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Surfaces Quiz

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This quiz tests the work covered in the lecture on Surfaces and corresponds to Section 12.5 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).
There are more web quizzes at Wiley, select Sections 4 and 5. This quiz has 14 questions on both this topic and the last.


 

Question 1

 
 
Use the catalogue on page 636 of Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.) to identify which of the statements below are correct.
a) x2 2 4z = 4 + y is a hyperbolic paraboloid.
b) 2 2 2 z - x - y = 0 is a cone.
c) 2x2 + 3y2 - z2 = 3 is not in the catalogue.
d) 4x2 +y2 + z2= - 4 4 is not in the catalogue.

 

There is at least one mistake.
For example, choice (a) should be false.
Making z the subject makes x2 y2 x = 16 + 4 which is an elliptical paraboloid.
There is at least one mistake.
For example, choice (b) should be true.
Rearranging into the form x2 + y2 - z2 = 0 is a cone with a = b = c = 1.
There is at least one mistake.
For example, choice (c) should be false.
This can be rearranged to --x2---+ y2 - -z2--= 1 (∘ -3)2 (√3)2 2 which is a hyperboloid of one sheet.
There is at least one mistake.
For example, choice (d) should be true.
4x2 +y2 + z2> 0 4 for all real x , y, z so this cannot represent a surface and is not in the catalogue.
Your answers are correct
  1. False. Making z the subject makes x2 y2 x = 16 + 4 which is an elliptical paraboloid.
  2. True. Rearranging into the form x2 + y2 - z2 = 0 is a cone with a = b = c = 1.
  3. False. This can be rearranged to --x2---+ y2 - -z2--= 1 (∘ -3)2 (√3)2 2 which is a hyperboloid of one sheet.
  4. True. 4x2 +y2 + z2> 0 4 for all real x , y, z so this cannot represent a surface and is not in the catalogue.
 

Question 2

 
 
Consider the function ∘ ---------16y2- f(x,y) = 16- 4x2 ----- . 9
Which of the following is the corresponding level surface g(x,y,z) = f(x,y)- z = 0?
a) The hyperboloid x2 y2 z2 4- + 9-- 16 = 1.   b) The top half of the ellipsoid x2 y2 z2 -- + -- + -- = 1. 4 9 16
c) The top half of the two sheet hyperboloid x2 y2 z2 4- + 9-- 16 = - 1 .   d) The top half of the cone x2 y2 z2 -4 + 9-- 16 = 0 .

 

Not correct. Choice (a) is false.
Try again, this would correspond to the function ∘ ------16y2----- f(x,y) = 4x2 + -9--- 16.
Your answer is correct.
Rearranging ∘ ---------16y2- 16- 4x2 ----- - z = 0 9 gives ∘ ------------2- z = 16- 4x2 - 16y . 9
Squaring both sides gives 2 z2 = 16 - 4x2 - 16y-. 9 Divide both sides by 16 and rearrange and we have the required level surface.
Not correct. Choice (c) is false.
Try again, this would correspond to the function -------------- ∘ 16y2 f(x,y) = 4x2 + -9--+ 16.
Not correct. Choice (d) is false.
Try again, this would correspond to the function ∘ ------16y2 f(x,y) = 4x2 + ----. 9
 

Question 3

 
 
Consider the function f(x,y,z) = (x + 1)2 + (y - 2)2 +(z + 3)2.
Which of the following most accurately describes the set of level surfaces?
a) A family of cylinders with radii 1,2 and 3.
b) A family of ellipsoids centre (0,0,0).
c) A family of spheres centre (-1,2,-3).
d) A family of spheres centre (1,-2,3).

 

Not correct. Choice (a) is false.
Try again, check the catalogue of surfaces.
Not correct. Choice (b) is false.
Try again, it is a special type of ellipsoid but check the centre.
Your answer is correct.
f(x,y,z) = (x- a)2 + (y- b)2 + (z - c)2 gives a family of spheres centre (a,b,c).
Not correct. Choice (d) is false.
Try again, you do not have the correct centre.
 

Question 4

 
 
Which of the following are the two functions which together form the two sheet hyperboloid x2 y2 z2 4-+ 9-- -4 = - 1 ?
a) ∘ ----------- 4y2 f (x,y) = x2 + -9-+ 4 and ∘ ----4y2---- g(x,y) = - x2 +--- +4 . 9
b) ∘ --------4y2 f(x,y) = 4 - x2 --9- and ∘ -------4y2- g(x,y) = - 4- x2 --9- .
c) ∘ ----------- x2 y2 f(x,y) = 4 1 - 4-- -9 and ∘ ----------- x2 y2 g(x,y) = - 4 1- 4-- -9 .
d) ∘ ----------- x2 y2 f(x,y) = 4 1 + 4-+ -9 and ∘ ----------- x2 y2 g(x,y) = - 4 1+ 4-+ -9 .

 

Your answer is correct.
You have successfully made z the subject of the equation and taken both the positive and negative square root.
Not correct. Choice (b) is false.
Make z the subject of the equation and take both the positive and negative square root.
Not correct. Choice (c) is false.
Make z the subject of the equation and take both the positive and negative square root.
Not correct. Choice (d) is false.
Make z the subject of the equation and take both the positive and negative square root.