Use your calculator to find the length of the arc from t = 0 to t = 3 of x = 2t + 1, y = t2 - 1.

Question

Use your calculator to find the length of the arc from t = 0 to t = 3 of x = 2t + 1, y = t² - 1.

Answer 1

Since x = 2t +1 then t = (x - 1)/2 and plugging that into y = t² - 1 we get y = (x - 1)²/4 - 1

y = 1/4(x² - 2x + 1) - 1

y = 1/4x² - 1/2x + 1/4 - 1

y = 1/4x² - 1/2x - 3/4

Arc length is _a∫^b√(1 + [f '(x)]²)dx

Since y = 1/4x² - 1/2x - 3/4, then y' = 1/2x - 1/2 and (y')² = 1/4x² - 1/2x + 1/4

Since we are integrating in "x" and our limits of integration are in "t" we must convert them. When t = 0, x = 1 (x = 2t + 1) and when t = 3, x = 7

So the arc length is ₁∫⁷√(1 +1/4x² - 1/2x + 1/4)dx = ₁∫⁷√(1/4x² - 1/2x + 5/4)dx

Plugging this into my calculator I get arc length = 24