Chapter 4: Problem 23
The parametric equations describe the motion of a particle. Find an equation of the curve along which the particle moves. $$\begin{aligned}&x=t+4\\\&y=t^{2}-3\end{aligned}$$
Short Answer
Expert verified
The particle moves along the curve \(y = x^2 - 8x + 13\).
Step by step solution
01
Identify the parametric equations
We are given the parametric equations \(x = t + 4\) and \(y = t^2 - 3\). Our goal is to find a relationship between \(x\) and \(y\) that eliminates \(t\).
02
Solve for t from x
Take the first equation \(x = t + 4\) and solve for \(t\): \[ t = x - 4 \]
03
Substitute t into the equation for y
Substitute the expression for \(t\) from Step 2 into the second parametric equation \(y = t^2 - 3\):\[ y = (x - 4)^2 - 3 \]
04
Simplify the equation
Expand and simplify the expression obtained in Step 3:\[ y = (x - 4)^2 - 3 = x^2 - 8x + 16 - 3 \]\[ y = x^2 - 8x + 13 \]
05
Write the equation of the curve
The equation of the curve in terms of \(x\) and \(y\) is:\[ y = x^2 - 8x + 13 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equation of Curve
In mathematics, an equation of a curve is crucial for simplifying complex parametric forms into more usable expressions. For a given particle's path described by parametric equations, we aim to find a relationship between coordinates that does not involve the parameter. Here, we derived the curve equation from the parametric forms by eliminating the parameter. This results in a functional relationship between the dependent variable, in this case, \( y \), and the independent variable, \( x \).
- The initial equation \( x = t + 4 \) informed us how \( t \) can be expressed in terms of \( x \).
- Replacing \( t \) in the second equation \( y = t^2 - 3 \) allowed us to solve for \( y \) purely in terms of \( x \).
Particle Motion
When examining particle motion through parametric equations, each equation typically represents one dimension of movement. In our exercise:- The first equation, \( x = t + 4 \), depicts horizontal motion.- The second equation, \( y = t^2 - 3 \), depicts vertical motion. Utilizing both, we can track the particle's path over time. This approach is incredibly beneficial because it allows:
- Visualization of complicated trajectories without initial plotting.
- Detailed insight into spatial dynamics, predicting where the particle will be at any given time.
Coordinate Geometry
Coordinate Geometry, or co-geom for short, involves examining curves, lines, and other geometrical figures using algebraic equations in a coordinate plane. The essence of coordinate geometry shines in exercises where you need to transform parametric forms into rectangular equations.Our given task involved such a transformation:
- From parametric equations, transferring the data into a single \( y = f(x) \) format provides insights into properties like intercepts and symmetries.
- This conversion helps in graphing the curve quickly on a Cartesian plane, enabling easier analysis.
Quadratic Equation
A Quadratic Equation is a polynomial equation of the form \( ax^2 + bx + c = 0 \). In the context of our problem, converting parametric equations into a quadratic helps link particle motion to a specific trajectory.The resultant equation \( y = x^2 - 8x + 13 \) is quadratic since the highest degree of \( x \) is 2. Properties of quadratic equations, such as:
- The shape of the graph being a parabola, open upwards in this case due to the positive leading coefficient.
- The vertex, axis of symmetry, and direction of opening can all be discerned from the standard form.