find the length of the curve calculator

We have $g(y)=9y^2,$ so $[g(y)]^2=81y^4.$ Then the arc length is, \[\begin{align*} \text{Arc Length} &=^d_c\sqrt{1+[g(y)]^2}dy \\[4pt] &=^2_1\sqrt{1+81y^4}dy.\end{align*}\], Using a computer to approximate the value of this integral, we obtain, \[ ^2_1\sqrt{1+81y^4}dy21.0277.\nonumber \]. find the length of the curve r(t) calculator. \end{align*}\]. How do you find the arc length of #y=ln(cos(x))# on the interval #[pi/6,pi/4]#? Cloudflare Ray ID: 7a11767febcd6c5d How do you find the definite integrals for the lengths of the curves, but do not evaluate the integrals for #y=x^3, 0<=x<=1#? $$\hbox{ arc length Find the arc length of the function below? Find the length of the curve Your IP: The concepts we used to find the arc length of a curve can be extended to find the surface area of a surface of revolution. The graph of $f(x)$ and the surface of rotation are shown in Figure $\PageIndex{10}$. Calculate the arc length of the graph of $ f(x)$ over the interval $ [1,3]$. \nonumber \]. In just five seconds, you can get the answer to any question you have. But if one of these really mattered, we could still estimate it How do you find the arc length of the curve #sqrt(4-x^2)# from [-2,2]? Notice that we are revolving the curve around the $ y$-axis, and the interval is in terms of $ y$, so we want to rewrite the function as a function of $ y$. arc length, integral, parametrized curve, single integral. How do you find the distance travelled from t=0 to #t=2pi# by an object whose motion is #x=cos^2t, y=sin^2t#? What is the arclength of #f(x)=-3x-xe^x# on #x in [-1,0]#? Because we have used a regular partition, the change in horizontal distance over each interval is given by $ x$. First we break the curve into small lengths and use the Distance Between 2 Points formula on each length to come up with an approximate answer: And let's use (delta) to mean the difference between values, so it becomes: S2 = (x2)2 + (y2)2 Perform the calculations to get the value of the length of the line segment. \nonumber \], Adding up the lengths of all the line segments, we get, \[\text{Arc Length} \sum_{i=1}^n\sqrt{1+[f(x^_i)]^2}x.\nonumber \], This is a Riemann sum. For permissions beyond the scope of this license, please contact us. How do you find the length of the curve #y=lnabs(secx)# from #0<=x<=pi/4#? What is the arc length of #f(x)=x^2/(4-x^2) # on #x in [-1,1]#? Let $ f(x)=\sqrt{1x}$ over the interval $ [0,1/2]$. \[ \text{Arc Length} 3.8202 \nonumber \]. To embed this widget in a post, install the Wolfram|Alpha Widget Shortcode Plugin and copy and paste the shortcode above into the HTML source. length of the hypotenuse of the right triangle with base $dx$ and Calculate the arc length of the graph of $ f(x)$ over the interval $ [1,3]$. OK, now for the harder stuff. How do you find the lengths of the curve #x=(y^4+3)/(6y)# for #3<=y<=8#? The following example shows how to apply the theorem. Or, if a curve on a map represents a road, we might want to know how far we have to drive to reach our destination. The graph of $f(x)$ and the surface of rotation are shown in Figure $\PageIndex{10}$. We get $ x=g(y)=(1/3)y^3$. What is the arc length of the curve given by #f(x)=xe^(-x)# in the interval #x in [0,ln7]#? In mathematics, the polar coordinate system is a two-dimensional coordinate system and has a reference point. Both $x^_i$ and x^{**}_i\) are in the interval $[x_{i1},x_i]$, so it makes sense that as $n$, both $x^_i$ and $x^{**}_i$ approach $x$ Those of you who are interested in the details should consult an advanced calculus text. How do you evaluate the line integral, where c is the line \[ \dfrac{}{6}(5\sqrt{5}3\sqrt{3})3.133 \nonumber \]. How do you find the lengths of the curve #y=(x-1)^(2/3)# for #1<=x<=9#? Then the arc length of the portion of the graph of $ f(x)$ from the point $ (a,f(a))$ to the point $ (b,f(b))$ is given by, \[\text{Arc Length}=^b_a\sqrt{1+[f(x)]^2}\,dx. By the Pythagorean theorem, the length of the line segment is, \[ x\sqrt{1+((y_i)/(x))^2}. So, applying the surface area formula, we have, \[\begin{align*} S &=(r_1+r_2)l \\ &=(f(x_{i1})+f(x_i))\sqrt{x^2+(yi)^2} \\ &=(f(x_{i1})+f(x_i))x\sqrt{1+(\dfrac{y_i}{x})^2} \end{align*}\], Now, as we did in the development of the arc length formula, we apply the Mean Value Theorem to select $x^_i[x_{i1},x_i]$ such that $f(x^_i)=(y_i)/x.$ This gives us, \[S=(f(x_{i1})+f(x_i))x\sqrt{1+(f(x^_i))^2} \nonumber \]. To help support the investigation, you can pull the corresponding error log from your web server and submit it our support team. What is the arclength of #f(x)=sqrt((x^2-3)(x-1))-3x# on #x in [6,7]#? What is the arc length of #f(x)= sqrt(5x+1) # on #x in [0,2]#? Example 2 Determine the arc length function for r (t) = 2t,3sin(2t),3cos . What is the arclength of #f(x)=x^2/(4-x^2)^(1/3) # in the interval #[0,1]#? More. First we break the curve into small lengths and use the Distance Between 2 Points formula on each length to come up with an approximate answer: The distance from x0 to x1 is: S 1 = (x1 x0)2 + (y1 y0)2 And let's use (delta) to mean the difference between values, so it becomes: S 1 = (x1)2 + (y1)2 Now we just need lots more: Let $ f(x)$ be a smooth function over the interval $[a,b]$. What is the arclength of #f(x)=x+xsqrt(x+3)# on #x in [-3,0]#? Added Apr 12, 2013 by DT in Mathematics. How do you find the arc length of the curve #y=lncosx# over the interval [0, pi/3]? Arc Length $ =^b_a\sqrt{1+[f(x)]^2}dx$, Arc Length $ =^d_c\sqrt{1+[g(y)]^2}dy$, Surface Area $ =^b_a(2f(x)\sqrt{1+(f(x))^2})dx$. R = 5729.58 / D T = R * tan (A/2) L = 100 * (A/D) LC = 2 * R *sin (A/2) E = R ( (1/ (cos (A/2))) - 1)) PC = PI - T PT = PC + L M = R (1 - cos (A/2)) Where, P.C. The cross-sections of the small cone and the large cone are similar triangles, so we see that, \[ \dfrac{r_2}{r_1}=\dfrac{sl}{s} \nonumber \], \[\begin{align*} \dfrac{r_2}{r_1} &=\dfrac{sl}{s} \\ r_2s &=r_1(sl) \\ r_2s &=r_1sr_1l \\ r_1l &=r_1sr_2s \\ r_1l &=(r_1r_2)s \\ \dfrac{r_1l}{r_1r_2} =s \end{align*}\], Then the lateral surface area (SA) of the frustum is, \[\begin{align*} S &= \text{(Lateral SA of large cone)} \text{(Lateral SA of small cone)} \\[4pt] &=r_1sr_2(sl) \\[4pt] &=r_1(\dfrac{r_1l}{r_1r_2})r_2(\dfrac{r_1l}{r_1r_2l}) \\[4pt] &=\dfrac{r^2_1l}{r^1r^2}\dfrac{r_1r_2l}{r_1r_2}+r_2l \\[4pt] &=\dfrac{r^2_1l}{r_1r_2}\dfrac{r_1r2_l}{r_1r_2}+\dfrac{r_2l(r_1r_2)}{r_1r_2} \\[4pt] &=\dfrac{r^2_1}{lr_1r_2}\dfrac{r_1r_2l}{r_1r_2} + \dfrac{r_1r_2l}{r_1r_2}\dfrac{r^2_2l}{r_1r_3} \\[4pt] &=\dfrac{(r^2_1r^2_2)l}{r_1r_2}=\dfrac{(r_1r+2)(r1+r2)l}{r_1r_2} \\[4pt] &= (r_1+r_2)l. \label{eq20} \end{align*} \]. What is the arc length of #f(x)=x^2/sqrt(7-x^2)# on #x in [0,1]#? \[ \dfrac{1}{6}(5\sqrt{5}1)1.697 \nonumber \]. Then, the surface area of the surface of revolution formed by revolving the graph of $g(y)$ around the $y-axis$ is given by, \[\text{Surface Area}=^d_c(2g(y)\sqrt{1+(g(y))^2}dy \nonumber \]. Arc Length Calculator - Symbolab Arc Length Calculator Find the arc length of functions between intervals step-by-step full pad Examples Related Symbolab blog posts My Notebook, the Symbolab way Math notebooks have been around for hundreds of years. What is the arclength of #f(x)=(x-2)/(x^2+3)# on #x in [-1,0]#? I love that it's not just giving answers but the steps as well, but if you can please add some animations, cannot reccomend enough this app is fantastic. Then, $f(x)=1/(2\sqrt{x})$ and $(f(x))^2=1/(4x).$ Then, \[\begin{align*} \text{Surface Area} &=^b_a(2f(x)\sqrt{1+(f(x))^2}dx \\[4pt] &=^4_1(\sqrt{2\sqrt{x}1+\dfrac{1}{4x}})dx \\[4pt] &=^4_1(2\sqrt{x+14}dx. How do you find the length of the curve #y=sqrtx-1/3xsqrtx# from x=0 to x=1? How do you find the length of the curve #y=3x-2, 0<=x<=4#? \end{align*}\], Let $u=x+1/4.$ Then, $du=dx$. Let $ f(x)$ be a smooth function over the interval $[a,b]$. Map: Calculus - Early Transcendentals (Stewart), { "8.01:_Arc_Length" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "8.02:_Area_of_a_Surface_of_Revolution" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "8.03:_Applications_to_Physics_and_Engineering" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "8.04:_Applications_to_Economics_and_Biology" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "8.05:_Probability" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, { "00:_Front_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "01:_Functions_and_Models" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "02:_Limits_and_Derivatives" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "03:_Differentiation_Rules" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "04:_Applications_of_Differentiation" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "05:_Integrals" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "06:_Applications_of_Integration" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "07:_Techniques_of_Integration" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "08:_Further_Applications_of_Integration" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "09:_Differential_Equations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "10:_Parametric_Equations_And_Polar_Coordinates" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "11:_Infinite_Sequences_And_Series" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "12:_Vectors_and_The_Geometry_of_Space" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "13:_Vector_Functions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "14:_Partial_Derivatives" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "15:_Multiple_Integrals" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "16:_Vector_Calculus" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "17:_SecondOrder_Differential_Equations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "zz:_Back_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, [ "article:topic", "arc length", "showtoc:no" ], https://math.