Thread: Calculus III Question

Let Τ = (t, (1+t)/(t), ((1-t^2)/(t)), t>0, and C=Trace(Τ).
Show that C lies on the plane Γ with equation: x - y + z +1 = 0

I got

T(1) = (1, 2, 0)

Plugged in:
x - y + z +1 = 0
(1-2+0+1) = 0
0=0

Then I sketched the graph x - y + z + 1 = 0 in (x,y,z)

Am I done?

can you clarify what you mean by Trace?

to sketch Τ = (t, (1+t)/(t), ((1-t^2)/(t)), t>0

yes you are done then.

Hi ErikFBueno1990!

Originally Posted by ErikFBueno1990

Let Τ = (t, (1+t)/(t), ((1-t^2)/(t)), t>0, and C=Trace(Τ).
Show that C lies on the plane Γ with equation: x - y + z +1 = 0

I got

T(1) = (1, 2, 0)

Plugged in:
x - y + z +1 = 0
(1-2+0+1) = 0
0=0

Then I sketched the graph x - y + z + 1 = 0 in (x,y,z)

Am I done?

It seems to me that you have to prove that the curve C lies in the plane Γ.
To do so for only T(1) would not suffice.
It would suffice if you also have a tangent that is perpendicular to the plane for every value of t.

Originally Posted by jakncoke

can you clarify what you mean by Trace?

Since C is a common symbol for curve and since T identifies a curve, I'm guessing that it means that C is the curve traced by T.