Informative line

Arc Length

Learn arc length formula and length of a curve calculus with solved examples. Find the Arc Length of the Curve between Two Points Whose Co-Ordinates are Given and arc length of curve given in integral form.

The Length of a Curve

Suppose a curve is defined by \(y=f(x)\) where \(f\) is continuous in \([a,\,b]\). We divide the interval \((a,\,b)\) in subintervals with end points \(x_0,\;x_1,\;x_2,\;.....x_n\) and equal width \(\Delta x\).

If \(y_i=f(x_i)\) then \(P_i(x_i,\;y_i)\) lies on the curve and the length of polygon with vertices \(P_0,\;P_1,\;P_2.....P_n\) is an approximation to the length of curve.

The approximation gets better as \(n\) increases.

\(\therefore\) Length of curve \(=\lim\limits_{n\to\infty}\,\sum\limits^n_{i=1}\left|P_{i-1}P_i\right|=L\)

  • It is similar to defining an area.

Now,  \(P_{i-1}P_i=\sqrt{(x_i-x_{i-1})^2+(y_i-y_{i-1})^2}\to\) distance formula.

\(=\sqrt{(\Delta x)^2+(\Delta y)^2}\)

  Now,  consider a small portion of the curve.

\(tan \theta=\dfrac{\Delta y}{\Delta x}\)

\(\Rightarrow\) \(f'(x_i^*)\)\(=\dfrac{\Delta y}{\Delta x}\)

\(\Rightarrow\;\Delta y=\Delta x×f'(x_i^*)\)

\(=\sqrt{(\Delta x)^2+(\Delta y)^2}\)

\(=\sqrt{(\Delta x)^2+\left(f'(x_i^*)\,\Delta x\right)^2}\to\) \(x_i^*\) lies in \(x_{i-1}\) and \(x_i\)

\(=(\Delta x)\sqrt{1+\left(f'(x_i^*)\right)^2}\)

\(\therefore\;L=\lim\limits_{n\to\infty}\,\sum\limits^n_{i=1}\left|P_{i-1}P_i\right|\)

\(=\lim\limits_{n\to\infty}\,\sum\limits^n_{i=1}\sqrt{1+\left(f'(x_i^*)\right)^2}\,(\Delta x)\)

\(=\displaystyle\int\limits^b_{a}\sqrt{1+\left(f'(x)\right)^2}dx\)

The Arc Length formula

If \(f'\) is continuous in \([a,\,b]\) then the length of curve \(y=f(x),\;a\leq x\leq b\) is

\(L=\displaystyle\int\limits^b_{a}\sqrt{1+\left(f'(x)\right)^2}dx\to(1)\)

\(=\underbrace{\displaystyle\int\limits^b_{a}\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}dx}_\text{Using Leibniz Notation}\;\;\;\;\;\;\;\;\to(2)\)

 

  • If the curve has the equation \(x=g(y)\) where \(c\leq y\leq d\) and \(g'(y)\) is continuous then the role of \(x\) and \(y\)  get interchanged in the formula

\(L=\displaystyle\int\limits^b_{a}\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}dx\;\) and it becomes 

\(L=\displaystyle\int\limits^d_{c}\sqrt{1+\left(\dfrac{dx}{dy}\right)^2}dy\;\)

 

Illustration Questions

Find the length of the curve \(y=x\sqrt x\) when \(1\leq x\leq2\).