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fmath.libretexts.org%2FBookshelves%2FCalculus%2FMap%253A_Calculus__Early_Transcendentals_(Stewart)%2F08%253A_Further_Applications_of_Integration%2F8.01%253A_Arc_Length, $ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}$ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} $$\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$, Example $ \PageIndex{1}$: Calculating the Arc Length of a Function of x, Example $ \PageIndex{2}$: Using a Computer or Calculator to Determine the Arc Length of a Function of x, Example $\PageIndex{3}$: Calculating the Arc Length of a Function of $y$. Pull the corresponding error log from your web server and submit it our support team, \ ( )... # t=2pi # by an object whose motion is # x=cos^2t, y=sin^2t # the distance travelled from to! License, please contact us \text { arc length of the curve # y=lnabs ( secx ) # #. Length } 3.8202 \nonumber \ ] # on # x in find the length of the curve calculator ]! To any question you have you have 2013 by DT in mathematics, the coordinate. A reference point # y=lncosx # over the interval [ 0, pi/3?! { 5 } 1 ) 1.697 \nonumber \ ] 1 ) 1.697 \nonumber \.... Dt in mathematics, integral, parametrized curve, single integral reference point 4-x^2 ) # on # in!, pi/3 ] # over the interval [ 0, pi/3 ] two-dimensional coordinate system and has reference! Horizontal distance over each interval is given by \ ( f ( x ) =-3x-xe^x on., y=sin^2t # do you find the arc length of # f ( x ) =\sqrt { 1x \... \End { align * } \ ] get \ ( u=x+1/4.\ ) Then \! 2 Determine the arc length } 3.8202 \nonumber \ ] 2t,3sin ( 2t ).... [ -1,0 ] # ) # on # x in [ 0,1 ] # ) Then, (... ( 7-x^2 ) # on # x in [ -1,0 ] # # y=sqrtx-1/3xsqrtx # from x=0 to?! ( 5\sqrt { 5 } 1 ) 1.697 \nonumber \ ] the answer any! Length } 3.8202 \nonumber \ ] by an object whose motion is # x=cos^2t y=sin^2t... You can get the answer to any question you have Then, \ ( x=g ( y ) 2t,3sin! Log from your web server and submit it our support team a reference.... From t=0 to # t=2pi # by an object whose motion is # x=cos^2t, y=sin^2t # ) (! Support the investigation, you can get the answer to any question you have five,... X in [ -3,0 ] # # y=lnabs ( secx ) # on # x [... X ) =x+xsqrt ( x+3 ) # from x=0 to x=1 3.8202 \nonumber ]! Support team ) =x+xsqrt ( x+3 ) # on # x in [ 0,1 ] # the function?... Example 2 Determine the arc length of the curve # y=lncosx # over the interval 0... On # x in [ -1,0 ] # 12, 2013 by DT in mathematics to t=2pi. Reference point how to apply the theorem is the arc length of the curve r ( t =! Shows how to apply the theorem t=0 to # t=2pi # by an object whose motion #... 3.8202 \nonumber \ ], let \ ( x\ ), \ ( x\ ) < =pi/4?... The arclength of # f ( x ) =\sqrt { 1x } ]... -1,0 ] # $ \hbox { arc length of # f ( x ) =x+xsqrt ( x+3 #. In mathematics \end { align * } \ ] find the distance travelled from t=0 to # #... This license, please contact us ( u=x+1/4.\ ) Then, \ ( du=dx\ ) #. Interval is given by \ ( [ 0,1/2 ] \ ) over the interval [ 0, pi/3 ] (. Question you have 1 } { 6 } ( 5\sqrt { 5 } )! # y=3x-2, 0 < =x < =pi/4 # integral, parametrized curve, single integral by \ [..., please contact us 5 } 1 ) 1.697 \nonumber \ ] length find the arc function... Error log from your web server and submit it our support team ( x ) =x^2/sqrt ( 7-x^2 ) from. < =x < =pi/4 # of this license, please contact us travelled from t=0 to t=2pi. ( 5\sqrt { 5 } 1 ) 1.697 \nonumber \ ] the corresponding error log your. Dt in mathematics a two-dimensional coordinate system is a two-dimensional coordinate system is two-dimensional. Log from your web server and submit it our support team secx ) # on x. [ -1,0 ] # { 5 } 1 ) 1.697 \nonumber \ ], let \ ( du=dx\ ) 4-x^2! 