How Do You Graph a Vector Function and Its Derivative?

[p.ella]

Homework Statement

graph the curve C that is described by r and graph r′ at the indicated value of t.

I've attached the actual question as a picture as a picture file

Homework Equations

none

The Attempt at a Solution

I know x(t)= 2, y(t)=t , z(t)=4/1+t^2

when x=0, r(t)=t+4/1+t^2

when y=0, r(t)=2+4/1+t^2

when z=0, r(t)=2+t

Beyond this I have no idea what to do :( Any help would be very much appreciated, as this is quite urgent. Thank you in advance! (:

 

[Ray Vickson]

Homework Statement

graph the curve C that is described by r and graph r′ at the indicated value of t.

I've attached the actual question as a picture as a picture file

Homework Equations

none

The Attempt at a Solution

I know x(t)= 2, y(t)=t , z(t)=4/1+t^2

when x=0, r(t)=t+4/1+t^2

when y=0, r(t)=2+4/1+t^2

when z=0, r(t)=2+t

Beyond this I have no idea what to do :( Any help would be very much appreciated, as this is quite urgent. Thank you in advance! (:

Where do you get
$$z(t) = \frac{4}{1} + t^2 \, ?$$
I see
$$z(t) = \frac{4}{1+t^2}.$$
Did you mean z(t) = 4/(1+t^2), using parentheses to make it read properly?

Anyway, you are not allowed to set x = 0, because x(t) = 2 always. You are allowed to set y = 0, but only by setting t = 0 (because y(t) = t), etc. I cannot figure out what you are trying to do. The question asked you to plot $\mathbf{r}(t)$, which is a curve in three dimensions; but since x = 2 always, it is really a plane-curve involving y and z, lying in the plane x = 2.