Answer:
a) [tex]x = 4\cdot \cos t[/tex], [tex]y = 1 + 4\cdot \sin t[/tex], b) [tex]x = 4\cdot \cos t[/tex], [tex]y = 1 + 4\cdot \sin t[/tex], c) [tex]x = 4\cdot \cos \left(t+\frac{\pi}{2} \right)[/tex], [tex]y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)[/tex].
Step-by-step explanation:
The equation of the circle is:
[tex]x^{2} + (y-1)^{2} = 16[/tex]
After some algebraic and trigonometric handling:
[tex]\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = 1[/tex]
[tex]\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = \cos^{2} t + \sin^{2} t[/tex]
Where:
[tex]\frac{x}{4} = \cos t[/tex]
[tex]\frac{y-1}{4} = \sin t[/tex]
Finally,
[tex]x = 4\cdot \cos t[/tex]
[tex]y = 1 + 4\cdot \sin t[/tex]
a) [tex]x = 4\cdot \cos t[/tex], [tex]y = 1 + 4\cdot \sin t[/tex].
b) [tex]x = 4\cdot \cos t[/tex], [tex]y = 1 + 4\cdot \sin t[/tex].
c) [tex]x = 4\cdot \cos t''[/tex], [tex]y = 1 + 4\cdot \sin t''[/tex]
Where:
[tex]4\cdot \cos t' = 0[/tex]
[tex]1 + 4\cdot \sin t' = 5[/tex]
The solution is [tex]t' = \frac{\pi}{2}[/tex]
The parametric equations are:
[tex]x = 4\cdot \cos \left(t+\frac{\pi}{2} \right)[/tex]
[tex]y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)[/tex]
