Problem 6
Question
Sketch the curve represented by the parametric equations (indicate the orientation of the curve), and write the corresponding rectangular equation by eliminating the parameter. $$ x=\sqrt[4]{t}, \quad y=3-t $$
Step-by-Step Solution
Verified Answer
The graph of the curve represents a function that starts at the point (0,3) and moves to the right and downwards. The corresponding rectangular equation for the given parametric equations is \(y = 3 - x^4\).
1Step 1: Understanding the Parametric Equations
Given the pair of parametric equations \(x = t^{1/4}\) and \(y = 3 - t\), where 'x' and 'y' are both expressed in terms of the parameter 't'. 'x' represents the fourth root of 't', demonstrating that 't' must be non-negative to achieve real values of 'x'. The equation for 'y' demonstrates that as 't' increases, 'y' decreases, indicating downward orientation of the curve.
2Step 2: Sketching the Curve
Begin at the point where \(t = 0\), which gives us the point \(x = 0\) and \(y = 3\). As 't' increases, 'x' increases due to the relation \(x = t^{1/4}\), while 'y' decreases because 'y' depends on the term \(3 - t\). As such, sketch a curve that originates at (0,3) and decreases moving to the right.
3Step 3: Elimination of the Parameter t
To find the rectangular equation that corresponds to the parametric equations, get rid of the parameter 't'. In this case, it is easier to solve the equation \(x = t^{1/4}\) for 't', giving \(t = x^4\). Then, replace 't' in the equation \(y = 3 - t\) with the newly found value to get \(y = 3 - x^4\). This is the required rectangular equation.
Key Concepts
Curve SketchingRectangular EquationEliminating the ParameterCalculus
Curve Sketching
Curve sketching is a method used to graphically represent the behavior of a curve based on its mathematical equations. In this case, the curve is defined by the parametric equations for both 'x' and 'y'.
The orientation of the curve shows how it moves on the plane as the parameter 't' changes.
The orientation of the curve shows how it moves on the plane as the parameter 't' changes.
- Start by placing a point at the initial value when the parameter 't' is zero. For these equations, the initial point is \( (0, 3) \).
- As 't' increases, the value of 'x'—calculated from \( t^{1/4} \)—also increases because the fourth root of 't' is always positive.
- The 'y' coordinate \( y = 3 - t \) decreases as 't' increases. This means that the curve moves downwards as it moves to the right, creating a rightward downward orientation.
Rectangular Equation
A rectangular equation is an algebraic expression that represents a curve using variables 'x' and 'y', without any parameters involved.
In many problems involving parametric equations, one goal is to translate them into a rectangular form. This form tends to be more familiar since it's what most students encounter in basic algebra courses.
In the current exercise, we translated the parametric equations from 'x = t^{1/4}' and 'y = 3 - t' into the rectangular form 'y = 3 - x^4'.
This simplification is crucial because:
In many problems involving parametric equations, one goal is to translate them into a rectangular form. This form tends to be more familiar since it's what most students encounter in basic algebra courses.
In the current exercise, we translated the parametric equations from 'x = t^{1/4}' and 'y = 3 - t' into the rectangular form 'y = 3 - x^4'.
This simplification is crucial because:
- It expresses the relationship directly between 'x' and 'y'.
- The behavior and path of the curve become easier to analyze and predict.
Eliminating the Parameter
Eliminating the parameter is the process of removing the variable that both parametric equations depend on, in this case 't'.
This technique is useful because it simplifies the analysis of curves by expressing them purely in terms of 'x' and 'y'.
To eliminate the parameter 't' here, begin with solving for 't' from one equation. In this scenario, from 'x = t^{1/4}', it's expressed as 't = x^4'.
Then, substitute 't = x^4' into the other equation 'y = 3 - t'. This provides us with:
This technique is useful because it simplifies the analysis of curves by expressing them purely in terms of 'x' and 'y'.
To eliminate the parameter 't' here, begin with solving for 't' from one equation. In this scenario, from 'x = t^{1/4}', it's expressed as 't = x^4'.
Then, substitute 't = x^4' into the other equation 'y = 3 - t'. This provides us with:
- A clear rectangular equation form: 'y = 3 - x^4'.
Calculus
Calculus plays a significant role in analyzing curves, particularly when they are written in their rectangular form.
With the equation 'y = 3 - x^4', calculus can help determine various properties such as rates of change, concavity, or even areas under the curve.
Using derivatives, one can find the slope of the curve at any given point, which aides in sketching more accurate graphs:
With the equation 'y = 3 - x^4', calculus can help determine various properties such as rates of change, concavity, or even areas under the curve.
Using derivatives, one can find the slope of the curve at any given point, which aides in sketching more accurate graphs:
- The first derivative of 'y = 3 - x^4' gives the slope of the tangent line at any point along the curve.
- The second derivative informs us about the concavity of the curve. Whether it is opening upwards or downwards.
Other exercises in this chapter
Problem 5
Find \(d y / d x\) and \(d^{2} y / d x^{2},\) and find the slope and concavity (if possible) at the given value of the parameter. $$ x=2 t, y=3 t-1 \quad t=3 $$
View solution Problem 5
Use the angle feature of a graphing utility to find the rectangular coordinates for the point given in polar coordinates. Plot the point. $$ (5,3 \pi / 4) $$
View solution Problem 6
Find the area of the region. One petal of \(r=6 \sin 2 \theta\)
View solution Problem 6
Find \(d y / d x\) and \(d^{2} y / d x^{2},\) and find the slope and concavity (if possible) at the given value of the parameter. $$ x=\sqrt{t}, y=3 t-1 \quad t
View solution