Calculus AB and Calculus BC
CHAPTER 7 Applications of Integration to Geometry
C. ARC LENGTH
If the derivative of a function y = f (x) is continuous on the interval a x b, then the length s of the arc of the curve of y = f (x) from the point where x = a to the point where x = b is given by
Here a small piece of the curve is equal approximately to
As Δx → 0, the sum of these pieces approaches the definite integral above.
If the derivative of the function x = g(y) is continuous on the interval c ≤ y ≤ d, then the length s of the arc from y = c to y = d is given by
If a curve is defined parametrically by the equations x = x(t) and y = y(t), if the derivatives of the functions x(t) and y(t) are continuous on |ta, tb], (and if the curve does not intersect itself), then the length of the arc from t = ta to t = tb is given by
The parenthetical clause above is equivalent to the requirement that the curve is traced out just once as t varies from ta to tb.
As indicated in Equation (4), formulas (1), (2), and (3) can all be derived easily from the very simple relation
and can be remembered by visualizing Figure N7–21.
Find the length, to three decimal places, of the arc of y = x3/2 from x = 1 to x = 8.
Find the length, to three decimal places, of the curve (x − 2)2 = 4y3 from y = 0 to y = 1.
SOLUTION: Since x − 2 = 2y3/2 and Equation (2) above yields
The position (x, y) of a particle at time t is given parametrically by x = t2 and Find the distance the particle travels between t = 1 and t = 2.
SOLUTION: We can use (4): ds2 = dx2 + dy2, where dx = 2 t dt and dy = (t2 − 1) dt. Thus,
Find the length of the arc of y = ln sec x from x = 0 to