Question

Consider the curve parameterized by x=6sin(θ),y=3cos(θ),0≤θ≤π. Eliminate the parameter to find a Cartesian equation for the curve: ______ where x∈ ___ and y∈ _____ (Use interval notation for the restrictions on x and y and include only those values of x and y that make up the curve.)

Answer 1

`y=(1)/(2)√(36-x^2)`

`x \in [-6,6]`

`y \in [0,3]`

Explanation:

Given parametric equations:

`x=6\sin\theta`

`y=3\cos\theta`

To eliminate the parameter θ, rewrite x in terms of cosθ, then substitute this into the second equation to find a Cartesian equation for the curve.

Rearrange the first equation to isolate sinθ:

`\sin \theta=(x)/(6)`

Substitute this into the trigonometric identity sin²θ + cos²θ = 1, then rearrange to isolate cosθ:

`\begin{aligned}\sin^2\theta+\cos^2\theta &=1\\\\\cos^2\theta &=1-\sin^2\theta\\\\\cos \theta &=\pm√(1-\sin^2\theta)\\\\\cos \theta &=\pm\sqrt{1-\left((x)/(6)\right)^2}\\\\\cos \theta &=\pm\sqrt{1-(x^2)/(36)}\\\\\cos \theta &=\pm\sqrt{(36-x^2)/(36)}\\\\\cos \theta &=\pm(√(36-x^2))/(6)\\\\\end{aligned}`

Substitute the expression for cosθ into the equation for y:

`\begin{aligned}y&=3 \cos \theta\\\\y&=3 \cdot \pm(√(36-x^2))/(6)\\\\y&=\pm (√(36-x^2))/(2)\\\\y&= \pm(1)/(2)√(36-x^2)\end{aligned}`

Since 0 ≤ θ ≤ π, we know that θ is in the first and second quadrants (above the x-axis). Therefore, the Cartesian equation for the curve is:

`\large\boxed{y=(1)/(2)√(36-x^2)}`

The expression under the square root sign should always be greater than or equal to zero. Therefore:

`\begin{aligned}√(36-x^2)&\geq 0\\\\36-x^2&\geq0\\\\36&\geq x^2\\\\x^2&\leq 36\end{aligned}`

`\textsf{For\;$u^2\leq a$,\;if\;$n$\;is\;even\;then\;$-√(a)\leq u\leq√(a)$.}`

Therefore:

`-√(36)\leq x \leq √(36)=-6 \leq x \leq 6`

So the domain (values of x) is limited to [-6, 6].

When x = ±6, y = 0. Therefore, the minimum value of y is zero.

The maximum value of y will be the value when the value under the square root sign is at its maximum, so when x = 0:

`\begin{aligned}y&=(1)/(2)√(36-0^2)\\\\y&=(1)/(2)√(36)\\\\y&=(1)/(2)\cdot 6\\\\y&=3\end{aligned}`

So the range (values of y) is limited to [0, 3].

Therefore:

`\large\boxed{\textsf{$y=(1)/(2)√(36-x^2)$\;\;where\;\;$x \in [-6,6]$\;\;and\;\;$y \in [0,3]$}}`