<p>Sketch the curves given parametrically by the following equations and find the Cartesian equations of these curves:</p>
<ol class="abc-list"><li><div class="list-item-paragraph">$x=1-t$, $y=2{{t}^{2}}$, where $t<1$</div></li><li><div class="list-item-paragraph">$x=\sin \theta +1$, $y=\frac{1}{2}\cos \theta $, where $-\frac{\pi }{2}\le \theta <0$</div></li><li><div class="list-item-paragraph">$x=\tan t$, $y=\sin t$, $0 < t < \frac{\pi }{2}$</div></li><li><div class="list-item-paragraph">$x=\frac{1}{2}\left( {{\mathrm{e}}^{3t}}+2{{\mathrm{e}}^{-3t}} \right)$, $y=\frac{1}{2}\left( {{\mathrm{e}}^{3t}}-2{{\mathrm{e}}^{-3t}} \right)$</div></li></ol>

Sketch the curves given parametrically by the following equations and find the Cartesian equations of these curves:

$x=\sin \theta +1$, $y=\frac{1}{2}\cos \theta $, where $-\frac{\pi }{2}\le \theta <0$

$x=\tan t$, $y=\sin t$, $0 < t < \frac{\pi }{2}$

$x=\frac{1}{2}\left( {{\mathrm{e}}^{3t}}+2{{\mathrm{e}}^{-3t}} \right)$, $y=\frac{1}{2}\left( {{\mathrm{e}}^{3t}}-2{{\mathrm{e}}^{-3t}} \right)$

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NYJC Graphing Techniques Tutorial Q8

$x = 1 - t$, $y = 2t^2$

When $t = 0$: $x = 1, y = 0 \Rightarrow (1, 0)$
When $t = 1$: $x = 0, y = 2 \Rightarrow (0, 2)$
When $y = 0$: $2t^2 = 0 \Rightarrow t = 0$, $x = 1$

From $x = 1 - t \Rightarrow t = 1 - x$ ... ①
$y = 2t^2$ ... ②

Substitute ① into ②:

Cartesian form: $y = 2(1 - x)^2$, $x \geq 0$

$x = \sin\theta + 1$, $y = \frac{1}{2}\cos\theta$

When $\theta = 0$: $(1, \frac{1}{2})$
When $\theta = -\frac{\pi}{2}$: $(0, 0)$

$\sin\theta = x - 1$, $\cos\theta = 2y$

$\sin^2\theta + \cos^2\theta = 1$
\[(x - 1)^2 + (2y)^2 = 1\]

Cartesian equation: $(x - 1)^2 + 4y^2 = 1$, $0 \leq x \leq 1$, $y \geq 0$

$x = \tan t$, $y = \sin t$

$x = \dfrac{\sin t}{\cos t}$
$x\cos t = \sin t$
$\cos t = \dfrac{y}{x}$ ... ①

$\sin^2 t + \cos^2 t = 1$ ... ②

Substitute $\sin t = y$ and ① into ②:
\[y^2 + \left(\frac{y}{x}\right)^2 = 1\]
\[y^2 + \frac{y^2}{x^2} = 1\]
\[y^2\left(1 + \frac{1}{x^2}\right) = 1\]
\[1 + \frac{1}{x^2} = \frac{1}{y^2}\]

$\therefore \dfrac{1}{y^2} - \dfrac{1}{x^2} = 1$, $0 < y < 1$, $x > 0$

The curve has a horizontal asymptote $y = 1$.

$x = \frac{1}{2}(e^{3t} + 2e^{-3t})$, $y = \frac{1}{2}(e^{3t} - 2e^{-3t})$

When $y = 0$: $\frac{1}{2}(e^{3t} - 2e^{-3t}) = 0$
$e^{3t} = 2e^{-3t}$
$e^{6t} = 2$
$6t = \ln 2$
$t = \frac{1}{6}\ln 2$
$x = \frac{1}{2}\left(\sqrt{2} + \frac{2}{\sqrt{2}}\right) = \frac{1}{2}(\sqrt{2} + \sqrt{2}) = \sqrt{2}$

So the curve passes through $(\sqrt{2}, 0)$.

Cartesian form:
$2x = e^{3t} + 2e^{-3t}$ ... ①
$2y = e^{3t} - 2e^{-3t}$ ... ②

① + ②: $2x + 2y = 2e^{3t}$
$x + y = e^{3t}$ ... ③

From ①: $x = \frac{1}{2}\left(e^{3t} + \frac{2}{e^{3t}}\right)$

Substitute ③:
$x = \frac{1}{2}\left((x + y) + \frac{2}{x + y}\right)$
$2x = (x + y) + \frac{2}{x + y}$
$2x(x + y) = (x + y)^2 + 2$
$2x^2 + 2xy = x^2 + 2xy + y^2 + 2$

$x^2 - y^2 = 2$, $x > 0$

NYJC Graphing Techniques Tutorial Q8

NYJC Graphing Techniques Tutorial Q8

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