Exercises 12.1

Exercises 7 - 14, give the domain and range of the multivariable functi????on.

  1. \(f(x; y) = \frac{1}{\sqrt{x2 + y2 -9???? }}\)

Exercises 15 - 22, describe in words and sketch the level curves for the function and given c values.

  1. \(f(x; y) = \sqrt{x^2 + 4y^2}; c = 1, 2, 3, 4\)

Solution

Consider first c = 1. The level curve for c = 1 is the set of all points (x,y) such that \(1 = \sqrt{x^2 + 4y^2}\) Squaring both sides quickly gives us \(1=x^2 + 4y^2\),an ellipse centered at (0,0) with horizontal major axis of length 2 and minor axis 0f length 1. Thus for any point (x; y) on this curve, f(x; y) = 0.

Consider first c = 2. The level curve for c = 2 is the set of all points (x,y) such that \(4 = \sqrt{x^2 + 4y^2}\) Squaring both sides quickly gives us \(1=x^2 + 4y^2\),an ellipse centered at (0,0) with horizontal major axis of length 4 and minor axis 0f length 2. Thus for any point (x; y) on this curve, f(x; y) = 0.

Consider first c = 3. The level curve for c = 3 is the set of all points (x,y) such that \(9 = \sqrt{x^2 + 4y^2}\) Squaring both sides quickly gives us \(1=x^2 + 4y^2\),an ellipse centered at (0,0) with horizontal major axis of length 6 and minor axis 0f length 3. Thus for any point (x; y) on this curve, f(x; y) = 0.

Consider first c = 4. The level curve for c = 4 is the set of all points (x,y) such that \(16 = \sqrt{x^2 + 4y^2}\) Squaring both sides quickly gives us \(1=x^2 + 4y^2\),an ellipse centered at (0,0) with horizontal major axis of length 8 and minor axis 0f length 4. Thus for any point (x; y) on this curve, f(x; y) = 0.

Use conture to draw the picture. The center is not at (0,0).

require(raster)
## Loading required package: raster
## Warning: package 'raster' was built under R version 3.4.3
## Loading required package: sp
## Warning: package 'sp' was built under R version 3.4.3
x <- seq(-4, 4, length=21)
f <- function(x,y){sqrt(x^2 + 4*y^2)}
z <- outer(x, x, f)
contour(z, levels = c(1, 2, 3, 4), lty = "dotted",
        xlim = range(0,1),
        ylim = range(0,1),
        )

Use matplot to draw the picture. The center is at (0,0).

for c=1,2,3,4:

require(graphics)
fun1 <- function(x){sqrt((1-x^2)/4)}
fun2 <- function(x){-sqrt((1-x^2)/4)}
fun3 <- function(x){sqrt((4-x^2)/4)}
fun4 <- function(x){-sqrt((4-x^2)/4)}
fun5 <- function(x){sqrt((9-x^2)/4)}
fun6 <- function(x){-sqrt((9-x^2)/4)}
fun7 <- function(x){sqrt((16-x^2)/4)}
fun8 <- function(x){-sqrt((16-x^2)/4)}
x<-seq(-4,4,0.01)

matplot(x,cbind(fun1(x),fun2(x),fun3(x),fun4(x),fun5(x),fun6(x),fun7(x),fun8(x)),ylim=c(-4,4),type="l")
## Warning in sqrt((1 - x^2)/4): NaNs produced

## Warning in sqrt((1 - x^2)/4): NaNs produced
## Warning in sqrt((4 - x^2)/4): NaNs produced

## Warning in sqrt((4 - x^2)/4): NaNs produced
## Warning in sqrt((9 - x^2)/4): NaNs produced

## Warning in sqrt((9 - x^2)/4): NaNs produced

  1. \(f(x, y) = x2 + 4y^2; c = 1, 2, 3, 4\)

Solution

Consider first c = 1. The level curve for c = 1 is the set of all points (x,y) such that \(1 = x^2 + 4y^2\) . It is an ellipse centered at (0,0) with horizontal major axis of length 2 and minor axis 0f length 1. Thus for any point (x; y) on this curve, f(x; y) = 0.

Consider first c = 2. The level curve for c = 2 is the set of all points (x,y) such that \(2 = x^2 + 4y^2\). It is an ellipse centered at (0,0) with horizontal major axis of length \(2\sqrt{2}\) and minor axis 0f length \(\sqrt{2}\). Thus for any point (x; y) on this curve, f(x; y) = 0.

Consider first c = 3. The level curve for c = 3 is the set of all points (x,y) such that \(3 = x^2 + 4y^2\). It is an ellipse centered at (0,0) with horizontal major axis of length \(2\sqrt{3}\) and minor axis 0f length \(sqrt{3}\). Thus for any point (x; y) on this curve, f(x; y) = 0.

Consider first c = 4. The level curve for c = 4 is the set of all points (x,y) such that \(4 = x^2 + 4y^2\). It is an ellipse centered at (0,0) with horizontal major axis of length 4 and minor axis 0f length 2. Thus for any point (x; y) on this curve, f(x; y) = 0.

Use conture to draw the picture. The center is not at (0,0).

require(raster)
x <- seq(-4, 4, length=21)
f <- function(x,y){x^2 + 4*y^2}
z <- outer(x, x, f)
contour(z, levels = c(1, 2, 3, 4), lty = "dashed",
        xlim = range(0,1),
        ylim = range(0,1),
        )

Use matplot to draw the picture. The center is at (0,0).

for c=1,2,3,4:

require(graphics)
fun1 <- function(x){sqrt((1-x^2)/4)}
fun2 <- function(x){-sqrt((1-x^2)/4)}
fun3 <- function(x){sqrt((2-x^2)/4)}
fun4 <- function(x){-sqrt((2-x^2)/4)}
fun5 <- function(x){sqrt((3-x^2)/4)}
fun6 <- function(x){-sqrt((3-x^2)/4)}
fun7 <- function(x){sqrt((4-x^2)/4)}
fun8 <- function(x){-sqrt((4-x^2)/4)}
x<-seq(-4,4,0.01)

matplot(x,cbind(fun1(x),fun2(x),fun3(x),fun4(x),fun5(x),fun6(x),fun7(x),fun8(x)),ylim=c(-4,4),type="l")
## Warning in sqrt((1 - x^2)/4): NaNs produced

## Warning in sqrt((1 - x^2)/4): NaNs produced
## Warning in sqrt((2 - x^2)/4): NaNs produced

## Warning in sqrt((2 - x^2)/4): NaNs produced
## Warning in sqrt((3 - x^2)/4): NaNs produced

## Warning in sqrt((3 - x^2)/4): NaNs produced
## Warning in sqrt((4 - x^2)/4): NaNs produced

## Warning in sqrt((4 - x^2)/4): NaNs produced

  1. Compare the level curves of Exercises 21 and 22. How are they similar, and how are they different? Each surface is a quadric surface; describe how the level curves are consistent with what we know about each surface.

Answer:

Compare the level curves of Exercises 21 and 2, we know that the level curves are all ellipses centered at (0,0). The major axis and minor axis of 21 are 2 times of those of 22 respectively. Each surface of 21 is 4 fold of that of 22.