Find the domain, the range, and describe the level curves for the function???

Question

Find the domain, the range, and describe the level curves for the function

 

f(x,y) = 1 - e^(-4x² -9y²)

 

I figure out that the domain is all real numbers

The level curve I believe  is 4x²+9y² = c is a concentric circle at the origin.

 

But I am really confused on how the range is determine.......

 

........Out of curiosity, what will the range be if it was like the other way around like

f(x,y) = 1 + e^(-4x² -9y²)

Answer 1

Let's examine the range.  4x² and 9y² are both ≥ 0, so their sum is ≥ 0.  Thus, -(4x² + 9y²) is ≤ 0. 

 

For any real N such that  -∞ < N ≤ 0 we have 0 < e^N ≤ 1.  Thus,

 

0 < e^(-4x²-9y²) ≤ 1

0 > -e^(-4x²-9y²) ≥ -1

1 > 1 -e^(-4x²-9y²) ≥ 0.  That's your range.

 

To get the shape of level curves, we assign the function equal to a constant value K, or

K = 1 - e^(-4x²-9y²).  We can rearrange terms and then take natural logs to get

ln[e^(-4x²-9y²)] = ln(1-K), or

-4x²-9y² = ln(1-K). 

 

Notice that the range of K found previously, 1 > K ≥ 0, is within the domain of the ln function, except for K=0.  We'll handle this exception at the end.

 

Now we can start working out the shape of the level curves:

 

4x²+9y²=-ln(1-K)

(4/36)x²+(9/36)y²=-(1/36)ln(1-K)

x²/9 + y²/4 = -ln(1-K)/36.  For our range of K, ln(1-K) < 0, so -ln(1-K)/36 > 0.  This is important for the next step:

 

Let C = -ln(1-K)/36.  Dividing both sides gives us

 

x²/(9C) + y²/(4C) = 1, which shows that the level surfaces are ellipses.  K=0 occurs only for x=y=0, and we refer back to the original equation to see that the corresponding level surface is a point at the origin.

 

 

 

Answer 2

Hi Stacy,

 

Working with multivariable functions is an exciting part of intermediate calculus studies.  As you learn to pick apart functions like the one in this problem, you will gain a better understanding of how these sorts of mappings correspond to physical relations between variables.

 

As you recall from your precalculus studies, a function is a mapping which assigns each value in the domain (a set of numbers) to one value in the range (another set of numbers).  When we talk about finding the domain and range of a function, we need to consider what values are assigned outputs (i.e. the domain), as well as which outputs can be produced (i.e. the range).

 

In this problem, the function f(x,y) takes in a two-dimensional input value (x, y) (which you can think of as a point on a plane) and assigns a scalar value z according to the equation:

f(x,y) = 1 - e^{(-4x² - 9y²)}

 

In considering which values on the ℜ² plane (the 2D plane consisting of points whose two coordinates are real numbers) are assigned values under f, we notice that all values will produce a valid z (since there is no (x, y) where f is undefined).  This means that the domain of f is ℜ².

 

Similarly, in analyzing the output values of the function, we notice that no matter what point (x, y) we plug in,  e^{(-4x² - 9y²)} will always yield a positive real value.  This means that f(x, y) can only produce real numbers z < 1.  For ℜ² values (x, y), e^{(-4x² - 9y²)} ≤ 1 since the exponent will never be a positive value.  This means that f(x, y) can only produce real numbers z ≥ 0.  These two constraints give us the range [0, 1).

 

Putting this together we can properly define the function in terms of its domain and range:

f : ℜ² → [0, 1).

 

As you pointed out in your question, the level curves will be centered on the origin and will be determined by the real values taken by the exponent 4x²+9y² = c.  Notice that this curve defines an ellipse, though, not a circle as you had thought, so the picture will be a little bit different.  The level curves can be described as concentric ellipses of eccentricity √(5/9) centered at the origin, with semimajor axes lying on the x-axis.

 

To answer your question about reversing the sign in the equation, that function is the same as 2 - f(x,y), which will have range (1, 2].

Answer 3

I agree the domain is all real numbers because there is no values for x and y that will cause the formula to "blow up." Consider all the possible values that 1/e^x can take. Starting x at negative infinity then we just have e^x which would be plus infinity. Therefore 1/e^x would go to zero. As x goes to 0 then we get 1/1 = 1. Then as x heads off to positive infinity 1/e^x starts heading off to zero. Therefore, 1/e^x goes from 0 to 1. Now, since we are subtracting this from 1, then f(x,y) would go from 0 to 1.

 

I have no idea what is meant by level of curves.