Ashley
Pagnotta & Hae Wong Yang a-pagnotta & h-yang1

Assignment #1
Problem 1.4.4

Joint Revision

Consider the curve β(t) = (t^{5}, t^{3}) in **R**^{2}.

(a) Show that at the point
where t = 0, the equations of this section define a tangent vector but not a
tangent line.

To find the tangent vector we first must take the derivative of
the vector-valued function β(t). Taking *dx/dt* and *dy/dt*, we get

β’(t)
= (5t^{4}, 3t^{2}).

By plugging in t = 0 to β’(t), we
can find the tangent vector at that point.
In this case, β’(0) = (0,0). However, when we plug this into the equation
normally used to find the tangent line [**r**
= **f**(t_{0})
+ (t-t_{0})**f**’(t_{0})],
we do not get a line, as shown here:

**r** = **β**(0) + (t-0)** β**’(0)

**= **(0,0)
+ (t-0)(0,0)

**= **(0,0)

Normally, this equation approximates the change in **f**(t) (in this
particular problem it is β(t) instead of **f**(t)); this is true here also.
Getting the zero vector for an answer tells us that at t = 0, the
function β(t) is approximately constant.

(b) Find a reparameterization of the curve
(define a new variable τ = ρ(t) via some
increasing function ρ) that enables the tangent line at the origin to be
constructed in the usual way.

With the current parameterization, the function is changing too
slowly to detect a tangent line. To
compensate for this, we can reparameterize β(t) using the function

τ
= t^{3}

and substituting this into β(t)
to get

β(τ)
= (τ^{5/3}, τ).

Following
the same procedure of Part (a), we take the derivative of β(τ)
to get

β’(τ)
= (5/3 τ^{2/3}, 1).

We
must now find what τ is equal to when t = 0 (the time at which we want to
find the tangent line). Since τ = t^{3},
τ = 0^{3} = 0. We plug in
τ = 0 to get β’(0) = (0,1). This enables us to find a tangent line at the
origin using the formula mentioned in Part (a).