Data 605 Discussion 15

Calculus section 12.1, question 14

Heather Geiger

Question

Give the domain and range of the multivariable function.

f(x,y) = 1/sqrt(x^2 + y^2 - 9)

Answer

For f(x,y) to be real and to not be undefined, the denominator must be a real number greater than 0.

For the denominator to be a real number, we remember that taking the square root of a negative number creates an imaginary number.

So, we are looking for the x and y such that x^2 + y^2 -9 > 0.

This simplifies to x^2 + y^2 > 9.

Draw the circle corresponding to x^2 + y^2 = 9 in red.

Then, draw two other circles corresponding to x^2 + y^2 > 9 in other colors.

library(plotrix)

plot(0,xlim=c(-5,5),ylim=c(-5,5),
    xlab="x",ylab="y",pch='',
    xaxp=c(-5,5,10),yaxp = c(-5,5,10),
    asp=1)

abline(h=-5:5,v=-5:5, col="gray", lty=3)
abline(h=0,v=0)

draw.circle(x=0,y=0,radius=3,angle=45,border="red",col="red",density=5)

draw.circle(x=0,y=0,radius=sqrt(10), border="black")
draw.circle(x=0,y=0,radius=5, border="blue")

legend("topright",
    legend=c("x^2 + y^2 = 9","x^2 + y^2 = 10","x^2 + y^2 = 25"),
    col=c("red","black","blue"),
    lwd=3,bty="n",cex=0.7)

The domain of f(x,y) is any set of x and y values NOT inside of the red circle. Examples of possible x,y coordinate pairs are any points inside of the black or blue circles shown, but not in the red circle. But really, any points not in the red circle, inside of any arbitrarily larger circle are possible.

The range of f(x,y) is based on 1/(any number larger than 0). If the denominator were a very small (but still positive value) like one quadrillionth, f(x,y) would be extremely large. If the denominator were a very large value, f(x,y) would be extremely small (but still positive).

So, the range of f(x,y) is [0,infinity]. Meaning it is necessarily larger than 0 (and cannot be exactly equal to 0). But can be infinitely large given that x^2 + y^2 can be infinitely close to 9.