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 Image x Image 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

Image

Here a small piece of the curve is equal approximately to Image

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 cyd, then the length s of the arc from y = c to y = d is given by

Image

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

Image

BC ONLY

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

Image

and can be remembered by visualizing Figure N7–21.

Image

FIGURE N7–21

EXAMPLE 13

Find the length, to three decimal places, of the arc of y = x3/2 from x = 1 to x = 8.

SOLUTION: Here Image

EXAMPLE 14

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 Image Equation (2) above yields

Image

EXAMPLE 15

The position (x, y) of a particle at time t is given parametrically by x = t2 and Image 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,

Image

BC ONLY

EXAMPLE 16

Find the length of the arc of y = ln sec x from x = 0 to Image

SOLUTION: Image