How To Draw Level Curves For A Differential Equation

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Section 1-5 : Functions of Several Variables

In this section we want to go over some of the basic ideas about functions of more than i variable.

First, remember that graphs of functions of two variables, $z = f\left( {x,y} \right)$ are surfaces in three dimensional space. For example, here is the graph of $z = 2{x^ii} + 2{y^two} - 4$.

$This is a graph with the standard 3D coordinate system. The positive z-axis is straight up, the positive x-axis moves off to the left and slightly downward and positive y-axis move off the right and slightly downward. This is a cup shaped object that is centered on the z-axis, starts at z=-4 and opens upward.$ $This is a boxed 3D coordinate system. The z-axis is right vertical edge of the box, the x-axis is the top back edge of the box and the y-axis is the bottom left edge of the box. This is a cup shaped object that is parallel to the z-axis, starts at z=-4 and opens upward.$

This is an elliptic paraboloid and is an instance of a quadric surface. Nosotros saw several of these in the previous section. We volition be seeing quadric surfaces adequately regularly subsequently on in Calculus Iii.

Another common graph that we'll be seeing quite a flake in this grade is the graph of a plane. Nosotros take a convention for graphing planes that will make them a little easier to graph and hopefully visualize.

Recollect that the equation of a plane is given past

\[ax + past + cz = d\]

or if nosotros solve this for $z$ we can write information technology in terms of function notation. This gives,

\[f\left( {ten,y} \right) = Ax + By + D\]

To graph a airplane we volition generally find the intersection points with the iii axes and and then graph the triangle that connects those three points. This triangle volition be a portion of the plane and it will give usa a fairly decent idea on what the plane itself should wait like. For instance, let'southward graph the plane given by,

\[f\left( {x,y} \right) = 12 - 3x - 4y\]

For purposes of graphing this it would probably be easier to write this every bit,

\[z = 12 - 3x - 4y\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\,\,\,3x + 4y + z = 12\]

Now, each of the intersection points with the three main coordinate axes is divers by the fact that 2 of the coordinates are nada. For instance, the intersection with the $z$-axis is divers by $x = y = 0$. So, the 3 intersection points are,

\[\begin{align*} & x - {\mbox{axis : }}\left( {iv,0,0} \right)\\ & y - {\mbox{axis : }}\left( {0,three,0} \correct)\\ & z - {\mbox{axis : }}\left( {0,0,12} \correct)\hspace{0.25in}\end{align*}\]

Here is the graph of the plane.

$This is a graph with the standard 3D coordinate system. The positive z-axis is straight up, the positive x-axis moves off to the left and slightly downward and positive y-axis move off the right and slightly downward. This is a triangle in the 1st octant (i.e. x,y and z are all positive) with vertices at (4,0,0), (0,3,0) and (0,0,12).$ $This is a boxed 3D coordinate system. The z-axis is right vertical edge of the box, the x-axis is the top back edge of the box and the y-axis is the bottom left edge of the box. This is a triangle in the 1st octant (i.e. x,y and z are all positive) with vertices at (4,0,0), (0,3,0) and (0,0,12).$

Now, to extend this out, graphs of functions of the course $west = f\left( {x,y,z} \right)$ would be four dimensional surfaces. Of grade, nosotros tin't graph them, but it doesn't hurt to bespeak this out.

We next want to talk about the domains of functions of more i variable. Recall that domains of functions of a single variable, $y = f\left( x \correct)$, consisted of all the values of $10$ that nosotros could plug into the function and get dorsum a existent number. At present, if we think nearly it, this means that the domain of a office of a single variable is an interval (or intervals) of values from the number line, or ane dimensional space.

The domain of functions of two variables, $z = f\left( {x,y} \correct)$, are regions from two dimensional infinite and consist of all the coordinate pairs, $\left( {ten,y} \correct)$, that we could plug into the part and get back a real number.

Example 1 Determine the domain of each of the following.

$f\left( {x,y} \right) = \sqrt {x + y} $
$f\left( {x,y} \right) = \sqrt x + \sqrt y $
$f\left( {x,y} \right) = \ln \left( {9 - {x^ii} - ix{y^2}} \correct)$

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a $f\left( {x,y} \correct) = \sqrt {x + y} $ Show Solution

In this case nosotros know that we tin't take the square root of a negative number so this means that we must require,

\[x + y \ge 0\]

Here is a sketch of the graph of this region.

$This is the graph of the line y=-x and the area above it is shaded.$

b $f\left( {10,y} \right) = \sqrt ten + \sqrt y $ Show Solution

This function is different from the function in the previous part. Here we must require that,

\[ten \ge 0\hspace{0.25in}{\mbox{and}}\,\,\,\,\,y \ge 0\]

and they really do need to exist separate inequalities. There is one for each foursquare root in the function. Here is the sketch of this region.

$This is the graph with a solid line on the positive x and positive y axis and the 1st quadrant is shaded.$

c $f\left( {x,y} \right) = \ln \left( {9 - {x^2} - 9{y^2}} \right)$ Show Solution

In this final part we know that we can't take the logarithm of a negative number or zero. Therefore, we need to require that,

\[ix - {ten^2} - ix{y^2} > 0\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\,\frac{{{x^2}}}{9} + {y^ii} < one\]

and upon rearranging nosotros see that we demand to stay interior to an ellipse for this function. Hither is a sketch of this region.

$This is an ellipse centered at the origin with right/left points of (3,0)/(-3,0) respectively and top/bottom points of (0,1)/(0,-1) respectively. The ellipse itself is dashed and the interior of the ellipse is shaded.$

Annotation that domains of functions of 3 variables, $west = f\left( {10,y,z} \right)$, will be regions in three dimensional infinite.