A \(\dfrac{8}{27}\left[\left(\dfrac{11}{2}\right)^{3/2}-\left(\dfrac{13}{4}\right)^{3/2}\right]\)

B \(\dfrac{1}{2}\)

C \(4\sqrt 2\)

D \(1\)

×

\(L=\displaystyle \int\limits^b_a\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}dx=\) length of the curve

In this problem,

 \(a=1,\;b=2,\;y=x\sqrt x=\;x^{3/2}\)

\(\Rightarrow\;\dfrac{dy}{dx}=\dfrac{3}{2}x^{1/2}\)

\(\therefore\;\displaystyle L=\int\limits^2_1\sqrt{1+\dfrac{9}{4}x}\,dx\)

\(=\Bigg[\dfrac{\left(1+\dfrac{9}{4}x\right)^{3/2}}{\dfrac{3}{2}×\dfrac{9}{4}}\Bigg]^2_1\)

\(=\dfrac{8}{27}\left[\left(1+\dfrac{9}{2}\right)^{3/2}-\left(1+\dfrac{9}{4}\right)^{3/2}\right]\)

\(=\dfrac{8}{27}\left[\left(\dfrac{11}{2}\right)^{3/2}-\left(\dfrac{13}{4}\right)^{3/2}\right]\)

Find the length of the curve \(y=x\sqrt x\) when \(1\leq x\leq2\).

A

\(\dfrac{8}{27}\left[\left(\dfrac{11}{2}\right)^{3/2}-\left(\dfrac{13}{4}\right)^{3/2}\right]\)

.

B

\(\dfrac{1}{2}\)

C

\(4\sqrt 2\)

D

\(1\)

Option A is Correct

Illustration Questions

Find the length of the curve \(y=\dfrac{1}{3}\sqrt x(x-3)\) when \(1\leq x\leq9\).

A \(16\)

B \(\dfrac{32}{3}\)

C \(\sqrt 2\)

D \(\dfrac{1}{2}\)

×

\(L=\displaystyle \int\limits^b_a\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}dx=\) length of the curve

In this case,

 \(a=1,\;b=9,\;\\y=\dfrac{1}{3}\sqrt x(x-3)=\;\dfrac{1}{3}\left(x^{3/2}-3x^{1/2}\right)\)

\(\therefore\;\dfrac{dy}{dx}=\dfrac{1}{3}\left[\dfrac{3}{2}x^{1/2}-\dfrac{3}{2}x^{-1/2}\right]\)

\(=\dfrac{1}{2}\left(\sqrt x-\dfrac{1}{\sqrt x}\right)\)

\(\left(\dfrac{dy}{dx}\right)^2=\dfrac{1}{4}\left(\sqrt x-\dfrac{1}{\sqrt x}\right)^2\)

\(=\dfrac{1}{4}\left(x+\dfrac{1}{x}-2\right)\)

\(=\dfrac{1}{4}x+\dfrac{1}{4x}-\dfrac{1}{2}\)

\(\therefore\;1+\left(\dfrac{dy}{dx}\right)^2\\=1+\dfrac{1}{4}x+\dfrac{1}{4x}-\dfrac{1}{2}\)

\(=\dfrac{1}{2}+\dfrac{1}{4}x+\dfrac{1}{4x}\)

\(=\dfrac{1}{4}\left(\sqrt x+\dfrac{1}{\sqrt x}\right)^2\)

\(\therefore\;L=\displaystyle\int\limits^9_1\sqrt{\dfrac{1}{4}\left(\sqrt x+\dfrac{1}{\sqrt x}\right)^2}dx\)

\(=\displaystyle\int\limits^9_1\dfrac{1}{2}\left(\sqrt x+\dfrac{1}{\sqrt x}\right)dx\)

\(=\dfrac{1}{2}\left[\dfrac{x^{3/2}}{\dfrac{3}{2}}+\dfrac{x^{1/2}}{\dfrac{1}{2}}\right]^9_1\)

 

\(=\Bigg[\dfrac{x^{3/2}}{3}+x^{1/2}\Bigg]^9_1\)

\(=\left(\dfrac{9^{3/2}}{3}+9^{1/2}\right)-\left(\dfrac{1}{3}+1\right)\)

\(=(9+3)-\left(\dfrac{4}{3}\right)\)

\(=12-\dfrac{4}{3}\\=\dfrac{32}{3}\)

Find the length of the curve \(y=\dfrac{1}{3}\sqrt x(x-3)\) when \(1\leq x\leq9\).

A

\(16\)

.

