Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

jrodriguez2315

  • 4 years ago

Let C represent the curve generated by the position function r(t)=((t^2)-2t, t+1,(t^2)+t-2) for -infinty<t<infinity. Is C a plane curve? Justify your answer analytically.

  • This Question is Closed
  1. JunkieJim
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    is \[ \frac{d\overrightarrow{r}(t)}{dt}\]a constant or a function of t? let's find out. \[\frac{d\overrightarrow{r}(t)}{dt}= (\frac{dx}{dt} , \frac{dy}{dt} , \frac{dz}{dt})\] \[=(2t-2 , 1 ,2t+t)\]which is still a function of t, so the slope of the curve is changing in the x direction and the z direction. because of this the curve cannot be a plane curve

  2. JunkieJim
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    woops, the z component of the derivative should be 2t + 1

  3. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy