The contours of the function [math: f(x,y)=e−x2−y2] consist of

Question

BlackTom AI · Accepted Answer

Level sets satisfy e^{-x^2-y^2} = c, so take logarithms to get -x^2 - y^2 = ln c.

1. Relevant knowledge points and concepts
- Contour (level) curves of a function f(x,y) are the sets where f(x,y) = constant.
- The exponential function is always positive, so allowable constants c must satisfy c > 0.
- Equations of the form x^2 + y^2 = R^2 represent circles centered at the origin with radius R.

2. Step-by-step reasoning
- Start with f(x,y) = e^{-x^2 - y^2}. Set f(x,y) = c (a contour at height c).
- e^{-x^2 - y^2} = c.

3. Intermediate calculation steps
- Taking natural log of both sides (valid since c > 0):
  ln(e^{-x^2-y^2}) = ln c
  -x^2 - y^2 = ln c
- Multiply by -1:
  x^2 + y^2 = - ln c
- Define R^2 = - ln c. Since c ∈ (0, ∞), ln c can be negative, zero, or positive; however for e^{-x^2-y^2} the maximum value is 1 (at x = y = 0) and values lie in (0,1], so c ∈ (0,1] and thus -ln c ≥ 0. Therefore R = sqrt(-ln c) is real and nonnegative.

4. Why the correct answer is correct
- The equation x^2 + y^2 = R^2 describes a circle centered at the origin. Different values of c produce different radii R, all sharing the same center. Hence the contour curves are concentric circles around the origin (with R = 0 when c = 1, and R → ∞ as c → 0+).

5. Why the other options are incorrect
- b. Parallel lines with gradient 1: level sets are not linear; dependence is on x^2 + y^2, not a linear combination x - y.
- c. Exponential curves: although f is an exponential function, the level-set equation reduces to a quadratic x^2 + y^2 = constant, i.e., circles, not exponential-shaped graphs.
- d. Lines parallel to the x-axis: these would be of form y = constant, which is not produced by x^2 + y^2 = constant except in trivial intersections.

6. Conclusion
- The contours of f(x,y) = e^{-x^2 - y^2} are concentric circles. This matches the provided correct answer: a. concentric circles.

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