B

\(\dfrac{32}{3}\)

C

\(\sqrt 2\)

D

\(\dfrac{1}{2}\)

Option B is Correct

Illustration Questions

Find the length of the curve \(x=\dfrac {y^2}{4}-\dfrac {1}{2}\ell n\,y\) when \(1\leq y\leq2\).

A \(\dfrac {1}{4}\)

B \(\dfrac{3}{4}+\dfrac {1}{2}\ell n2\)

C \(\ell n \,2+1\)

D \(\dfrac {1}{\sqrt 2}\)

×

If \(x = g(y)\) where \(c\leq y \leq d\) and  \(g'(y)\) is continuous, then the length of curve 

\(L=\int\limits _c^d \sqrt {1+\left ( \dfrac {dx}{dy}\right)^2}\,dy\)

In this case,

 \(c=1,\;d=2,\;x=\dfrac{y^2}{4}-\dfrac{\ell n\,y}{2}\)

\(\Rightarrow \;\dfrac{dx}{dy}=\dfrac{2y}{4}-\dfrac{1}{2y}\\=\dfrac{y}{2}-\dfrac {1}{2y}\)

\(\therefore \;\Big(\dfrac{dx}{dy}\Big)^2=\left (\dfrac{y}{2}-\dfrac{1}{2y}\right)^2\\=\dfrac{y^2}{4}+\dfrac {1}{4y^2}-\dfrac {1}{2}\)

\(\therefore \;1+\left(\dfrac{dx}{dy}\right)^2\\=\dfrac{y^2}{4}+\dfrac{1}{4y^2}-\dfrac{1}{2}+1\\=\dfrac{y^2}{4}+\dfrac{1}{4y^2}+\dfrac{1}{2}\)

\(=\left(\dfrac{y}{2}+\dfrac{1}{2y}\right)^2\)

 

\(\therefore\;L=\int\limits _1^2 \sqrt {1+\left ( \dfrac {dx}{dy}\right)^2}\,dy\)

\(\therefore\;L=\displaystyle\int\limits^2_1\sqrt{\left(\dfrac{y}{2}+\dfrac{1}{2y}\right)^2}dy\)

\(=\Bigg[\dfrac{y^2}{4}+\dfrac{1}{2} \ell n \,y\Bigg]^2_1\)

\(=\left(\dfrac{4}{4}+\dfrac {1}{2}\,\ell n2\right)-\left(\dfrac{1}{4}+ \dfrac{1}{2}\ell n\,1 \right)\)

\(=1+\dfrac{1}{2}\ell n \,2-\dfrac{1}{4}\\=\dfrac{3}{4}+\dfrac{1}{2}\,\ell n\,2\)

Find the length of the curve \(x=\dfrac {y^2}{4}-\dfrac {1}{2}\ell n\,y\) when \(1\leq y\leq2\).

A

\(\dfrac {1}{4}\)

.

B

\(\dfrac{3}{4}+\dfrac {1}{2}\ell n2\)

C

\(\ell n \,2+1\)

D

\(\dfrac {1}{\sqrt 2}\)

Option B is Correct

Illustration Questions

Find the exact of the curve \(x=\sqrt {y-y^2}+sin^{-1}\sqrt y\) where \(0\leq y\leq \dfrac {1}{2}\).

A \(5\)

B \(1\)

C \(\sqrt 2\)

D \(\sqrt 3\)

×

If \(x = g(y)\) where \(c\leq y \leq d\) and  \(g'(y)\) is continuous the length of curve 

\(L=\int\limits _c^d \sqrt {1+\left ( \dfrac {dx}{dy}\right)^2}\,dy\)

In this case,

 \(c=0,\;d=\dfrac {1}{2},\;x=\sqrt {y-y^2}+sin^{-1}\sqrt y\)

\(\Rightarrow \;\dfrac{dx}{dy}=\dfrac{1}{2\sqrt {y-y^2}}\dfrac{d}{dy}(y-y^2)+\dfrac{1}{\sqrt {1-(\sqrt y)^2}}×\dfrac {d}{dx}(\sqrt y)\rightarrow\)  (Chain Rule)

