Solutions - 0265-01

(a) From the first equation, $t=x-3$.

Plug that in for $t$ in $y(t)$:

$$\begin{array}{c}y=4t\gg \gg y=4(x-3)\\ \\ \gg \gg y=4x-12\end{array}$$

The sketched curve should be the portion of the line with $x\in (\mathrm{3,4})$.

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(b)

For this one, best not to solve for $t$. Instead, notice the trig identity:

$$x^2+y={(\cos t)}^2+{\sin }^{2}t=1$$

Therefore the points on the curve satisfy the equation $x^2+y=1$. Solve this for the function:

$$y=1-x^2$$