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

Assignment #1                                               Problem 1.4.4

Joint Revision

Consider the curve β(t) = (t5, t3) in R2.

(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) = (5t4, 3t2).

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(t0) + (t-t0)f’(t0)], 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 = t3

and substituting this into β(t) to get

β(τ) = (τ5/3, τ).

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

β’(τ) = (5/3 τ2/3, 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 = t3, τ = 03 = 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).