Length of Curve

Calculus Level 3

A parabolic function is defined by f ( x ) = − 1 2 x 2 + 3 x + 1 f(x) = -\dfrac{1}{2} x^2 + 3 x + 1 . It is plotted in the figure above over the interval [ 0 , 5 ] [0, 5] . Find the length of the curve of f ( x ) f(x) over this interval.


The answer is 8.6105.

Hosam Hajjir by Hosam Hajjir , 56, Canada

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3 solutions

Chew-Seong Cheong
Dec 26, 2020

The curve is given by y = 1 + 3 x − x 2 2 y=1 + 3x - \dfrac {x^2}2 . The length of the curve for x ∈ [ 0 , 5 ] x \in [0,5] is given by:

s = ∫ 0 5 1 + ( d y d x ) 2 d x = ∫ 0 5 1 + ( 3 − x ) 2 d x Let 3 − x = sinh ⁡ t ⟹ − d x = cosh ⁡ t d t = ∫ sinh ⁡ − 1 ( − 2 ) sinh ⁡ − 1 3 1 + sinh ⁡ 2 t cosh ⁡ t d t Since cosh ⁡ 2 z − sinh ⁡ 2 z = 1 = ∫ sinh ⁡ − 1 ( − 2 ) sinh ⁡ − 1 3 cosh ⁡ 2 t d t = ∫ sinh ⁡ − 1 ( − 2 ) sinh ⁡ − 1 3 cosh ⁡ 2 t + 1 2 d t = sinh ⁡ 2 t 4 + t 2 ∣ sinh ⁡ − 1 ( − 2 ) sinh ⁡ − 1 3 Note that sinh ⁡ − 1 z = ln ⁡ ( z + z 2 + 1 ) = sinh ⁡ t cosh ⁡ t 2 ∣ sinh ⁡ − 1 ( − 2 ) sinh ⁡ − 1 3 + 1 2 ln ⁡ ( 3 + 10 5 − 2 ) and sinh ⁡ 2 z = 2 sinh ⁡ z cosh ⁡ z = 3 10 + 2 5 2 + 1 2 ln ⁡ ( 3 + 10 5 − 2 ) and cosh ⁡ 2 z − sinh ⁡ 2 z = 1 ≈ 8.61 \begin{aligned} s & = \int_0^5 \sqrt{1+\left(\frac {dy}{dx} \right)^2} dx \\ & = \int_0^5 \sqrt{1+ \blue{(3-x)}^2} dx & \small \blue{\text{Let }3-x = \sinh t \implies - dx = \cosh t \ dt} \\ & = \int_{\sinh^{-1}(-2)}^{\sinh^{-1} 3} \blue{\sqrt{1+\sinh^2 t}} \cosh t \ dt & \small \blue{\text{Since }\cosh^2 z - \sinh^2 z = 1} \\ & = \int_{\sinh^{-1}(-2)}^{\sinh^{-1} 3} \cosh^2 t \ dt \\ & = \int_{\sinh^{-1}(-2)}^{\sinh^{-1} 3} \frac {\cosh 2t + 1}2 dt \\ & = \frac {\sinh 2t}4 + \frac t2 \ \bigg|_{\sinh^{-1}(-2)}^{\sinh^{-1} 3} & \small \blue{\text{Note that }\sinh^{-1} z = \ln (z+\sqrt{z^2+1})} \\ & = \frac {\sinh t \cosh t}2 \ \bigg|_{\sinh^{-1}(-2)}^{\sinh^{-1} 3} + \frac 12 \ln \left(\frac {3+\sqrt{10}}{\sqrt 5 - 2} \right) & \small \blue{\text{and }\sinh 2z = 2\sinh z \cosh z} \\ & = \frac {3\sqrt{10}+2\sqrt 5}2+ \frac 12 \ln \left(\frac {3+\sqrt{10}}{\sqrt 5 - 2} \right) & \small \blue{\text{and }\cosh^2 z - \sinh^2 z = 1} \\ & \approx \boxed{8.61} \end{aligned}

4 Helpful 3 Interesting 2 Brilliant 0 Confused

Nice substitution with sinh, upvoted! Didn't come to my mind - very elegant solution. I substituted with tan.

Veselin Dimov - 5 months, 1 week ago

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Yes, elegant and I learned it from @Pi Han Goh . Check out the reference .

