Chapter 3: Problem 24
Eliminate the parameter \(t\), write the equation in Cartesian coordinates, then sketch the graphs of the vector-valued functions. (Hint: Let \(x=2 t\) and \(y=t^{2} .\) Solve the first equation for \(x\) in terms of \(t\) and substitute this result into the second equation.) \(\mathbf{r}(t)=2(\sinh t) \mathbf{i}+2(\cosh t) \mathbf{j}, t>0\)
Short Answer
Step by step solution
Setup and Variable Substitution
Express \(t\) in terms of \(x\)
Express \(y\) in terms of \(t\)
Cartesian Equation Derivation
Simplify and Sketch
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Hyperbolic functions
- Hyperbolic sine, \(\sinh(t) = \frac{e^t - e^{-t}}{2}\).
- Hyperbolic cosine, \(\cosh(t) = \frac{e^t + e^{-t}}{2}\).
Cartesian coordinates
- Each point in this system is identified by a pair of numerical coordinates (x, y).
- This system allows us to convert complex vector equations into simpler equations, like those of a parabola or hyperbola, that can be easily graphed and understood.
Parameter elimination
- We start with equations \(x = 2 \sinh(t)\) and \(y = 2 \cosh(t)\).
- The goal is to solve one equation—for instance, solve for \(t\) in terms of \(x\) using \(x = 2 \sinh(t)\).
- Once \(t\) is expressed in terms of \(x\), substitute it into the other equation \(y = 2 \cosh(t)\).
Graphing hyperbolas
- Start with the standard form of the hyperbola equation: \(\frac{y^2}{4} - \frac{x^2}{4} = 1\). This indicates a hyperbola that opens vertically because the term involving \(y\) is positive.
- The center of this hyperbola is at the origin, (0,0).
- Identify the asymptotes, which can be found by ignoring the constant on the right side: these are the lines \(y = \pm x\).