0
$\begingroup$

Sketch the curve described by these parametric equations. $$\begin{align} x &=3\cos t+2 \\ y &=3\sin t -3 \end{align}$$ for $0 \leq t < 2\pi$.

I found the equation to be $$\left(\frac{y+3}{3}\right)^2+\left(\frac{x-2}{3}\right)^2 = 1$$ hence centre to be $(2,-3)$ and can sketch the graph on the cartesian plane but I don't know how to sketch it with this: $0 \leq t < 2\pi$.

Can someone explain and send an image of what it would look like? Thanks!

$\endgroup$
2
  • $\begingroup$ Do you the cartesian equation of a circle? $\endgroup$ – hamam_Abdallah May 1 '20 at 2:57
  • $\begingroup$ What? @hamam_Abdallah $\endgroup$ – Maths May 1 '20 at 2:58
1
$\begingroup$

Very straightforward, plug in values of $t$: just notice the offset:

enter image description here

Here are some points and the $t$ values that led to them:

enter image description here

Clear now??

$\endgroup$
3
  • $\begingroup$ But what do I do for the 0 ≤ t < 2π I don't quite get how that will affect the graph and what changes I make. $\endgroup$ – Maths May 1 '20 at 3:00
  • $\begingroup$ I have gotten that graph; that's easy but I don't get how to use the 0 ≤ t < 2π and how it's going to affect which parts of the graph are accepted and which aren't $\endgroup$ – Maths May 1 '20 at 3:01
  • $\begingroup$ You USE $0 \leq t \leq 2 \pi$ to get the circle. Plug in $t=0$ to find one point on the plane; then plug in $t = 0.1$, then plug in $t=.2$... well, you get the idea. $\endgroup$ – David G. Stork May 1 '20 at 3:05
0
$\begingroup$

Your equation can be written as

$$(x-2)^2+(y+3)^2=9=3^2$$

your curve is then a circle with the point $C(2,-3) $ as center and Radius $ R=3$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.