Arc Length of a Curve

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Arc Length of a Curve
If f′(x)f'\left(x\right)f′(x) is continuous on [a,b][a,b][a,b], then the length of the curve y=f(x)y=f(x)y=f(x) where a≤x≤ba\le x\le ba≤x≤b is L=∫ab1+[f′(x)]2dxL=
\displaystyle\int_a^b
\sqrt{1+\left[f'\left(x\right)\right]^2}dxL=∫ab​1+[f′(x)]2​dx.

If g′(y)g'\left(y\right)g′(y) is continuous on [c,d][c,d][c,d], then the length of the curve x=g(y)x=g(y)x=g(y) where c≤y≤dc\le y\le dc≤y≤d is L=∫cd1+[g′(y)]2dyL=
\displaystyle\int_c^d
\sqrt{1+\left[g'\left(y\right)\right]^2}dyL=∫cd​1+[g′(y)]2​dy.

Arc Length Function
For any smooth curve C with equation y=f(x)y=f\left(x\right)y=f(x), the arc length (distance) from a starting point P0(a,f(a))P_0\left(a,f\left(a\right)\right)P0​(a,f(a)) to any given point Q(x,f(x))Q\left(x,f\left(x\right)\right)Q(x,f(x)) is given by the function s(x)=∫ax1+[f′(t)]2dts\left(x\right)=
\displaystyle \int_a^x
\sqrt{1+\left[f'\left(t\right)\right]^2}dts(x)=∫ax​1+[f′(t)]2​dt

Practice: Arc Length
Find the exact length of the curve x=y(y−2)6x=\frac{\sqrt{y}\left(y-2\right)}{\sqrt{6}}x=6​y​(y−2)​, 1≤y≤41\le y\le41≤y≤4.

Practice Question
Find the arc length function for the curve y=2x32y=2x^{\frac{3}{2}}y=2x23​ with starting point P0(0,3)P_0\left(0,3\right)P0​(0,3).

13(1+9x)32+13\frac{1}{3}\left(1+9x\right)^{\frac{3}{2}}+\frac{1}{3}31​(1+9x)23​+31​

13(1+9x)32−13\frac{1}{3}\left(1+9x\right)^{\frac{3}{2}}-\frac{1}{3}31​(1+9x)23​−31​

13(1+9x)32−13x\frac{1}{3}\left(1+9x\right)^{\frac{3}{2}}-\frac{1}{3}x31​(1+9x)23​−31​x

227(1+9x)32\frac{2}{27}\left(1+9x\right)^{\frac{3}{2}}272​(1+9x)23​

227(1+9x)32−227\frac{2}{27}\left(1+9x\right)^{\frac{3}{2}}-\frac{2}{27}272​(1+9x)23​−272​

I don't know

Wize University Calculus 2 Textbook > Applications of Integrals

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