ARC LENGTH |

**Arc Length**

The best way to approach this topic is to learn the "*recipe*" and do the calculations. The formula comes from dividing the arc segment into teeny right triangles and suming the lengths of their hypoteni which lie on the arc. Here's the picture.

The length of the hypotenuse of each triangle is which, with some algebraic manipulation becomes since we divide through by (*dx*)² to get (* dy/dx *)²

In some cases, the function rule makes it easier to integrate with *y *instead of *x*. In such a case, we use the second formula with the *dy* in it.

The length of the arc from x = a to x = b along y = f (x) isThe length of the arc from |

**Finding Arc Length**

- 1st step:

- 2nd step: simplify the sum -- take the square root.

- 3rd step: determine

- 4th step: integrate over the interval

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**Examples**

If *f *(*x*) = 3*x*^{ 2 / 3} – 10, find the length of the arc along *f* from A(8, 2) to B(27, 17)

- First we find

Then we integrate over the values of the corresponding variable.

- Since

- The arc length is the integral from

Now we make the substituion *u* = *x*^{ 2 / 3} + 4, which makes *du* = 2/3(*x*^{ – 1 / 3}) *dx* , and

changes the limits of integration to 8 and 13 using *x*^{ 2 / 3} + 4.

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Now let's do the same question using the other formula where we integrate with respect to *y*.

First, we have to find *x* in terms of *y*.

We find

Now we differentiate, square it, and add 1:

Now we integrate the square root of this expression over the values of *y*.

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**Practice**

1)Find the arc length of the graph of (*x* + 3)^{ 2} = 8* *(*y – *1)^{ 3} from A(–2, 3/2) to B(5, 3)*.*

2) Find the length of the arc along *y* = *x*^{ 3 / 2} from A(1, 1) to B(4, 8).

3) Find the length of the arc along (* y* + 1)² =* *(* x *– 4 )³ from **A*** *(5, 0) to **B*** *(8, 7).

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**Solutions**

1) When we solve (*x + 3*)^{ 2} = *8 *(*y – 1*)^{ 3} for *y*, we get *f *( *x *)* = *½( *x + 3* )^{ 2 / 3} + 1

The arc length formula is , so we need [ *f '*(*x*)] ²

Now,

We need *1 + *[*f '(x)*] ², which is

Notice that the denominator is a perfect square, so when we take the root, and integrate, we get

Using the substitution * u* = 1 + 9(*x* + 3)^{ 2 / 3} and *du* = 6(*x* + 3)^{ – 1 / 3}* dx*, we get

2)

3)

( Calculus II MathRoom Index )

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