Chew-Seong Cheong - 5 months, 1 week ago

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Or you can try Euler's substitution . (which is not the best substitution here)

Using one of your intermediate steps, we have ∫ 0 5 1 + ( 3 − x ) 2 d x = z = 3 − x ∫ − 2 3 1 + z 2 d z \int_0^5 \sqrt{1+ (3-x)^2} \, dx \stackrel{z=3-x}{=} \int_{-2}^3 \sqrt{1+z^2} \, dz

Let 1 + z 2 = z + t \sqrt{1+z^2} = z + t

Squaring both sides gives 1 + z 2 = z 2 + 2 z t + t 2 ⇒ z = 1 − t 2 2 t ⇒ d z d t = − t 2 + 1 2 t 2 1 + \cancel{z^2} = \cancel{z^2} + 2zt + t^2 \quad \Rightarrow \quad z = \dfrac{1-t^2}{2t} \quad \Rightarrow \quad \dfrac{dz}{dt} = - \dfrac{t^2+1}{2t^2}

When z = − 2 , t = 2 + 5 z = -2, t = 2 + \sqrt5 .
When z = 3 , t = 10 − 3 z = 3, t = \sqrt{10} - 3 .

The integral transforms to ∫ 2 + 5 10 − 3 ( z + t ) ⋅ − t 2 + 1 2 t 2 d t = ∫ 2 + 5 10 − 3 ( 1 − t 2 2 t + t ) ⋅ − t 2 + 1 2 t 2 d t = − 1 4 ∫ 2 + 5 10 − 3 ( t 2 + 1 ) 2 t 3 d t \int_{2 + \sqrt5}^{\sqrt{10}-3} (z+t) \cdot -\dfrac{t^2+1}{2t^2} \, dt \quad = \quad \int_{2 + \sqrt5}^{\sqrt{10}-3} \left(\dfrac{1-t^2}{2t} + t \right) \cdot -\dfrac{t^2+1}{2t^2} \, dt \quad= \quad -\frac14 \int_{2 + \sqrt5}^{\sqrt{10}-3} \dfrac{ (t^2+1)^2}{t^3} \, dt

Continuing on,

1 4 ∫ 10 − 3 2 + 5 ( t + 2 t + 1 t 3 ) d t = 1 4 [ t 2 2 − 2 ln ⁡ ∣ t ∣ − 1 2 t 2 ] 2 + 5 10 − 3 = ( plenty of tedious simplication ) = ( the same answer ) \frac14\int_{\sqrt{10} - 3}^{2 + \sqrt5}\left(t + \dfrac2t + \dfrac1{t^3} \right) \, dt = \frac14 \left[ \dfrac{t^2}2 - 2 \ln |t| - \dfrac1{2t^2} \right]_{2 + \sqrt5}^{\sqrt{10} - 3 } = (\text{plenty of tedious simplication}) = (\text{the same answer})

HAPPY NEW YEAR

Pi Han Goh - 5 months, 1 week ago

Those integrations are too extreme for me <3

Valentin Duringer - 5 months, 1 week ago
Karan Chatrath
Dec 25, 2020

The arc length is given by:

S = ∫ x i n i t i a l x f i n a l 1 + ( d y d x ) 2 d x ⟹ S = ∫ 0 5 1 + ( 3 − x ) 2 d x S = \int_{x_\mathrm{initial}}^{x_\mathrm{final}}\sqrt{1 + \left(\frac{dy}{dx}\right)^2} \ dx \implies S = \int_{0}^{5} \sqrt{1 + (3-x)^2} \ dx

Taking z = 3 − x z = 3-x transforms the integral to: S = ∫ − 2 3 1 + z 2 d z S = \int_{-2}^{3} \sqrt{1 + z^2} \ dz Integrating by parts or by trigonometric substitution yields (steps left out): S = 1 2 ( z 1 + z 2 + ln ⁡ ( z + z 2 + 1 ) ) ∣ − 2 3 ≈ 8.6105 S = \frac{1}{2} \left(z\sqrt{1+z^2} + \ln\left(z + \sqrt{z^2+1}\right)\right) \biggr\rvert_{-2}^{3} \approx 8.6105

3 Helpful 0 Interesting 0 Brilliant 1 Confused
Vijay Simha
Dec 25, 2020

In our case f'(x) = -x+3.

0 Helpful 0 Interesting 0 Brilliant 3 Confused

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