Question

the ellipse \\(\frac{x^2}{6^2} + \frac{y^2}{3^2} = 1\\) can be drawn with parametric equations where \\(x(t)\\) is written in the form\\(x(t) = r\cos(t)\\) with \\(r = \square\\)and \\(y(t) = \square\\)question help: \\(\boxed{\text{video}}\\)

Explanation:

Step1: Recall Ellipse Parametric Form

The standard parametric equations for an ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) are \(x = a\cos(t)\) and \(y = b\sin(t)\).

Step2: Identify \(r\) for \(x(t)\)

Given the ellipse \(\frac{x^2}{6^2}+\frac{y^2}{3^2}=1\), comparing with the parametric form \(x(t)=r\cos(t)\), we see that \(r = a\) (where \(a = 6\) from the ellipse equation). So \(r = 6\).

Step3: Determine \(y(t)\)

Using the parametric form for the ellipse, with \(b = 3\) (from \(\frac{y^2}{3^2}\)), the equation for \(y(t)\) is \(y(t)=3\sin(t)\) (since the standard parametric equation for \(y\) in an ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) is \(y = b\sin(t)\)).

Answer:

For \(r\), the value is \(6\). For \(y(t)\), the expression is \(3\sin(t)\). So \(r=\boxed{6}\) and \(y(t)=\boxed{3\sin(t)}\)