Problem 32
Question
Sketch the graphs of the given curves and compare them. Do they differ and if so, how? (a) \(x=t, \quad y=t^{2}\) (b) \(x=\sqrt{t}, \quad y=t\) (c) \(x=e^{t}, \quad y=e^{2 t}\)
Step-by-Step Solution
Verified Answer
Answer: Yes, all three curves differ from each other in terms of their shape, orientation, and growth rates. Curve (a) is a parabolic curve opening upwards, curve (b) starts from the origin and increases in a slower pace, and curve (c) is an exponential curve growing along both the x and y directions.
1Step 1: Identifying the given curves
Here are the curves given in their parametric form:
(a) \(x=t, \quad y=t^{2}\)
(b) \(x=\sqrt{t}, \quad y=t\)
(c) \(x=e^{t}, \quad y=e^{2 t}\)
2Step 2: Sketch the graphs of the curves
For each equation in (a), (b), and (c), sketch the graphs.
(a) \(x=t, \quad y=t^{2}\)
To sketch this curve, simply plot the points \((t, t^2)\) for various values of \(t\). This will result in a parabolic curve with its vertex at the origin.
(b) \(x=\sqrt{t}, \quad y=t\)
To sketch this curve, plot the points \((\sqrt{t}, t)\) for various values of \(t\). This will result in a curve that starts from the origin and increases in a slower pace.
(c) \(x=e^{t}, \quad y=e^{2 t}\)
To sketch this curve, plot the points \((e^{t}, e^{2t})\) for various values of \(t\). This curve will start from the point \((1, 1)\) and grow exponentially along the x and y directions.
3Step 3: Compare the graphs
Now, compare the graphs of the given curves to determine if they differ or not.
(a) and (b) differ in their shapes as (a) is a parabolic curve, and (b) is a curve that starts from the origin and increases in a slower pace. Furthermore, their orientations in the Cartesian plane are also different. In (a), the parabolic curve opens upwards, while in (b), the curve is increasing more in the x direction.
(a) and (c) also differ, as both their shapes and orientations are distinct. While (a) is a parabolic curve opening upwards, (c) is an exponential curve growing along both the x and y directions.
(b) and (c) differ as well, considering that (b) is a curve that starts from the origin and increases in a slower pace, while (c) is an exponential curve. Their growth rates and the orientation of the curves on the Cartesian plane are distinct.
In conclusion, all three curves differ from each other in terms of their shape, orientation, and growth rates.
Key Concepts
Parametric CurvesParabolic Curve GraphingExponential Curve Graphing
Parametric Curves
Parametric equations provide a convenient way to represent curves by expressing the coordinates of the points on the curve as functions of a single parameter, typically denoted by t. Unlike the traditional y as a function of x approach, parametric equations encapsulate more complex relationships and motions.
For instance, consider the parametric equations in the textbook exercise. In equation (a), both x and y are defined in terms of t, allowing us to trace the shape of a curve as t varies, rather than as x varies independently of y. This reveals a parabolic path. Such a representation is superb for illustrating paths of projectiles or the contours of objects in motion. Conceptually, parametric curves can encompass a vast array of different shapes – from straight lines to intricate spirals – simply by altering the defining functions of x(t) and y(t).
For instance, consider the parametric equations in the textbook exercise. In equation (a), both x and y are defined in terms of t, allowing us to trace the shape of a curve as t varies, rather than as x varies independently of y. This reveals a parabolic path. Such a representation is superb for illustrating paths of projectiles or the contours of objects in motion. Conceptually, parametric curves can encompass a vast array of different shapes – from straight lines to intricate spirals – simply by altering the defining functions of x(t) and y(t).
Parabolic Curve Graphing
Parabolic curves are a specific type of curve with the classic 'u-shaped' appearance that result from quadratic functions. When graphing parametric equations that yield a parabolic curve, such as the equation (a) x=t and y=t^2, the relationship between x and y is not direct, but rather a function of t.
Plotting a Parabola
By choosing a range of t values and computing the corresponding x and y values, one can plot individual points that lay on the curve. When t is zero, we begin at the origin. As t increases or decreases, we trace the symmetric arms of the parabola. Notably, the vertex of this parabola is at the origin, a detail that helps in sketching the complete graph. The parabolic graphing method illustrates the important concept of how parametric equations can neatly describe even the pathways of objects acted upon by constant acceleration, such as gravity.Exponential Curve Graphing
Exponential curves represent rapid growth or decay and are common in modeling phenomena in physics, biology, finance, and many other fields. The textbook’s equation (c) x=e^t and y=e^(2t) describes such a curve. Here e represents the mathematical constant approximately equal to 2.71828, and the exponentiation indicates that the value of y grows much faster than that of x.
Understanding Exponential Growth
As you plot points for increasing values of t, the distance between successive points on the curve will rapidly increase. This is because exponential functions compress a vast range of y values into a relatively small range of t values. Additionally, the curve will never touch the x-axis, illustrating the concept of asymptotic behavior – where the function approaches but never reaches zero. Exponential graphing helps to visualize concepts such as population growth, radioactive decay, or any process where a quantity increases by a consistent percentage.Other exercises in this chapter
Problem 31
In Exercises \(29-34,\) find the latus rectum of the parabola whose equation is given. [Hint: Examples 3 and 4 may be help. ful in Exercises \(29-30.1\) $$y^{2}
View solution Problem 31
Identify the conic whose equation is given and find its graph. If it is an ellipse, list its center, vertices, and foci. If it is a hyperbola, list its center,
View solution Problem 32
Calculus can be used to show that the area of the ellipse with equation \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is \(\pi\)ab. Use this fact to find the ar
View solution Problem 32
In Exercises \(29-34,\) find the latus rectum of the parabola whose equation is given. [Hint: Examples 3 and 4 may be help. ful in Exercises \(29-30.1\) $$y=2 x
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