Is the tangent line really a point when the slope is zero?

[hellsingfan]

Homework Statement

I was given parametric equations.

x(t) = a(t)
y(t) = b(t)
z(t) = c(t)

where a, b, and c are functions that depend on t.

I was supposed to find equation of the tangent line at t = f given:

x(f)= m
y(f) = n
z (f) = o

where m,n,o are some constant numbers

and given
x'(f)=y'(f)=z'(f)= 0

Homework Equations

I think this is the relevant equation though its not given.
r(t) = <Px,Py,Pz>+tv

The Attempt at a Solution

I'm super confused here. Given that the derivative of x(t),y(t),z(t) are all 0 at t=f. Then There is no slope whatsoever. That means its basically 3 intersecting planes where x=m, y=n, and z=o. Which isn't a line but a point.

If I use the equation for a line r(t) = <Px,Py,Pz>+tv then,

<Px,Py,Pz> = <m,n,o>
and v = <0,0,0>

so r(t) = <m,n,o>+t<0,0,0> = <m,n,o>

But <m,n,o> is a vector with direction, while I already know that the tangent is only a point (m,n,o) given by the intersection of planes x=m, y=n, z=o.

Is <m,n,o> vector really the tangent line, I feel like there is no 'line' as its a point.

 

[Simon Bridge]

I think you need to infer that the surface is not discontinuous at t=f - or the problem does not make sense.
To help you understand the situation - have a play around with the conditions:
i.e. x=t-f+m, y=(t-f)^2+n, z=(t-f)^3+o ... what's the tangent for t=f?

If you have a 3D plotting program, pick some values for f,m,n,o and see what the surface looks like.