Calculus....Length of a Plane Curve?

Question

Consider the parametric curve given by the equations

x(t) = t^2 + 23t - 8
y(t) = t^2 + 23t + 33

How many units of distance are covered by the point
P(t) = (x(t), y(t)) between t = 0, and t = 2?

Answer 1

x(t) = t^2 + 23t - 8
So:
dx = 2tdt +23dt

y(t) = t^2 +23t + 33
So:
dy = 2tdt + 23dt

Therefore,
dx^2 + dy^2 = 2*(2t + 23)^2 dt^2

So ds = sqrt(dx^2 + dy^2) = sqrt(2) (2t+23) dt

Length = Integral (t = 0, 2) [sqrt(2)(2t + 23)dt]
= sqrt(2) * Integral (t = 0, 2) [(2t + 23)dt]
= sqrt(2)*[t^2 + 23t] (t=0, 2)
= sqrt(2)*[2^2 +23*2]
= sqrt(2) (4 + 46)
= 50 sqrt(2)

Now that we've gotten the answer, it occurs to me there's a much easier way to see this:
Let A(t) = t^2 + 23t
Then notice that:
x(t) = A(t) - 8
y(t) = A(t) + 33

We shift coordinates:
x'(t) = x(t) + 8 = A(t)
y'(t) = y(t) - 33 = A(t)

Because x' and y' are just shifted from x and y, they are just as good for calculating the length of the curve.

But now look what's happening: x'(t) = y'(t), so the curve is a straight line. So, regardless of the parameterization in terms of t, the curve is going from the beginning point at (0,0) to the end point at (50, 50) [because (2^2 + 23*2) = 4 + 46 = 50 ], in a straight line.
So the curve is the hypotenuse of a right triangle with two equal sides of length 50, so the answer is of course 50*sqrt(2).

Answer 2

dy = (2t+23)dt
dx= (2t+23)dt
so dy/dx=1
L = Int(0,2) sqrt(1+1^2)dt) = 2sqrt(2)