\(=\dfrac {1-2y}{2\sqrt {y-y^2}}+\dfrac {1}{2\sqrt y \;\sqrt {1-y}}\)

\(=\dfrac {2-2y}{2\sqrt y \;\sqrt {1-y}}\)

\(=\dfrac {1-y}{\sqrt y \;\sqrt {1-y}}=\sqrt {\dfrac {1-y}{y}}\)

\(\therefore \;\Big(\dfrac{dx}{dy}\Big)^2=\left (\dfrac{1-y}{y}\right)\) and \(1+\Big(\dfrac{dx}{dy}\Big)^2=1+\dfrac{1-y}{y}=\dfrac {1}{y}\)

 

\(\therefore\;L=\displaystyle\int\limits _0^{1/2} \sqrt {\dfrac {1}{y}}\,dy\)

\(=\displaystyle\int\limits _0^{1/2} {\dfrac {1}{\sqrt y}}\,dy\)

\(=\Bigg[\dfrac{y^{1/2}}{1/2}\Bigg]_0^{1/2}\)

\(=2\left [ \Big (\dfrac {1}{2}\Big)^{1/2} -0\right]\\=2×\dfrac {1}{\sqrt 2}\\=\sqrt 2\)

 

Find the exact of the curve \(x=\sqrt {y-y^2}+sin^{-1}\sqrt y\) where \(0\leq y\leq \dfrac {1}{2}\).

A

\(5\)

.

B

\(1\)

C

\(\sqrt 2\)

D

\(\sqrt 3\)

Option C is Correct

Finding the Length of the Curve Between Two Points Whose Co-Ordinates are Given

  • Let \(y= f(x)\) be a curve, then the length of this curve between the two points  \(A(a,\,f(a))\) and  \(B(b,\,f(b))\)  (both \(A,\;B\) are on the curve) is given by

\(L=\displaystyle\int\limits^b_a\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}dx\)

Illustration Questions

Find the length of the arc of curve \(y=ln\,(cos\,x)\) from point \(P(0,\,0)\) to \(Q\left(\dfrac{\pi}{3},\,ln\dfrac{1}{2}\right)\).

A \(ln(2+\sqrt3)\)

B \(ln\,5\)

C \(ln\sqrt2\)

D \(ln\sqrt3\)

×

Length of the curve of \(y=f(x)\) from \(P(a,f(a))\) to \((b,\,f(b))=L=\displaystyle\int\limits^b_a\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}dx\)

In this case,

 \(a=0,\;b=\dfrac{\pi}{3},\;y=ln(cos\,x)\)

\(\Rightarrow\;\dfrac{dy}{dx}=\dfrac{1}{cos\,x}×\dfrac{d}{dx}(cos\,x)\)

\(=-\dfrac{sin\,x}{cos\,x}\\=-tan\,x\)

\(\therefore\;\left(\dfrac{dy}{dx}\right)^2=tan^2x\)

\(\Rightarrow\;1+\left(\dfrac{dy}{dx}\right)^2\\=1+tan^2x\\=sec^2x\)

\(\therefore\;L=\displaystyle\int\limits^{\dfrac{\pi}{3}}_0\sqrt{sec^2x}\;dx\)

\(=\displaystyle\int\limits^{\dfrac{\pi}{3}}_0(sec\,x)\;dx\)

\(=\Bigg[ln|sec\,x+tan\,x|\Bigg]^{\pi/3}_0\)

\(=ln\left(sec\dfrac{\pi}{3}+tan\dfrac{\pi}{3}\right)-ln(sec\,0+tan\,0)\)

\(=ln(2+\sqrt3)-ln(1+0)\)

\(=ln(2+\sqrt3)\)

Find the length of the arc of curve \(y=ln\,(cos\,x)\) from point \(P(0,\,0)\) to \(Q\left(\dfrac{\pi}{3},\,ln\dfrac{1}{2}\right)\).

