Find the length of the graph of y=1/3x^3/2-x^1/2 from (1, -2/3) to (4,2/3)?

Answer 1

You don't even have to solve the equation. The coordinates are given to you (if you do plug 1 into the equation, you will see that y = -2/3; and if you do plug 4 into the equation, you will see that y equals 2/3).

Subtract the x's from each other and subtract the y's from each other:

x= 4 - 1 = 3
y = 2/3 - (-2/3) = 4/3

Plug into the Pythagorean Theorem:

3^2 + (4/3)^2 = d^2
9 + 16/9 = d^2
97/9 = d^2
sqrt (97)/3 =d

That's approximately 3.28

Answer 2

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POINTS:
(1 , -2 / 3)
(4 , 2 / 3)

Use pythagorean theorem to find the length:

x^2 = (4 - 1)^2 + (2/3 + 2/3)
x^2 = 3^2 + (4/3)^2
x^2 = 9 + 16/9
x^2 = 97/9
x = 3.28

The length of the graph y=1/3x^3/2-x^1/2 from (1, -2/3) to (4,2/3) is 3.28.

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Answer 3

L = ∫[from x = 1 to 4] {√(1 + (dy/dx)²)} dx

= ∫[from x = 1 to 4] {√[1 + (½x^½ - ½x^(-½))²]}dx

= ∫[from x = 1 to 4]{√[1 + ¼(x - 2 + 1/x)]}dx

= ½∫[from x = 1 to 4]{√[4 + x - 2 + 1/x]}dx

= ½∫[from x = 1 to 4]{√[x + 2 + 1/x]} dx

= ½∫[from x = 1 to 4]{√[(x^½ + x^(-½))²}dx

= ½∫[from x = 1 to 4][(x^½ + x^(-½))dx

= ½[ 2/3x^(3/2) + 2x^½][from x = 1 to 4]

= ½{[2/3 * 8 + 2*2] - [2/3 + 2]}

= ½ (14/3 + 2)
= 10/3

Check By Pythagoras Rule

Straight line distance = √[(Δx)² + (Δy)²]
= √[(3)² + (4/3)²]

= √(9 + 16/9)
= √(97/9)

= ⅓ √97 ≈3.283 (≈10/3 answer above for exact length)

Answer 4

when did they start puttin letters with it?