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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the length of curve `y = ln((e^x - 2)/(e^x + 2))` , over the interval `[1,2]` , using the following formula, such that:

`L = int_1^2 sqrt(1 + ((dy)/(dx))^2)`

You need to differentiate the function `y = ln((e^x - 2)/(e^x + 2))` , with respect to x, using the chain rule, such that`:(e^x - 2)/(e^x + 2)`

`(dy)/(dx) = 1/((e^x - 2)/(e^x + 2))*((e^x - 2)/(e^x + 2))'`

`(dy)/(dx) = 1/((e^x - 2)/(e^x + 2))*(e^x(e^x + 2) - e^x(e^x - 2))/((e^x + 2)^2)`

`(dy)/(dx) = (e^x + 2)/(e^x - 2)*(e^x(e^x + 2 - e^x + 2))/((e^x + 2)^2)`

`(dy)/(dx) = (e^x + 2)/(e^x - 2)*(4e^x)/((e^x + 2)^2)`

`(dy)/(dx) = (4e^x)/((e^x + 2)(e^x - 2))`

`(dy)/(dx) = (4e^x)/(e^(2x) - 4)`

Raising to square yields:

`((dy)/(dx))^2 = (16e^(2x))/((e^(2x) - 4)^2)`

Adding 1 both sides yields:

`1 + ((dy)/(dx))^2 = 1 + (16e^(2x))/((e^(2x) - 4)^2)`

`1 + ((dy)/(dx))^2 = ((e^(2x) - 4)^2 + 16e^(2x))/((e^(2x) - 4)^2)`

`1 + ((dy)/(dx))^2 = (e^(4x) - 8e^(2x) + 16 + 16e^(2x))/((e^(2x) - 4)^2)`

`1 + ((dy)/(dx))^2 = (e^(4x) + 8e^(2x) + 16)/((e^(2x) - 4)^2)`

`1 + ((dy)/(dx))^2 = ((e^(2x) + 4)^2)/((e^(2x) - 4)^2)`

Taking square root both sides, yields:

`sqrt(1 + ((dy)/(dx))^2) = sqrt(((e^(2x) + 4)^2)/((e^(2x) - 4)^2))`

`sqrt(1 + ((dy)/(dx))^2) = ((e^(2x) + 4)/|(e^(2x) - 4|`

`sqrt(1 + ((dy)/(dx))^2) = ((e^(2x) + 4)/(4 - e^(2x))` if `x in [1,2]`

You need to evaluate the length of the given curve, such that:

`L = int_1^2 sqrt(1 + ((dy)/(dx))^2) dx `

`L = int_1^2 ((e^(2x) + 4)/(4 - e^(2x)) dx`

You should use integration by substitution, such that:

`4 - e^(2x) = t => -2e^(2x)dx = dt => dx = (dt)/(2(t - 4))`

`int ((e^(2x) + 4)/(4 - e^(2x))) dx = int (4 - t + 4)/(2t(t - 4)) dt`

`int (4 - t + 4)/(2t(t - 4)) dt = (1/2)int (8 - t)/(t(t - 4)) dt`

You need to use partial fraction decomposition, such that:

`(8 - t)/(t(t - 4)) = a/t + b/(t - 4)`

`8 - t = a(t - 4) + bt`

`8 - t = at - 4a + bt => -t + 8 = t(a + b) - 4a`

Equating the coefficients of like powers yields:

`{(a + b = -1),(-4a = 8):} => {(a + b = -1),(a = -2):} => {(-2 + b = -1),(a = -2):} => {(b = 1),(a = -2):}`

`(8 - t)/(t(t - 4)) = -2/t + 1/(t - 4)`

Integrating both sides yields:

`int (8 - t)/(t(t - 4)) dt = int -2/t dt + int 1/(t - 4) dt`

`int (8 - t)/(t(t - 4)) dt = -2 ln|t| + ln|t - 4|`

`(1/2)int (8 - t)/(t(t - 4)) dt = -ln|t| + (1/2)ln|t - 4|`

Replacing back `4 - e^(2x)` for t yields:

`int ((e^(2x) + 4)/(4 - e^(2x)) dx = -ln|4 - e^(2x)| + (1/2)ln|4 - e^(2x) - 4|`

`int ((e^(2x) + 4)/(4 - e^(2x)) dx = -ln|4 - e^(2x)| + (1/2)ln (e^(2x))`

`int ((e^(2x) + 4)/(4 - e^(2x)) dx = x - ln|4 - e^(2x)|`

Taking the definite integral, yields:

`int_1^2 ((e^(2x) + 4)/(4 - e^(2x)) dx = (x - ln|4 - e^(2x)|)|_1^2`

Using the fundamental theorem of calculus, yields:

`int_1^2 ((e^(2x) + 4)/(4 - e^(2x))) dx = 2 - ln(e^4 - 4) - 1 + ln(e^2 - 4)`

`int_1^2 ((e^(2x) + 4)/(4 - e^(2x))) dx = 1 + ln ((e^2 - 4)/(e^4 - 4))`

Hence, evaluating the length of the given curve, yields `int_1^2 ((e^(2x) + 4)/(4 - e^(2x))) dx = 1 + ln ((e^2 - 4)/(e^4 - 4)).`

(e^x - 2)/(e^x + 2)

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