Case two Determine the domain of the following role, \[f\left( {x,y,z} \right) = \frac{1}{{\sqrt {{x^2} + {y^2} + {z^2} - 16} }}\]

Show Solution

In this instance nosotros accept to deal with the square root and division by zero issues. These volition require,

\[{x^2} + {y^2} + {z^2} - 16 > 0\hspace{0.25in}\,\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,{ten^2} + {y^2} + {z^two} > 16\]

So, the domain for this function is the set of points that lies completely exterior a sphere of radius 4 centered at the origin.

The next topic that we should look at is that of level curves or contour curves. The level curves of the function $z = f\left( {x,y} \right)$ are two dimensional curves we go by setting $z = g$, where $g$ is any number. So the equations of the level curves are $f\left( {x,y} \right) = k$. Note that sometimes the equation will exist in the form $f\left( {x,y,z} \right) = 0$ and in these cases the equations of the level curves are $f\left( {x,y,k} \right) = 0$.

You've probably seen level curves (or contour curves, whatever y'all desire to call them) earlier. If you've ever seen the tiptop map for a slice of state, this is nix more than the profile curves for the role that gives the elevation of the land in that area. Of class, we probably don't have the function that gives the pinnacle, but we can at least graph the contour curves.

Let'south do a quick example of this.

Example 3 Place the level curves of $f\left( {10,y} \right) = \sqrt {{10^2} + {y^2}} $. Sketch a few of them.

Prove Solution

Offset, for the sake of practice, permit's identify what this surface given by $f\left( {x,y} \right)$ is. To practise this allow'south rewrite it as,

\[z = \sqrt {{x^2} + {y^2}} \]

Think from the Quadric Surfaces section that this the upper portion of the "cone" (or hour glass shaped surface).

Note that this was not required for this problem. Information technology was done for the practice of identifying the surface and this may come in handy down the road.

At present on to the existent problem. The level curves (or contour curves) for this surface are given by the equation are found by substituting $z = k$. In the case of our instance this is,

\[k = \sqrt {{10^2} + {y^2}} \hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}{x^2} + {y^2} = {yard^2}\]

where $thousand$ is any number. So, in this case, the level curves are circles of radius $thousand$ with center at the origin.

We can graph these in one of two means. We can either graph them on the surface itself or nosotros can graph them in a two dimensional axis system. Hither is each graph for some values of $k$.

$This is a graph with the standard 3D coordinate system. The positive z-axis is straight up, the positive x-axis moves off to the left and slightly downward and positive y-axis move off the right and slightly downward. This is a cone centered on the z-axis and opening upwards. On the cone surface there are series of circles at z=1, z=2, z=3, z=4 and z=5.$ $This is a boxed 3D coordinate system. The z-axis is right vertical edge of the box, the x-axis is the top back edge of the box and the y-axis is the bottom left edge of the box. This is a cone parallel to the z-axis and opening upwards. On the cone surface there are a series of circles at z=1, z=2, z=3, z=4 and z=5.$

$This is a series of circles centred at the origin with radius 1, 2, 3, 4, and 5.$

Note that nosotros tin can recollect of contours in terms of the intersection of the surface that is given by $z = f\left( {x,y} \right)$ and the plane $z = k$. The contour volition correspond the intersection of the surface and the plane.

For functions of the grade $f\left( {x,y,z} \right)$ we will occasionally expect at level surfaces. The equations of level surfaces are given by $f\left( {x,y,z} \right) = one thousand$ where $k$ is any number.

The final topic in this section is that of traces. In some ways these are like to contours. Equally noted above we can call back of contours as the intersection of the surface given by $z = f\left( {x,y} \right)$ and the aeroplane $z = k$. Traces of surfaces are curves that represent the intersection of the surface and the aeroplane given past $x = a$ or $y = b$.

Let'southward accept a quick look at an example of traces.

Example four Sketch the traces of $f\left( {x,y} \right) = 10 - 4{ten^2} - {y^2}$ for the plane $x = 1$ and $y = two$.

Show Solution

We'll showtime with $x = 1$. We can go an equation for the trace by plugging $x = 1$ into the equation. Doing this gives,

\[z = f\left( {1,y} \correct) = ten - 4{\left( 1 \right)^2} - {y^ii}\hspace{0.25in}\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,z = 6 - {y^2}\]

and this will exist graphed in the aeroplane given past $x = 1$.

Below are ii graphs. The graph on the left is a graph showing the intersection of the surface and the aeroplane given by $x = 1$. On the right is a graph of the surface and the trace that nosotros are after in this part.

$This is a cup shaped object centered on the z-axis, starting at z=10 and opening downwards. There is a plane parallel to the yz-plane that is cutting into the object at x=1.$ $This is a cup shaped object centered on the z-axis, starting at z=10 and opening downwards. There is a line on the object at x=1 that represents where the plane that is in the image to the left cuts into the object.$

For $y = 2$ we will do pretty much the same affair that nosotros did with the first part. Here is the equation of the trace,

\[z = f\left( {x,ii} \right) = x - 4{x^2} - {\left( 2 \right)^2}\hspace{0.25in}\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,z = 6 - 4{ten^2}\]

and here are the sketches for this example.

$This is a cup shaped object centered on the z-axis, starting at z=10 and opening downwards. There is a plane parallel to the xz-plane that is cutting into the object at y=2.$ $This is a cup shaped object centered on the z-axis, starting at z=10 and opening downwards. There is a line on the object at y=2 that represents where the plane that is in the image to the left cuts into the object.$