Sketching level curves and the surface of a sphere

In summary, the conversation discusses how to sketch the level curves of a given equation in 2D and how to sketch a sphere in 3D. It also discusses finding a point on the surface of the sphere and determining its center and radius. The steps for sketching the level curves and the sphere are explained, including setting z values, solving for variables, and determining the location of the center and radius. The conversation ends with a question about how to arrange the axes to sketch the sphere.
  • #1
Cottontails
33
0

Homework Statement


(a) Sketch the level curves of z = (x^2 - 2y +6)/(3x^2 + y) at heights z = 0 and z =1.
(b) Sketch the surface (x−1)^2 + (y+2)^2 + z^2 = 2 in R^3. Write down a point which is on the surface.

Homework Equations


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The Attempt at a Solution


(a) From the question, I assume that you would draw the equation in 2D on the xy-axis and then mark the curves z = 0 and z =1. However, what I am unsure of how is to how to actually sketch the level curves and then find at what values would z = 0 and z = 1 be drawn.
(b) From the equation, I recognise that the surface will be a sphere. But is R^3 the xyz-plane? If so, then how would I be able to sketch this as a sphere (on paper)? Would sketching it within a cube be appropriate? Moreover, how would I find a point which is on the surface? Would a point on the surface be an integer solution? (For example, x=0, y=-3, z=0 so, (0,-3,0)).
 
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  • #2
Cottontails said:

Homework Statement


(a) Sketch the level curves of z = (x^2 - 2y +6)/(3x^2 + y) at heights z = 0 and z =1.
(b) Sketch the surface (x−1)^2 + (y+2)^2 + z^2 = 2 in R^3. Write down a point which is on the surface.

Homework Equations


--

The Attempt at a Solution


(a) From the question, I assume that you would draw the equation in 2D on the xy-axis and then mark the curves z = 0 and z =1.
No, you have the steps reversed. First, set z = 0, and then graph that equation. That will be the level curve for z = 0.
Second, set z = 1, and then graph that curve. You will have a different equation for each level curve.
Cottontails said:
However, what I am unsure of how is to how to actually sketch the level curves and then find at what values would z = 0 and z = 1 be drawn.
(b) From the equation, I recognise that the surface will be a sphere. But is R^3 the xyz-plane?
There is no xyz plane, not is there an xy-axis, as you said above. You can talk about the x-axis or y-axis or z-axis - these axes are straight lines. You can also talk about the xy-plane or xz-plane or yz-plane. These are coordinate planes.
Cottontails said:
If so, then how would I be able to sketch this as a sphere (on paper)? Would sketching it within a cube be appropriate? Moreover, how would I find a point which is on the surface? Would a point on the surface be an integer solution? (For example, x=0, y=-3, z=0 so, (0,-3,0)).

Can you draw the part of the sphere that is in the first octant?

Does the point (0, -3, 0) satisfy the equation (x−1)^2 + (y+2)^2 + z^2 = 2? If so, the point is on the surface of this sphere. If not, it isn't. Since you have three variables and only one equation, you can arbitrarily set any two variables and then solve for the third.

Since you know that this equation is of a sphere, do you also know the other things of interest, such as the location of its center and its radius? You can pretty much pick this information off the equation.
 
  • #3
Mark44 said:
No, you have the steps reversed. First, set z = 0, and then graph that equation. That will be the level curve for z = 0.
Second, set z = 1, and then graph that curve. You will have a different equation for each level curve.

Okay, thanks. I set z = 0 and then sketched the level curve. I got a positive parabola, intersecting the y-axis at 3 (no roots). Is that correct?
For z = 1, I got a negative parabola, intersecting the y-axis at 2, with roots -2 and 2. Is that correct? Also, would I draw each level curve on separate axes or on the same one?

Mark44 said:
Can you draw the part of the sphere that is in the first octant?

Sorry, I don't understand what you mean by this. How would you be able to figure out the arc of the circle that would be in the first octant? (Sorry, I have never sketched a sphere before.)

Mark44 said:
Does the point (0, -3, 0) satisfy the equation (x−1)^2 + (y+2)^2 + z^2 = 2? If so, the point is on the surface of this sphere. If not, it isn't. Since you have three variables and only one equation, you can arbitrarily set any two variables and then solve for the third.

Okay, so (0, -3, 0) does not satisfy the equation then. I set two variables, x = 2, y = -2. With z = 1. I then substituted them into the equation and the equal to 2. So, does that therefore mean (2, -2, 1) is a point on the surface of this sphere then?

Mark44 said:
Since you know that this equation is of a sphere, do you also know the other things of interest, such as the location of its center and its radius? You can pretty much pick this information off the equation.

Would the center be (1, -2, 0) and the radius √2?
 
  • #4
Cottontails said:
I set two variables, x = 2, y = -2. With z = 1. I then substituted them into the equation and the equal to 2. So, does that therefore mean (2, -2, 1) is a point on the surface of this sphere then?

Yes.

Cottontails said:
Would the center be (1, -2, 0) and the radius √2?

Yes.


ehild
 
  • #5
Okay, thank you. However, I am confused with actually sketching the sphere. So, I understand that the center must be (1, -2, 0) with radius √2. Yet, how would I arrange the axes to sketch the sphere? Since y=-2, it exists on the negative y axis. & I assume you would arrange the axes (appropriately and correctly) so that it forms a cube for the sphere to be within, right? That, I'm unsure of how to do as would all the axes still meet at the origin?
 
