Problem 3
Question
Could we use the parametric equations \(x=3 t, y=3 t\) to represent the straight line through the origin and making an angle of \(45^{\circ}\) with the \(x-\) axis? If so, and if \(t\) represents time, what would these equations say about motion along this line as compared with the motion represented by the parametric equations of Exercise \(1 ?\)
Step-by-Step Solution
Verified Answer
The provided parametric equations represent a line making a \(45^{\text{\textdegree}}\) angle with the \(x-\) axis as they have the same coefficient (3), resulting in a slope of 1. With respect to time \( t \), the object moves along the line at a constant speed, with its position in \( x \) and \( y \) increasing at the same rate.
1Step 1: Analyze the given parametric equations
Consider the given parametric equations for the line: \( x = 3t \) and \( y = 3t \). This indicates that for every value of the parameter \( t \), both \( x \) and \( y \) increase by equal amounts, as they are both directly proportional to \( t \) with the same coefficient of proportionality.
2Step 2: Determine the slope of the line
The slope of the parametric line can be determined by taking the derivative \( \frac{dy}{dx} \). Since \( y=3t \) and \( x=3t \), we have \( \frac{dy}{dt} = 3 \) and \( \frac{dx}{dt} = 3 \). Thus, the slope \( \frac{dy}{dx} \) is \( \frac{3}{3} = 1 \), which corresponds to a \( 45^{\circ} \) angle with the positive \( x- \) axis.
3Step 3: Compare with the motion along the line from Exercise 1
Without the specifics of Exercise 1, the comparison can only be generalized. Assuming motion along the line in Exercise 1 was also at a constant rate, the difference lies in the rate of change of \( x \) and \( y \) with respect to \( t \). If the coefficients were different in Exercise 1, the rate at which position changes along the line would differ as well. The representation gives constant speed along the line with an equal rate of change in both directions.
Key Concepts
Derivatives in Parametric EquationsSlope of a LineMotion Along a Straight Line
Derivatives in Parametric Equations
Understanding derivatives in parametric equations is crucial when dealing with the motion of objects and changes within systems described parametrically. Parametric equations represent a set of equations where each output variable, such as x and y, is expressed in terms of an independent parameter, often time (t).
To find the rate of change of one variable with respect to another, we use derivatives. For example, consider the equations x=3t and y=3t. To find the rate at which y changes with respect to x, you compute the derivatives of both with respect to t: dx/dt and dy/dt. Then, you take the ratio of these derivatives to find dy/dx, which simplifies to 1 in this case. This derivative tells us how steep the path is that an object follows along this line, which is essential for understanding its motion.
To find the rate of change of one variable with respect to another, we use derivatives. For example, consider the equations x=3t and y=3t. To find the rate at which y changes with respect to x, you compute the derivatives of both with respect to t: dx/dt and dy/dt. Then, you take the ratio of these derivatives to find dy/dx, which simplifies to 1 in this case. This derivative tells us how steep the path is that an object follows along this line, which is essential for understanding its motion.
Slope of a Line
The slope of a line is a measure of its steepness, often interpreted as the rise over the run. In traditional (x, y) coordinates, it's calculated by (change in y)/(change in x). However, in parametric equations like x=3t and y=3t, where t is the parameter, the process is slightly different.
The slope can be found by computing the derivatives of both y and x with respect to t and then dividing these derivatives to get dy/dx. A slope of 1, or a 45° angle with the x-axis, indicates a line that rises one unit for every unit it runs horizontally. This is precisely the case for the given parametric equations, confirming that they indeed represent a line through the origin with the mentioned orientation.
The slope can be found by computing the derivatives of both y and x with respect to t and then dividing these derivatives to get dy/dx. A slope of 1, or a 45° angle with the x-axis, indicates a line that rises one unit for every unit it runs horizontally. This is precisely the case for the given parametric equations, confirming that they indeed represent a line through the origin with the mentioned orientation.
Motion Along a Straight Line
When analyzing motion along a straight line in a parametric context, we are often interested in understanding the speed and direction of an object's movement. In the provided parametric equations, x=3t and y=3t, if t represents time, it implies that the object moves along the line at a constant speed. Both x and y coordinates increase at the same rate, maintaining the 1:1 ratio that corresponds to a 45° angle.
As both variables depend linearly on time, their rate of change is constant, indicating uniform motion. This differs from scenarios where the rate of change for x and y might not be the same, which would result in a curved path, non-uniform speed, or a different trajectory. The equations suggest that the motion represented is straightforward and predictable, with equal changes in position along the x and y axes over time.
As both variables depend linearly on time, their rate of change is constant, indicating uniform motion. This differs from scenarios where the rate of change for x and y might not be the same, which would result in a curved path, non-uniform speed, or a different trajectory. The equations suggest that the motion represented is straightforward and predictable, with equal changes in position along the x and y axes over time.
Other exercises in this chapter
Problem 3
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A particle moves along a curve with constant speed. Show that the acceleration is perpendicular to the velocity.
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Suppose that a particle moves along a cycloid in accordance with \(\theta=\) \(\omega t\), where \(\omega\) is constant: (a) When is the magnitude of the veloci
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