How to find a parametric equation for the circle with center at the point $(4,4)$ and radius $4$ starting on the $x$-axis when $t=0$?
Answer:
Parametric equations in vector form is
$\vec{r}$$(t)$=$(4 + 4 \sin t)$$\hat{i}$+$(4-4 \cos t)$$\hat{j}$
Can someone explain how to solve this problem?
We know that $(x-a)^2 + (y-b)^2 = r^2$ describes a circle in $xy$-plane with center $(a,b)$. Also we know that $\sin^2 t + \cos^2 t = 1$ holds for all $t\in \mathbb{R}$. If we set $$x = a+r\sin t \\ y = b + r\cos t$$ we have $$(x-a)^2 +(y-b)^2 = r^2(\sin^2 t + \cos^2 t) = r^2$$ which is a circle. Note that because $\sin t$ and $\cos t$ are periodic with period $T = 2\pi$, it's enough to consider $t\in [0,2\pi)$. Using the mentioned parameterization, we can control the $x-$axis and $y-$axis coordinates simultaneously with the single variable $t$.
We can get the equation of an ellipse with a small modification in the equation of the circle. Using different factors for $x$ and $y$, $$x = a+r\sin t \\ y = b + r'\cos t$$ we have $$(\frac{x-a}{r})^2 + (\frac{y-b}{r'})^2 = 1$$ which is the equation of an ellipse. Parametric equations can give us really beautiful curves, for example Lissajous curve and Butterfly curve. Also take a look at this. This example shows a helix: