Question

Find real numbers a and b such that equations \(A\) and \(B\) describe the same curve. \(A: x=t+t^{3}, y=3+t^{2} ;-2 \leq t \leq 2\) \(B: x=t^{1 / 3}+t, y=3+t^{2 / 3} ; a \leq t \leq b\)

Short answer

The values of a and b are: a = -1 and b = 1.

Step-by-step solution

Understand equation A

Equations A are given as \(x = t + t^3\) \(y = 3 + t^2\) with the parameter constraints \(-2 \leq t \leq 2\). These equations describe a parametric curve, where x and y are expressed as functions of the parameter t.

Understand equation B

Equations B are given as \(x = t^{\frac{1}{3}} + t\) \(y = 3 + t^{\frac{2}{3}}\) with the parameter constraints \(a \leq t \leq b\). Like in equation A, these equations describe a parametric curve, where x and y are expressed as functions of the parameter t.

Find the relationship between the parameters

We need to find the relationship between the parameters in both sets of equations, i.e., find a transformation that takes us from one set of equations to the other. For this, we examine the x and y expressions in both A and B: For the x-equations: \(t + t^3 = t^{\frac{1}{3}} + t\) Next, we eliminate t: \(t^3 - t^{\frac{1}{3}} + t - t = t^3 - t^{\frac{1}{3}}\) Thus, we have \(t^3 = t^{\frac{1}{3}}\) Now, coming to the y-equations: \(3 + t^2 = 3 + t^{\frac{2}{3}}\) Subtract 3 from both sides: \(t^2 = t^{\frac{2}{3}}\)

Solve for the transformation relationship

Since we have found the relationship \(t^3 = t^{\frac{1}{3}}\) and \(t^2 = t^{\frac{2}{3}}\), we can relate the parameter t in equation A to the parameter t in equation B by solving the following equation: \(t^3 = t^{\frac{1}{3}} \rightarrow t^{\frac{9}{3}} = t^{\frac{1}{3}} \rightarrow t^9 = t\) Dividing both sides by t gives us \(t^8 =1\). To find t, take the 8th root on both sides: \(t = \pm1\). So, we found that when \(t_A = \pm1\), the corresponding \(t_B\) values are also \(\pm1\).

Find the values of a and b

Now that we have found the transformation relationship between the parameter t in equation A and B, we can find the values of a and b. Since \(t_A = \pm1\), by the transformation, \(t_B = \pm1\) as well. Also, we know that in equation B, \(a \leq t \leq b\). So, we have \(a \leq (-1) \leq b\) and \(a \leq 1 \leq b\). Thus, \(a = -1\) and \(b = 1\).

Conclusion

The real numbers a and b such that equations A and B describe the same curve are \(a = -1\) and \(b = 1\).

Key concepts

Transformation of Parameters

In parametric equations, a transformation of parameters allows us to understand how two seemingly different sets of equations may actually describe the same curve. By exploring how one parameter in set A can relate to a parameter in set B, we unravel the hidden connection between them. In the given problem, both equation sets A and B describe a curvilinear path but use different expressions for x and y, signifying a transformation.

• For curve A: Parameters expressed as \(x = t + t^3\) and \(y = 3 + t^2\).
• For curve B: Parameters expressed as \(x = t^{\frac{1}{3}} + t\) and \(y = 3 + t^{\frac{2}{3}}\).

The goal is to align these parameters. The transformation is essentially finding equivalence: we found that \(t^3 = t^{\frac{1}{3}}\) and \(t^2 = t^{\frac{2}{3}}\). Solving these gives insights into how the t-values morph from one description to another. Ultimately, it helps to redefine the interval for parameters a and b to maintain the equivalence for the same curve.

Solving Equations

Solving equations is a crucial step in understanding how different parametric equations can describe the same curve. It involves using algebraic manipulations to find the relationships between different expressions of the parameters involved.When we solved the given equations:- We equated the expressions from both equations A and B: \(t + t^3 = t^{\frac{1}{3}} + t\) for x, which simplifies to \(t^3 = t^{\frac{1}{3}}\) after simplification.- Similarly, for the y component: \(3 + t^2 = 3 + t^{\frac{2}{3}}\) simplifies to \(t^2 = t^{\frac{2}{3}}\).Solving these gives us insight into how each t maps from one equation to another. Specifically, the step of dividing and taking root showed that \(9 = 1\), since \(t^8 = 1\), revealing solutions like \(t = \pm 1\). This detailed process of ruling out and verifying potential solutions ensures accuracy in finding the equivalent parameter values needed for identical parametric curves.

Parametric Curves

Parametric curves allow us to represent geometric shapes through equations dependent on a parameter, usually denoted as t. These curve equations express both x and y coordinates in terms of the parameter, enabling us to plot complex paths more flexibly than standard Cartesian equations.For the given exercise, we have two parametric curves:

• Curve A: defined by \(x = t + t^3\) and \(y = 3 + t^2\), which describes a curve parameterized by -2 to 2.
• Curve B: defined by \(x = t^{\frac{1}{3}} + t\) and \(y = 3 + t^{\frac{2}{3}}\), requiring identification of suitable limits \(a\) and \(b\).

Parametric curves are particularly useful in scenarios where the movement along the curve is not uniform or straight. Their ability to illustrate flexible trajectories makes them invaluable for accurately capturing curves that standard forms may not easily express. In our solution, by aligning parameters for A and B, we achieved a seamless depiction of the same curve by defining exact parameter ranges: \(a = -1\) and \(b = 1\). This effectively uses the power of parametric representation to unearth the shared identity of different mathematical descriptions.