George Wong
28 Feb 2012
Calc 3
Sections 11.1 and 11.2
Problem 1 Describe the level surfaces of the function.
f (x, y, z) = x2 + 2y 2 + 3z 2
The equation fits the form ax2 + by 2 + cz 2 = k, which indicates that it is in the family of ellipsoids.
Problem 2 Describe the level surfaces of the function.
f (x, y, z) = x2 y 2
The surface family given by x2 − y 2 are cylindrical (as there is no specific z parameter(s)) and are
noticably hyperbolic as can be seen by the ax2 − by 2 , a, b > 0.
Problem 3 Graph the functions below. (Do this on paper. Your instructor may ask you to turn
in this work.)
Each function was graphed by hand and is below shown. In addition, the plot generated by Wolfram
Alpha is attached.
p
f (x, y) = x2 + y 2
√
f (x, y) = e
x2 +y 2
f (x, y) = ln(
p
x2 + y 2 )
1
f (x, y) = sin(
f (x, y) = √
p
x2 + y 2 )
1
x2 +y 2
In general, if g is a function of one variable, how is the graph of f defined below obtained from the
graph of g? f (x, y) = g(sqrt(x2 + y 2 ))
In such a case, the graph z = g(x) is simply rotated about the z-axis.
Problem 4
(
f (x, y) =
Consider the function
(x+y)r
x2 +y 2
0
(x, y) 6= (0, 0)
(x, y) = (0, 0)
where r is a constant. For what values of r is f continuous? In order for the function f to have
a limit as (x, y) → (0, 0), the value of the function must approach zero as x and y both approach
zero. We note that when r = 2, the numerator takes the form of x2 + 2xy + y 2 , which ≈ x2 + y 2
when x, y → 0. This leads to an overall value of 1 for the fraction. Anything greater than that has
some number ≈ 0 multiplied by the top, which is at the same time canceling out with the bottom
giving 1 × 0 = 0. We can therefore conclude that f is continuous ∀ r > 2.
We write this as 2 < r < ∞
2