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MATH1111 Quizzes

Graphs of Functions of Two Variables Quiz
Web resources available Questions

This quiz tests the work covered in the lecture on graphs of functions of two variables and corresponds to Section 12.2 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).

There is not as much information on the web on functions of two variables as there is on functions of one variable. There are a few interesting sites. You can start at http://www.ucl.ac.uk/Mathematics/geomath/level2/pdiff/pd5.html. It refers to an earlier example which you can find at http://www.ucl.ac.uk/Mathematics/geomath/level2/pdiff/pd21.html.

There is a nice applet at http://www-math.mit.edu/18.02/applets/FunctionsTwoVariables.html which allows you to type in a function or select one from the drop down list and it plots the graph.

There are more web quizzes at Wiley, select Section 2. It now only has 5 questions instead of the usual 15.

Which of the following statements are correct? (Zero or more options can be correct)
a)
(1,2,5) is on the graph f(x,y) = x2 + y2 2xy 3.
b)
(2,2,13) is on the graph f(x,y) = x2 + y2 2xy 3.
c)
(1,π,0) is on the graph f(x,y) = xsiny.
d)
(0,0,1) is on the graph f(x,y) = e(x2+y2) .
e)
(1,5,7) is on the graph f(x,y) = 2x.

There is at least one mistake.
For example, choice (a) should be False.
f(1,2) = (1)2 + 22 2 ×1 × 2 3 = 1 + 4 + 4 3 = 6 so the point (1,2,6) is on the graph and (1,2,5) is not on the graph.
There is at least one mistake.
For example, choice (b) should be True.
f(2,2) = 22 + (2)2 2 × 2 ×2 3 = 4 + 4 + 8 3 = 13 so (2,2,13) is on the graph.
There is at least one mistake.
For example, choice (c) should be True.
f(1,π) = 1 × sinπ = 1 × 0 = 0 so (1,π,0) is on the graph.
There is at least one mistake.
For example, choice (d) should be True.
f(0,0) = e0 = 1 so (0,0,1) is on the graph.
There is at least one mistake.
For example, choice (e) should be False.
f(1,5) = 2 × 1 = 2 so so the point (1,5,2) is on the graph and (1,5,7) is not on the graph.
Correct!
  1. False f(1,2) = (1)2 + 22 2 ×1 × 2 3 = 1 + 4 + 4 3 = 6 so the point (1,2,6) is on the graph and (1,2,5) is not on the graph.
  2. True f(2,2) = 22 + (2)2 2 × 2 ×2 3 = 4 + 4 + 8 3 = 13 so (2,2,13) is on the graph.
  3. True f(1,π) = 1 × sinπ = 1 × 0 = 0 so (1,π,0) is on the graph.
  4. True f(0,0) = e0 = 1 so (0,0,1) is on the graph.
  5. False f(1,5) = 2 × 1 = 2 so so the point (1,5,2) is on the graph and (1,5,7) is not on the graph.
Which of the following is the shape of the cross-section of z = f(x,y) = 2x2 y2 4 when y = 4? Exactly one option must be correct)
a)
A line.
b)
A parabola.
c)
A circle.
d)
None of the above.

Choice (a) is incorrect
Try again, substitute y = 4 into z = f(x,y) = 2x2 y2 4 and see what sort of function you get.
Choice (b) is correct!
z = 2x2 16 4 = 2x2 20 when y = 4 so the shape is a parabola.
Choice (c) is incorrect
Try again, substitute y = 4 into z = f(x,y) = 2x2 y2 4 and see what sort of function you get.
Choice (d) is incorrect
Try again, substitute y = 4 into z = f(x,y) = 2x2 y2 4 and see what sort of function you get, it is one of the above.
Which of the following is the shape of the cross-section of z = f(x,y) = 2x + 3y3 5 when x = 2? Exactly one option must be correct)
a)
A line.
b)
A parabola.
c)
A circle.
d)
None of the above.

Choice (a) is incorrect
Try again, substitute x = 2 into z = f(x,y) = 2x + 3y3 5 and see what sort of function you get.
Choice (b) is incorrect
Try again, substitute x = 2 into z = f(x,y) = 2x + 3y3 5 and see what sort of function you get.
Choice (c) is incorrect
Try again, substitute x = 2 into z = f(x,y) = 2x + 3y3 5 and see what sort of function you get.
Choice (d) is correct!
z = 4 + 3y3 5 = 3y3 1 so the shape is a cubic.
The function z = f(x,y) = x2 y2 is a saddle-shaped surface with its ‘saddle point’ at the the origin.
Which one of the following describes the function z = g(x,y) = (x 1)2 (y + 2)2 + 4 most accurately? Exactly one option must be correct)
a)
g is a saddle-shaped surface with its ‘saddle point’ at (1,-2,4).
b)
g is a saddle-shaped surface with its ‘saddle point’ at (1,-2,-4).
c)
g is a saddle-shaped surface with its ‘saddle point’ at (-1,2,4).
d)
None of the above.

Choice (a) is correct!
g(x,y) = f(x 1,y + 2) + 4 so we have shifted the saddle shape 1 unit in the x-direction, -2 units in the y-direction and up 4 units in the z-direction.
Choice (b) is incorrect
Try again, you have not moved the ‘saddle point’ correctly.
Choice (c) is incorrect
Try again, you have not moved the ‘saddle point’ correctly.
Choice (d) is incorrect
Try again, we have shifted the saddle-shaped surface.