A

\(ln(2+\sqrt3)\)

.

B

\(ln\,5\)

C

\(ln\sqrt2\)

D

\(ln\sqrt3\)

Option A is Correct

Illustration Questions

Find the length of arc of the curve \(y=2+\dfrac{1}{4}\left(e^{2x}+e^{-2x}\right)\) between the point \(P\left(0,\,\dfrac{5}{2}\right)\) to \(Q\left(1,\,\dfrac{8+e^2+e^{-2}}{4}\right)\).

A \(\dfrac{1}{4}\left(\dfrac{e^4-1}{e^2}\right)\)

B \(e-1\)

C \(ln\,2\)

D \(e^2\)

×

Length of the curve of \(y=f(x)\) from \(P(a,f(a))\) to \((b,\,f(b))=L=\displaystyle\int\limits^b_a\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}dx\)

In this problem,

 \(a=0,\;b=1,\;y=2+\dfrac{1}{4}\left(e^{2x}+e^{-2x}\right)\)

\(\Rightarrow\;\dfrac{dy}{dx}=\dfrac{1}{4}\left(e^{2x}×2+e^{-2x}×-2\right)\)

\(=\dfrac{e^{2x}-e^{-2x}}{2}\)

\(\left(\dfrac{dy}{dx}\right)^2=\left(\dfrac{e^{2x}-e^{-2x}}{2}\right)^2\)

\(\Rightarrow\;1+\left(\dfrac{dy}{dx}\right)^2\\=1+\left(\dfrac{e^{2x}-e^{-2x}}{2}\right)^2\)

\(=1+\dfrac{e^{4x}+e^{-4x}-2}{4}\)

\(=\dfrac{e^{4x}+e^{-4x}+2}{4}\)

\(=\dfrac{\left(e^{2x}+e^{-2x}\right)^2}{4}\)

\(\therefore\;L=\displaystyle\int\limits^{1}_0\sqrt{\dfrac{\left(e^{2x}+e^{-2x}\right)^2}{4}}\,dx\)

\(=\dfrac{1}{2}\,\displaystyle\int\limits^{1}_0\left(e^{2x}+e^{-2x}\right)\;dx\)

\(=\dfrac{1}{2}\left[\dfrac{e^{2x}}{2}-\dfrac{e^{-2x}}{2}\right]^1_0\)

\(=\dfrac{1}{2}\left[\left(\dfrac{e^{2}}{2}-\dfrac{e^{-2}}{2}\right)-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)\right]\)

\(=\dfrac{1}{4}\left(e^2-e^{-2}\right)\)

\(=\dfrac{1}{4}\left(\dfrac{e^4-1}{e^2}\right)\)

Find the length of arc of the curve \(y=2+\dfrac{1}{4}\left(e^{2x}+e^{-2x}\right)\) between the point \(P\left(0,\,\dfrac{5}{2}\right)\) to \(Q\left(1,\,\dfrac{8+e^2+e^{-2}}{4}\right)\).

A

\(\dfrac{1}{4}\left(\dfrac{e^4-1}{e^2}\right)\)

.

B

\(e-1\)

C

\(ln\,2\)

D

\(e^2\)

Option A is Correct

The Arc Length Function

  • Sometimes it is useful to construct a function that measure the arc length of curve from particular starting point to any other variable point of the curve.
  • Let a curve \(C\) has the function expression \(y=f(x)\;(a\leq x\leq b)\). Let \(s(x)\) be the distance along  \(C\) from initial point \(P_0(a,\,f(a))\) to \(Q(x,\,f(x))\). We define the arc length function

\(s(x)=\displaystyle\int\limits^x_a\sqrt{1+(f'(t))^2}dt\)

Illustration Questions

Find the arc length function for the curve \(y=(1-x^{2/3})^{3/2}\) with starting point \(P_0(0,\,1)\).

