Chapter 11: Problem 66
Find parametric equations for the following curves. Include an interval for the parameter values. The left half of the parabola \(y=x^{2}+1\), originating at (0,1)
Short Answer
Expert verified
Answer: The parametric equations for the left half of the parabola are x(t) = t and y(t) = t^2 + 1, with t in the interval \((-\infty, 0]\).
Step by step solution
01
Introduce a new parameter
To begin, we can introduce a new parameter t and rewrite the equation in terms of t instead of x. We will choose t to represent x so that the relationship between x, y, and t is clearly defined.
x = t
y = t^2 + 1
02
Rewrite the parametric equations
Now we have the parametric equations for the entire parabola, not just the left half. These are:
x(t) = t
y(t) = t^2 + 1
By changing t, we can generate different points on the curve, which demonstrates how the equation can be written in parametric form.
03
Identify the interval for the left half of the parabola
Now we need to figure out the interval for the parameter values such that it only represents the left half of the parabola. Notice that when x is negative, we will be on the left half of the parabola. So, we need to restrict t to be non-positive, i.e., t ≤ 0.
04
Write the final solution with the interval
Now we can combine our parametric equations and the interval for t to answer the given question. The parametric equations for the left half of the parabola \(y=x^2+1\), originating at (0,1), are:
x(t) = t
y(t) = t^2 + 1
with t in the interval \((-\infty, 0]\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parabola
A parabola is a symmetrical, U-shaped curve that you often encounter in mathematics. The simplest form of a parabolic equation is quadratic: \(y = ax^2 + bx + c\). In our specific case, the given parabola is \(y = x^2 + 1\).
This parabola opens upwards because the coefficient of \(x^2\), which is 1, is positive. The "+1" indicates that the entire parabola is shifted one unit up along the y-axis.
Parabolas have a vertex, which represents their highest or lowest point. For \(y = x^2 + 1\), the vertex is at \((0, 1)\), meaning it starts at the point (0,1) and opens symmetrically from this point along the x-axis. An interesting property of parabolas is their symmetry about the vertical line that originates from the vertex. Thus, understanding parabolas in a parametric form is crucial as it helps in controlling the traversal along the curve.
This parabola opens upwards because the coefficient of \(x^2\), which is 1, is positive. The "+1" indicates that the entire parabola is shifted one unit up along the y-axis.
Parabolas have a vertex, which represents their highest or lowest point. For \(y = x^2 + 1\), the vertex is at \((0, 1)\), meaning it starts at the point (0,1) and opens symmetrically from this point along the x-axis. An interesting property of parabolas is their symmetry about the vertical line that originates from the vertex. Thus, understanding parabolas in a parametric form is crucial as it helps in controlling the traversal along the curve.
Parameter Interval
In the context of parametric equations, the parameter interval is crucial as it defines the range over which the parameter, commonly denoted as \(t\), varies. This interval essentially specifies which part of the curve the parametric equations will represent.
For the parabola \(y = x^2 + 1\), when we express it in parametric form as \(x(t) = t\) and \(y(t) = t^2 + 1\), \(t\) acts as our parameter. To find the left half of the parabola, it's important to remember that as \(t\) is non-positive, \(x(t)\) will also be non-positive, covering just the left side.
For the parabola \(y = x^2 + 1\), when we express it in parametric form as \(x(t) = t\) and \(y(t) = t^2 + 1\), \(t\) acts as our parameter. To find the left half of the parabola, it's important to remember that as \(t\) is non-positive, \(x(t)\) will also be non-positive, covering just the left side.
- The interval \((-\infty, 0]\) is chosen for \(t\) because, within this range, \(t\) (and hence \(x\)) is non-positive, perfectly representing the left half of the parabola.
- By restricting \(t\) in this way, the parametric equation captures only the desired portion of the parabola.
Curve Sketching
Curve sketching involves drawing reliable representations of curves based on their equations. For parametric equations, understanding this task is a bit different from regular functions.
In parametric form, both \(x\) and \(y\) are expressed in terms of a third variable \(t\). This means to sketch the left half of the parabola \(y = x^2 + 1\), we need to follow this procedure:
This approach allows for a clearer visualization of the curve without needing to plug in a multitude of \(x\) values. The curve's symmetry can also be considered to make the sketch more precise and aesthetically accurate. Curve sketching with parametric equations opens up the possibility of drawing complex curves that are otherwise difficult to represent using simple Cartesian equations.
In parametric form, both \(x\) and \(y\) are expressed in terms of a third variable \(t\). This means to sketch the left half of the parabola \(y = x^2 + 1\), we need to follow this procedure:
- Start by plotting the vertex, which in our example is (0,1).
- Determine the direction in which the parameter \(t\) will drive the curve. Since \(t\) is negative progressing to zero, the curve will traverse from left to right until it reaches the vertex.
- It's practical in parametrics to sketch a few specific points from the parameter interval, like \(t = -1, -0.5,\) and \(0\). This can act as anchors or guide points.
This approach allows for a clearer visualization of the curve without needing to plug in a multitude of \(x\) values. The curve's symmetry can also be considered to make the sketch more precise and aesthetically accurate. Curve sketching with parametric equations opens up the possibility of drawing complex curves that are otherwise difficult to represent using simple Cartesian equations.