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3 years ago

Y = (1/4)x^2 - (1/2)lnx..over the interval (1, 7e) ...what is the arc length ?

1 answer:

6 0

So, f[x] = 1/4x^2 - 1/2Ln(x)
thus f'[x] = 1/4*2x - 1/2*(1/x) = x/2 - 1/2x 
thus f'[x]^2 = (x^2)/4 - 2*(x/2)*(1/2x) + 1/(4x^2) = (x^2)/4 - 1/2 + 1/(4x^2) 
thus f'[x]^2 + 1 = (x^2)/4 + 1/2 + 1/(4x^2) = (x/2 + 1/2x)^2 
thus Sqrt[...] = (x/2 + 1/2x)

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Round to the nearest tenth if necessary.

Andrei [34K]

Answer:

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General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Coordinates (x, y)

Algebra II

Distance Formula: $\displaystyle d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Step-by-step explanation:

Step 1: Define

Identify

Point (-1, -8)

Point (-4, -4)

Step 2: Find distance d

Simply plug in the 2 coordinates into the distance formula to find distance d

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