Hyperboloid Radius: Find Circle Radius At Z=3
Hey guys! Let's dive into the fascinating world of hyperboloids and figure out how to pinpoint the radius of a circle formed when a plane slices through it. Specifically, we'll tackle the challenge of finding the radius of the circle on a one-sheet hyperboloid at the plane z=3. Grab your thinking caps, and let's get started!
Understanding the Hyperboloid Equation
Before we jump into calculations, let’s make sure we’re all on the same page about what a hyperboloid is and what its equation represents. The general equation of a one-sheet hyperboloid is given by:
(x²/a²) + (y²/b²) - (z²/c²) = -1
In this equation:
- x, y, and z are the Cartesian coordinates.
- a, b, and c are constants that determine the shape and size of the hyperboloid along the x, y, and z axes, respectively.
The negative sign in front of the 1 on the right side of the equation is crucial. It indicates that we're dealing with a hyperboloid of one sheet. Imagine a shape that looks like a saddle that extends infinitely. The constants a, b, and c essentially stretch or compress this saddle along the respective axes. A key thing to note is that cross-sections of this hyperboloid, parallel to the xy-plane, are ellipses (or circles if a = b). Our goal is to find the specific ellipse (or circle) formed when z = 3.
To truly understand this, picture the hyperboloid sitting in 3D space. The a and b values dictate how wide the hyperboloid spreads along the x and y axes, while c affects how quickly it opens up along the z-axis. If a and b are equal, the cross-sections are perfect circles. If they are different, you get ellipses. Now, think about slicing this shape with a flat plane at z = 3. The intersection of the plane and the hyperboloid will form a circle or an ellipse, and we want to figure out the radius (or semi-major and semi-minor axes) of that shape. This involves substituting z = 3 into the general equation and then manipulating the equation to reveal the dimensions of the resulting ellipse (or circle). It’s a fun exercise in spatial reasoning and algebraic manipulation!
Steps to Find the Radius at z=3
Alright, let's break down the process step-by-step. Our mission is to find the radius of the circle formed on the hyperboloid when z = 3.
Step 1: Substitute z=3 into the Hyperboloid Equation
First, we're going to substitute z = 3 into the general equation of the hyperboloid. This will give us an equation that describes the cross-section of the hyperboloid at that specific plane. So, here we go:
(x²/a²) + (y²/b²) - (3²/c²) = -1
(x²/a²) + (y²/b²) - (9/c²) = -1
Step 2: Rearrange the Equation
Next, we need to rearrange the equation to get it into a more recognizable form – specifically, the equation of an ellipse. We want to isolate the terms with x and y on one side of the equation. To do this, we'll add (9/c²) to both sides:
(x²/a²) + (y²/b²) = -1 + (9/c²)
(x²/a²) + (y²/b²) = (9/c²) - 1
To make things cleaner, let's denote the right-hand side of the equation as R²: R² = (9/c²) - 1. This simplifies the equation to:
(x²/a²) + (y²/b²) = R²
Step 3: Normalize the Equation
To truly see the equation of an ellipse (or circle), we need the right side to be equal to 1. We achieve this by dividing both sides of the equation by R²:
(x² / (a² * R²)) + (y² / (b² * R²)) = 1
This is now in the standard form of an ellipse equation:
(x² / A²) + (y² / B²) = 1
where A² = a² * R² and B² = b² * R².
Step 4: Determine the Radius (or Semi-Axes)
Now, let's consider two scenarios:
-
Scenario 1: a = b
If a = b, we have a circle. In this case, A = B, and the radius of the circle, r, is simply:
r = a R = a √((9/c²) - 1)
-
Scenario 2: a ≠ b
If a ≠ b, we have an ellipse. Here, A and B are the semi-major and semi-minor axes of the ellipse, respectively. So:
Semi-major axis, A = a R = a √((9/c²) - 1)
Semi-minor axis, B = b R = b √((9/c²) - 1)
Practical Example
Let's solidify our understanding with a practical example. Suppose we have the equation:
(x²/4) + (y²/9) - (z²/16) = -1
Here, a² = 4, b² = 9, and c² = 16. Thus, a = 2, b = 3, and c = 4.
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Substitute z=3:
(x²/4) + (y²/9) - (9/16) = -1
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Rearrange:
(x²/4) + (y²/9) = -1 + (9/16)
(x²/4) + (y²/9) = (9/16) - 1
(x²/4) + (y²/9) = -7/16
Uh oh! We've hit a snag. Notice that the right-hand side of the equation, -7/16, is negative. This implies there is no real solution for x and y, meaning the plane z=3 does not intersect the hyperboloid in this particular case. This can happen if the plane is too far from the 'center' of the hyperboloid.
Let’s tweak the equation slightly to make it work. Consider:
(x²/4) + (y²/9) - (z²/25) = -1
So now, *a* = 2, *b* = 3, and *c* = 5.
1. **Substitute z=3:**
**(x²/4) + (y²/9) - (9/25) = -1**
2. **Rearrange:**
**(x²/4) + (y²/9) = -1 + (9/25)**
**(x²/4) + (y²/9) = -16/25**
Again, we get a negative value on the right-hand side, indicating no intersection. Let's adjust the original hyperboloid equation once more to:
**(x²/4) + (y²/9) - (z²/1) = -1**
Here, *a* = 2, *b* = 3, and *c* = 1.
1. **Substitute z=3:**
**(x²/4) + (y²/9) - (9/1) = -1**
2. **Rearrange:**
**(x²/4) + (y²/9) = -1 + 9**
**(x²/4) + (y²/9) = 8**
3. **Normalize:**
**(x² / (4 * 8)) + (y² / (9 * 8)) = 1**
**(x² / 32) + (y² / 72) = 1**
Now we have a valid ellipse equation. The semi-major axis *A* = √72 ≈ 8.49, and the semi-minor axis *B* = √32 ≈ 5.66.
Important Considerations
- R² Value: If R² = (9/c²) - 1 is negative, it means the plane z = 3 does not intersect the hyperboloid. The plane is too far away from the vertex. In this case, there is no real solution for the radius.
- a = b: When a = b, the cross-section is a circle, making the calculation of the radius straightforward.
- Units: Ensure that all your units are consistent to avoid errors in your calculations.
Conclusion
So, there you have it! By substituting z = 3 into the general equation of the hyperboloid, rearranging the equation, and normalizing it, we can find the radius (or semi-axes) of the resulting circle (or ellipse). Remember to check if the plane actually intersects the hyperboloid by verifying that R² is positive. Keep practicing, and you'll become a hyperboloid master in no time! This method provides a solid foundation for understanding how planes intersect with hyperboloids, giving valuable insights into their geometric properties. Keep exploring and happy calculating!