A \(s(x)=\dfrac{3}{2}x^{4/3}\)

B \(s(x)=\dfrac{3}{2}x^{2/3}\)

C \(s(x)=x^2\)

D \(s(x)=ln\,4\)

×

Are length function\(=s(x)=\displaystyle\int\limits^x_a\sqrt{1+(f'(t))^2}\,dt\)

In this problem,

 \(a=0,\;f(t)=\left(1-t^{2/3}\right)^{3/2}\)

\(\Rightarrow\,f'(t)=\dfrac{3}{2}\left(1-t^{2/3}\right)^{1/2}×-\dfrac{2}{3}t^{-1/3}\)

\(=-\dfrac{(1-t^{2/3})^{1/2}}{t^{1/3}}\)

\(\Rightarrow\;(f'(t))^2=\dfrac{1-t^{2/3}}{t^{2/3}}\)

\(\Rightarrow\;1+(f'(t))^2\\=1+\dfrac{1-t^{2/3}}{t^{2/3}}\)

\(=\dfrac{1}{t^{2/3}}=t^{-2/3}\)

\(\therefore\;s(x)=\displaystyle\int\limits^x_0\sqrt{t^{-2/3}}dt\)

\(=\displaystyle\int\limits^x_0t^{-1/3}\;dt=\Bigg[\dfrac{t^{2/3}}{{2}/{3}}\Bigg]^x_0\)

\(=\dfrac{3}{2}\left[x^{2/3}-0\right]\)

\(=\dfrac{3}{2}x^{2/3}\)

Find the arc length function for the curve \(y=(1-x^{2/3})^{3/2}\) with starting point \(P_0(0,\,1)\).

A

\(s(x)=\dfrac{3}{2}x^{4/3}\)

.

B

\(s(x)=\dfrac{3}{2}x^{2/3}\)

C

\(s(x)=x^2\)

D

\(s(x)=ln\,4\)

Option B is Correct

Length of the Arc of Curve Given in the Integral Form

  • Suppose  \(y=\displaystyle\int\limits^x_0f(t)\,dt\)  is the curve  \(a\leq x\leq b\)  given then \(\dfrac{dy}{dx}=f(x)\to\) by Leibniz's formula

\(\Rightarrow\;L=\text{length of curve}=\displaystyle\int\limits^b_a\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\,dx\)

\(=\displaystyle\int\limits^b_a\sqrt{1+(f'(x))^2}\,dx\)

Illustration Questions

Find the length of the curve \(y=\displaystyle\int\limits^x_1\sqrt{t^2-1}\,dt\) when \(1\leq x\leq6\,\,\,.\)

A \(4\)

B \(\dfrac{215}{3}\)

C \(\dfrac{81}{5}\)

D \(\dfrac{27}{4}\)

×

Length of the curve,  \((L)=\displaystyle\int\limits^b_a\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\,dx\)

In this problem,

 \(a=1,\;b=6,\;y=\displaystyle\int\limits^x_1\sqrt{t^2-1}\,dt\)

 By Leibniz's Rule,

\(\Rightarrow\;\dfrac{dy}{dx}=\sqrt{x^2-1}\)                   

\(\therefore\;\left(\dfrac{dy}{dx}\right)^2=x^2-1\)

\(\Rightarrow\;1+\left(\dfrac{dy}{dx}\right)^2\\=1+x^2-1\\=x^2\)

\(\therefore\;L=\displaystyle\int\limits^6_1x^2\,dx\\=\Bigg[\dfrac{x^3}{3}\Bigg]^6_1\)

\(=\dfrac{6^3}{3}-\dfrac{1}{3}\)

\(=\dfrac{215}{3}\)

Find the length of the curve \(y=\displaystyle\int\limits^x_1\sqrt{t^2-1}\,dt\) when \(1\leq x\leq6\,\,\,.\)

A

\(4\)

.

B

\(\dfrac{215}{3}\)

C

\(\dfrac{81}{5}\)

D

\(\dfrac{27}{4}\)

Option B is Correct

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