Arc Length 11/18/2011 Suppose you want to know what the length of a - - PowerPoint PPT Presentation

Arc Length

11/18/2011

Suppose you want to know what the length of a curve y = f (x) is from the point (a, f (a)) to the point (b, f (b)):

a b y=f(x)

Suppose you want to know what the length of a curve y = f (x) is from the point (a, f (a)) to the point (b, f (b)):

x0 xn y=f(x) x1 x2 ∆x . . .

Slice! ` = lim

n!1 n

X

i=1

(little length)i

Suppose you want to know what the length of a curve y = f (x) is from the point (a, f (a)) to the point (b, f (b)):

x0 xn y=f(x) x1 x2 ∆x . . .

• ne

piece

− →

xi xi+1

Slice! ` = lim

n!1 n

X

i=1

(little length)i

Suppose you want to know what the length of a curve y = f (x) is from the point (a, f (a)) to the point (b, f (b)):

x0 xn y=f(x) x1 x2 ∆x . . .

• ne

piece

− →

∆y xi xi+1 ∆x

∆ l

Slice! ` = lim

n!1 n

X

i=1

(∆`)i

Suppose you want to know what the length of a curve y = f (x) is from the point (a, f (a)) to the point (b, f (b)):

x0 xn y=f(x) x1 x2 ∆x . . .

• ne

piece

− →

∆y xi xi+1 ∆x

∆ l

Slice!

Let n go to 1

• ` = lim

n!1 n

X

i=1

(∆`)i = Z x=b

x=a

d`

dy dx

dl

Suppose you want to know what the length of a curve y = f (x) is from the point (a, f (a)) to the point (b, f (b)):

x0 xn y=f(x) x1 x2 ∆x . . .

• ne

piece

− →

∆y xi xi+1 ∆x

∆ l

Slice!

Let n go to 1

• ` = lim

n!1 n

X

i=1

(∆`)i = Z x=b

x=a

d` d` = p dx2 + dy2

dy dx

dl

Manipulating into something we can actually calculate...

dy dx

dl

Remember, y = f (x). d` = p dx2 + dy2

Manipulating into something we can actually calculate...

dy dx

dl

Remember, y = f (x). d` = p dx2 + dy2 = p dx2 + dy2 dx dx

Manipulating into something we can actually calculate...

dy dx

dl

Remember, y = f (x). d` = p dx2 + dy2 = p dx2 + dy2 dx dx = r dx2 + dy2 dx2 dx

Manipulating into something we can actually calculate...

dy dx

dl

Remember, y = f (x). d` = p dx2 + dy2 = p dx2 + dy2 dx dx = r dx2 + dy2 dx2 dx = r dx2 dx2 + dy2 dx2 dx

Manipulating into something we can actually calculate...

dy dx

dl

Remember, y = f (x). d` = p dx2 + dy2 = p dx2 + dy2 dx dx = r dx2 + dy2 dx2 dx = r dx2 dx2 + dy2 dx2 dx = s✓dx dx ◆2 + ✓dy dx ◆2 dx

Manipulating into something we can actually calculate...

dy dx

dl

Remember, y = f (x). d` = p dx2 + dy2 = p dx2 + dy2 dx dx = r dx2 + dy2 dx2 dx = r dx2 dx2 + dy2 dx2 dx = s✓dx dx ◆2 + ✓dy dx ◆2 dx = q 1 + (f 0(x))2 dx

Manipulating into something we can actually calculate...

dy dx

dl

Remember, y = f (x). d` = p dx2 + dy2 = p dx2 + dy2 dx dx = r dx2 + dy2 dx2 dx = r dx2 dx2 + dy2 dx2 dx = s✓dx dx ◆2 + ✓dy dx ◆2 dx = q 1 + (f 0(x))2 dx So ` = Z b

x=a

q 1 + (f 0(x))2 dx

Example

Find the length of the arc y = x3/2, from x = 0 to x = 1.

1 2 1 2 3

Find the length of the curve y = x4 +

1 32x2 from x = 1 to x = 2.

1 2 5 10 15

Find the length of the curve y = x4 +

1 32x2 from x = 1 to x = 2.

1 2 5 10 15

f (x) = x4 + 1 32x2 = ⇒ f 0(x) = 4x3 − 1 16x3 = 64x6 − 1 16x3

Find the length of the curve y = x4 +

1 32x2 from x = 1 to x = 2.

1 2 5 10 15

f (x) = x4 + 1 32x2 = ⇒ f 0(x) = 4x3 − 1 16x3 = 64x6 − 1 16x3 Keeping the algebra tame: Let A = (2x)3 = 8x3 and so A2 = 64x6, and f 0(x) = A21

2A .

Find the length of the curve y = x4 +

1 32x2 from x = 1 to x = 2.

1 2 5 10 15

f (x) = x4 + 1 32x2 = ⇒ f 0(x) = 4x3 − 1 16x3 = 64x6 − 1 16x3 Keeping the algebra tame: Let A = (2x)3 = 8x3 and so A2 = 64x6, and f 0(x) = A21

2A .

So 1+(f 0(x))2 = 1+ ✓A2 − 1 2A ◆2

Most of the time, the resulting integral is “hard” (not elementary)

Set up (but do not integrate) the integrals which compute the length of the following functions:

• 1. f (x) = x2 from x = −3 to 2
• 2. f (x) = x2 + 5 from x = −3 to 2
• 3. f (x) = −x2 + ⇡ from x = −3 to 2
• 4. f (x) = sin(x) from x = 0 to π

2

• 5. f (x) = ex from x = 0 to 1
• 6. f (x) =

√ 1 − x2 from x = −1 to 1