Answer to Question #114589 in Calculus for Bibhuti bhusan dehury

Question

Trace the curve x = a sin 2t (1+cos2t),
y=acos2t(1-cos2t), 0<=t<=π

Answer 1

Let us determine the properties of the curve. First, we rewrite the formulae in form

$\begin{cases} x(t) =2a\sin(2t)\cos^2t, \\ y(t) =2a\cos(2t)\sin^2t. \end{cases}$

Or in form

$\begin{cases} x(t) =4a\sin(t)\cos^3t, \\ y(t) =2a(1-2\sin^2t)\sin^2t. \end{cases}$

We know that $\cos(\pi-t) = -\cos t , \;\; \sin(\pi-t) = \sin t$. Therefore,

$\begin{cases} x(\pi-t) = -x(t), \\ y(\pi-t) = y(t). \end{cases}$

We see that the curve is is symmetrical about y-axis, and we should determine the shape of curve on $0≤t≤\dfrac{π}{2}$ .

The curve passes through (0,0) if $t=0.$

Let us determine maximum and minimum values of y. Minimum value of y will be at $t=\dfrac{\pi}{2},$ because $\sin^2t$ has the maximum value. So, $y(\frac{\pi}{2}) = -2a, \; x(\frac{\pi}{2})=0.$

If we take a derivative of $y(t)$ we get

$y'(t) = -4a\cos(t)\sin(t)(2\sin(t)-1)(2\sin(t)+1)$.

On $0\le t \le \dfrac{\pi}{2}$ extrema are located at $0, \dfrac{\pi}{2}$ (see above) and $\dfrac{\pi}{6}$ (maximum, $x=\dfrac{3\sqrt3}{4}a, \; y = \dfrac{a}{4}$ ), because at these points $y'(t)=0.$

Next let us determine maximum and minimum values of x on $0≤t≤\dfrac{π}{2}$ . We know that the curve passes through (0,0), so minimum value of x is 0. If we take the derivative of $x(t)$, we get

$x'(t) = 4a\cos^2t(\cos^2t - 3\sin^2t) = 4a\cos^2t(1-4\sin^2t).$

$x'(t)=0$ at $t=\dfrac{\pi}{2}$ (minimum, x=0) and $t=\dfrac{\pi}{6}.$ We see that for $t=\dfrac{\pi}{6}$ both x and y have their maximum values.

Next, let us determine $x$ values when $y=0.$ We can see that $y(t)=0$ if $\sin t = 0$ or $\sin^2 = \dfrac12.$ On $0\le t \le \dfrac{\pi}{2}$ this means that $t=0$ or $t=\dfrac{\pi}{4}.$ If $t=0$ , we get $x=y=0$ . If $t=\dfrac{\pi}{4}$ , we get $x=a.$

Now let us write points we have calculated above:

$x \quad \quad \quad y \\ -----\\ 0 \quad \quad \quad 0 \\ a \quad \quad \quad 0\\ \dfrac{3\sqrt3}{4}a \quad \dfrac{a}{4} \\ 0 \quad \quad -2a$

We may also calculate the derivative $\dfrac{dy}{dx} = \dfrac{-4a\cos t\sin t(4\sin^2t-1)}{4a\cos^2t(1-4\sin^2t)} = \tan t.$ The derivative if positive on $0\le t \le \dfrac{\pi}{2}$ and it is an increasing function and equals to 0 at $(x,y) = (0,0).$ We'll use this information when creating a graph.