The arc of the parabola y=1/4(x-2)^2 + 1 from the point (-2,5) to the point (4,2).length of arc of the graph of a function addtional applications of the definite integral

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll use the following formula to calculate the length of the arc of the graph of parabola:

b

`int` sqrt[1+|f'(x)|^2]dx

a

Therefore, we need to determine the derivative of the function, using the quotient rule:

f'(x) = -8(x-2)/16(x-2)^4

f'(x) = -1/2(x-2)^3

The arc length of of f from x = -2 to x = 4 is:

4                                            4

`int` sqrt[1 + 1/4(x-2)^6]dx  = `int` {sqrt[(x-2)^6 + 1]}dx/2(x-2)^3

-2                                        -2

To calculate the definite integral above, we'll use the following formula:

`int` rdt/t = r - a*ln|(a+r)/t|, where r = sqrt (t^2 + a^2)

If t = (x-2)^3 and a = 1, we'll have:

`int` {sqrt[(x-2)^6 + 1]}dx/2(x-2)^3 = (1/2)*{sqrt[(x-2)^6 + 1] - ln|{1+sqrt[(x-2)^6 + 1]}/(x-2)^3|} + C

To evaluate the definite integral, we'll use Leibniz Newton formula:

4

`int` {sqrt[(x-2)^6 + 1]}dx/2(x-2)^3= F(4) - F(2)

-2

F(4) = (1/2)*{sqrt[(2)^6 + 1] - ln|{1+sqrt[(2)^6 + 1]}/(2)^3|}

F(4) = (1/2)*[sqrt65 - ln|(1+sqrt65)/8|]

F(4) = (1/2)*(8.062-0.123)

F(4) = 3.969

F(-2) =  (1/2)*{sqrt[(-4)^6 + 1] - ln|{1+sqrt[(-4)^6 + 1]}/(-4)^3|}

F(-2) = 32.003

4

`int` {sqrt[(x-2)^6 + 1]}dx/2(x-2)^3=28.034

-2

The arc length of of y = 1/4(x-2)^2 + 1, from x = -2 to x = 4 is of 28.034.

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