Question

Need to brush up on calc 2, have a few questions that I don't remember how to do. Please oblige :)

Answer 1

If C is a curve with parametric equations:
x=2-e^t
y=t^3+2t
z=t
-1<=t<=2, find parametric equations of the tangent line to C at t=1

Answer 2

@KingGeorge

Answer 3

Let me review up on my parametric curves real quick.

Answer 4

The equation of a line in parametric form is:
\[ Y = y(1)+y'(1)t\]
\[ X = x(1)+x'(1)t\]
\[ Z = z(1)+z'(1)t\]

Answer 5

But the parameter is offset, so for the tangent, t=0 is where it intersects the curve. I should have used a different parameter.

Answer 6

and
\[y^{\prime}=\frac{dy}{dt}=3t^2+2\ \ etc.\]

Answer 7

That \(z=t\) is throwing me off. I'm used to graphs in 2D.

Answer 8

But the parameter is offset, so for the tangent, t=0 is where it intersects the curve. I should have used a different parameter.
what do you mean by this?

Answer 9

So if t is time, then the original curve starts at time = -1 below the xy plane and moves up as it curves around. At t=1 we take a snapshot of the point on the curve we are at and draw a tangent line in 3D. That tangent line can also be described parametrically. If I use the parameter s, say, we could describe a line that intersects where s=0 and the same time and place where t=1 on the first curve.

Answer 10

By using t as my parameter for the second curve, I give the impression that both the curve and the tanget line are being drawn at the same time, so to speak.

Answer 11

but i still just do X=x(1)+x'(1)t, etc. right?

Answer 12

Right, I just would have rather said X=x(1)+x'(1)s, etc. to emphasize that we are using a different parameter.

Answer 13

Alternatively, we could have done it like:
X=x(1)-x'(1)+x'(1)t etc. and the curve and the tangent would have coincided at the same time and place.

Answer 14

okk i see, thanks a lot.