## 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 Δ*x* → 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 |*t _{a}*,

*t*], (and if the curve does not intersect itself), then the length of the arc from

_{b}*t*=

*t*to

_{a}*t*=

*t*is given by

_{b}**BC ONLY**

The parenthetical clause above is equivalent to the requirement that the curve is traced out just once as *t* varies from *t _{a}* to

*t*.

_{b}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.

**FIGURE N7–21**

**EXAMPLE 13**

Find the length, to three decimal places, of the arc of *y* = *x*^{3/2} from *x* = 1 to *x* = 8.

**SOLUTION:** Here

**EXAMPLE 14**

Find the length, to three decimal places, of the curve (*x* − 2)^{2} = 4*y*^{3} from *y* = 0 to *y* = 1.

**SOLUTION:** Since *x* − 2 = 2*y*^{3/2} and Equation (2) above yields

**EXAMPLE 15**

The position (*x*, *y*) of a particle at time *t* is given parametrically by *x* = *t*^{2} and Find the distance the particle travels between *t* = 1 and *t* = 2.

**SOLUTION:** We can use (4): *ds*^{2} = *dx*^{2} + *dy*^{2}, where *dx* = 2 *t* *dt* and *dy* = (*t*^{2} − 1) *dt*. Thus,

**BC ONLY**

**EXAMPLE 16**

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

**SOLUTION:**