0 , so the first factor gives us 3x + 2 = 0 so , and (from the second, identical factor) 3x + 2 = 0 so again.

Any time you end up with zero inside the square root of the Quadratic Formula, you'll only get one solution to the equation, in the sense of getting one number that solves the equation. But you'll get two solutions, in the sense of the one value being counted twice. In other words, a discriminant (that is, the expression b 2 &ndash 4ac ) with a value of zero means that you'll get one "repeated" solution value.

Below is what the graph of the associated function, y = 9x 2 + 12x + 4 , looks like:

The parabola only just touches the x -axis at it doesn't actually cross. This relationship is always true: if you have a root that appears exactly twice (or, which is the same thing, if you get zero inside the square root), then the graph will "kiss" the axis at the solution value, but it will not pass through the axis.

Solve 3x 2 + 4x + 2 = 0

Since there are no factors of (3)(2) = 6 that add up to 4 , this quadratic does not factor. But the Quadratic Formula always works in this case, I'll be plugging in the values a = 3, b = 4, and c = 2 :

At this point, I have a negative number inside the square root. If you haven't learned about complex numbers yet, then you would have to stop here, and the answer would be "no solution" if you do know about complex numbers, then you can continue the calculations:

Thus, depending upon your level of study, your answer will be one of the following:

real-number solutions: no solution

But whether or not you know about complexes, you know that you cannot graph your answer, because you cannot graph the square root of a negative number on the regular Cartesian place. There are no such values on the x -axis. Since you can't find a graphable solution to the quadratic, then reasonably there should not be any x -intercepts (because you can graph an x -intercept).

Here's the graph of the associated function, y = 3x 2 + 4x + 2 :

As you can see, the graph does not cross, or even touch, the x -axis. This relationship is always true: If you get a negative value inside the square root, then there will be no real number solution, and therefore no x -intercepts. In other words, if the the discriminant (being the expression b 2 &ndash 4ac ) has a value which is negative, then you won't have any graphable zeroes.

(The relationship between the discriminant (being the value inside the square root), the type of solutions (two distinct solutions, one repeated solution, or no graphable solutions), and the number of x -intercepts on the graph (two, one, or none) is summarized in a chart on the next page.)

Watch the video: Quadratic Equation, unit 1f, CA foundation,Technical Maths by Pradeep Soni (December 2021).