  • #6
The axes meet at the origin, the centre of the sphere is at (1,-2,0). You can use the level lines to make the sphere look 3-dimensional. Perhaps something like in my figure.

ehild
 

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  • #7
Okay, thanks.

Cottontails said:
Okay, thanks. I set z = 0 and then sketched the level curve. I got a positive parabola, intersecting the y-axis at 3 (no roots). Is that correct?
For z = 1, I got a negative parabola, intersecting the y-axis at 2, with roots -2 and 2. Is that correct? Also, would I draw each level curve on separate axes or on the same one?

Moreover, is this correct for part (a)?
 
  • #8
Cottontails said:
Okay, thanks. I set z = 0 and then sketched the level curve. I got a positive parabola, intersecting the y-axis at 3 (no roots). Is that correct?
"Positive" doesn't have much meaning here. It would be better to say that it opens up or down or to the left of right. In those terms how would you describe the parabola?
Cottontails said:
For z = 1, I got a negative parabola, intersecting the y-axis at 2, with roots -2 and 2. Is that correct? Also, would I draw each level curve on separate axes or on the same one?
The parabola opens downward, and has a y-intercept at (0, 2), but your x-intercepts are wrong.

If you graph both level curves on the same axis, label each one as z = 0 or z = 1. For this problem the two level curves don't give much insight as to the shape of the three-dimensional surface in the problem.

Topographical maps are essentially the same thing as level curves. On such a map high points and low points are shown with closed curves that are at the same height. Hikers can use these maps to determine how steep a trail will be.
 
  • #9
Mark44 said:
"Positive" doesn't have much meaning here. It would be better to say that it opens up or down or to the left of right. In those terms how would you describe the parabola?

Okay. Then, the parabola opens upward, with the y-intercept at (0,3). It has no x-intercepts but the opening (with how it expands/terms of width) of the curve is within x=-4, x =4.

Mark44 said:
The parabola opens downward, and has a y-intercept at (0, 2), but your x-intercepts are wrong.

Yes, I realized that my x-intercepts are wrong. The x-intercepts are √3 and -√3, right?

Mark44 said:
If you graph both level curves on the same axis, label each one as z = 0 or z = 1. For this problem the two level curves don't give much insight as to the shape of the three-dimensional surface in the problem.

So, should I draw both level curves on the same axis then? Or is it also fine to draw them both on separate axes? With the level curves as well, it would be the parabola on the xy-axis with "z=0" or "z=1" next to it, right? & That would answer the question?
Also, the question is just asking for the level curves and not for the shape of the surface itself, right (with the shape of the surface more focused in part (b))?
 
  • #10
Cottontails said:
Okay. Then, the parabola opens upward, with the y-intercept at (0,3). It has no x-intercepts but the opening (with how it expands/terms of width) of the curve is within x=-4, x =4.
No, the parabola is not confined to the interval [-4, 4].
Cottontails said:
Yes, I realized that my x-intercepts are wrong. The x-intercepts are √3 and -√3, right?
Right.
Cottontails said:
So, should I draw both level curves on the same axis then?
That's probably best. (BTW, it would be the same axis system. I mistakenly wrote "axis" when I should have written "axes" or "axis system". You can't graph a parabola on just a single axis.
Cottontails said:
Or is it also fine to draw them both on separate axes? With the level curves as well, it would be the parabola on the xy-axis with "z=0" or "z=1" next to it, right? & That would answer the question?
Again, there is no such thing as "xy-axis". I would advise that you draw both level curves on the same axis system, and close to each curve write z = 0 or z = 1 to identify them.

Cottontails said:
Also, the question is just asking for the level curves and not for the shape of the surface itself, right (with the shape of the surface more focused in part (b))?
Right.
 

Related to Sketching level curves and the surface of a sphere

1. What is the purpose of sketching level curves and the surface of a sphere?

The purpose of sketching level curves and the surface of a sphere is to visually represent the mathematical concept of a scalar field, where each point on the surface has a corresponding numerical value. This helps to better understand and analyze the behavior of the field.

2. How do you sketch level curves on a sphere?

To sketch level curves on a sphere, you first need to identify the equation or function that represents the scalar field. Then, choose a few specific values for the field and plot points on the surface of the sphere corresponding to those values. Finally, connect the points to create level curves.

3. What is the process for sketching the surface of a sphere?

To sketch the surface of a sphere, you can start by plotting the center point and then marking points around it at equal distance, creating a circle. Then, you can use a compass or ruler to plot more points around the circle at the same distance. Finally, connect the points to form the surface of the sphere.

4. Can level curves intersect on a sphere?

Yes, level curves can intersect on a sphere. This usually occurs when the scalar field has multiple critical points, where the gradient is equal to zero. In this case, the level curves will intersect at the critical points.

5. How can sketching level curves and the surface of a sphere be applied in real-life situations?

Sketching level curves and the surface of a sphere can be applied in various fields such as physics, engineering, and geography. For example, in physics, it can be used to visualize electric or magnetic fields. In engineering, it can be used to understand the behavior of temperature or pressure in a specific area. In geography, it can be used to map out elevation or ocean currents.

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