- Difficulty
- Introductory
- Suggested time
- 3 min
- Marks
- 2
Functions of two variables and partial derivatives practice question 1
Let . Which statement gives and the domain of correctly?
Test each choice on two counts: the value at , and which points give the square root a number it can take.
Substitute and , squaring with care. For the domain, ask which points make .
Theory UBC Math 100 §14.2Key results
Worked solution
Strategy
Evaluate by substituting both coordinates. The domain is every point (x, y) where the formula makes sense: here, where the number under the square root is not negative.
Working
x = 2 and y = −1; the square of −1 is +1.
A square root needs an input that is not negative; zero is allowed.
The points within distance 3 of the origin: the disc of radius 3, its edge included.
Answer
The answer is f(2, −1) = 2, and the domain is the disc x² + y² ≤ 9: every point on or inside the circle of radius 3 about the origin..
Check
As 4 + 1 + 4 = 9, the point (2, −1, 2) is on the sphere x² + y² + z² = 9, and f(3, 0) = 0 is defined, so the circle is included.
Where this goes wrong
Leaving the circle out of the domain. The square root of 0 is 0, so the points with x² + y² = 9 are in; only the points outside the circle are excluded.
Leans on
Domains of functions, from Functions, graphs and inverses
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