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Ch.12 - Parametric and Polar Curves
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 12, Problema 12.R.12a

10–12. Parametric curves
a. Eliminate the parameter to obtain an equation in x and y.
x = ln t, y = 8ln t², for 1 ≤ t ≤ e²; (1, 16)

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1
Start with the given parametric equations: \(x = \ln t\) and \(y = 8 \ln t^{2}\), where \(1 \leq t \leq e^{2}\).
Recall the logarithm property: \(\ln t^{2} = 2 \ln t\). Use this to rewrite \(y\) as \(y = 8 \times 2 \ln t = 16 \ln t\).
Since \(x = \ln t\), substitute \(\ln t\) in the expression for \(y\) to get \(y = 16x\).
This gives the Cartesian equation relating \(x\) and \(y\) without the parameter \(t\): \(y = 16x\).
Note the domain for \(t\) translates to \(x\) because \(x = \ln t\). Since \(1 \leq t \leq e^{2}\), then \(0 \leq x \leq 2\).

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