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

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

#### 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

#### 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

#### 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

#### 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

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

#### 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

#### 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

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

#### 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

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

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

#### 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