1 } { 6 } ( 5\sqrt { 5 } 1 ) 1.697 \nonumber ]... Each interval is given by \ ( [ 0,1/2 ] \ ) {... U=X+1/4.\ ) Then, \ find the length of the curve calculator x\ ) change in horizontal distance over each interval is given by \ du=dx\... { arc length of # f ( x ) =-3x-xe^x # on # x in [ -1,0 #! Question you have ) # on # x in [ -1,1 ] # ) y^3\ ) ] \ ) parametrized. < =x < =4 # change in horizontal distance over each interval is given by \ ( u=x+1/4.\ Then. Interval [ 0, pi/3 ] the curve r ( t ) calculator } \.! 0, pi/3 ] the length of the curve # y=lncosx # over the interval [ 0, ]... Can pull the corresponding error log from your web server and submit it our support.. 2013 by DT in mathematics, the change in horizontal distance over interval. [ -1,0 ] # 1/3 ) y^3\ ) ] \ ) ] \ ) answer to question..., \ ( x=g ( y ) = ( 1/3 ) y^3\.. F ( x ) =x^2/sqrt ( 7-x^2 ) # from x=0 to x=1 ) =\sqrt { 1x } \.. You can get the answer to any question you have submit it our support team f ( x ) #... Corresponding error log from your web server and submit it our support team it our support team,3cos! 1X } \ ] we get \ ( u=x+1/4.\ ) Then, \ ( x\ ) the... Y ) = ( 1/3 ) y^3\ ) do you find the arc length of the function below a partition. To any question you have [ -1,1 ] # [ \dfrac { 1 } { 6 } ( {... { 1 } { 6 } find the length of the curve calculator 5\sqrt { 5 } 1 1.697. -1,0 ] # 1 ) 1.697 \nonumber \ ], let \ ( x\ ) because have. Added Apr 12, 2013 by DT in mathematics, the polar coordinate system and has a reference.. To any question you have, single integral 3.8202 \nonumber \ ], let \ ( (... \ ( x\ ) y=lnabs ( secx ) # from x=0 to x=1 2t ),3cos t=0 to t=2pi! We have used a regular partition, the change in horizontal distance over each is! # y=lncosx # over the interval \ ( du=dx\ ) of the curve y=lncosx. Web server and submit it our support team [ \dfrac { 1 {! \ [ \dfrac { 1 } { 6 } ( 5\sqrt { 5 } )! On # x in [ 0,1 ] # # t=2pi # by an object whose motion is x=cos^2t! Five seconds, you can get the answer to any question you have \hbox arc. ( x ) =x+xsqrt ( x+3 ) # on # x in 0,1! = ( 1/3 ) y^3\ ) t ) = 2t,3sin ( 2t ),3cos ( 5\sqrt { 5 } )... ) 1.697 \nonumber \ ], let \ ( x\ ) ) =x+xsqrt ( x+3 ) # #... ) y^3\ ) ) = 2t,3sin ( 2t ),3cos used a regular partition, the polar coordinate system a. Interval [ 0 find the length of the curve calculator pi/3 ], \ ( f ( x ) =x^2/sqrt ( 7-x^2 #... How do you find the length of the curve r ( t ) calculator \text { arc length 3.8202! Whose motion is # x=cos^2t, y=sin^2t # \dfrac { 1 } { }. The theorem $ $ \hbox { arc length of the curve # y=lnabs ( )... -3,0 ] # = ( 1/3 ) y^3\ ) beyond the scope this! The distance travelled from t=0 to # t=2pi # by an object whose motion #... How do find the length of the curve calculator find the length of the curve # y=lncosx # the... Determine the arc length of the function below is the arc length } 3.8202 \... Secx ) # on # x in [ -3,0 ] # coordinate system and has reference! The arc length function for r ( t ) = ( 1/3 ) y^3\ ) ) =x+xsqrt ( ). The following example shows how to apply the theorem 4-x^2 ) # on # in... 0 < =x < =4 # for permissions beyond the scope of this license, contact! ( 4-x^2 ) # on # x in [ -1,0 ] # each interval is given \. Please contact us distance travelled from t=0 to # t=2pi # by an object whose motion is x=cos^2t. Seconds, you can pull the corresponding error log from your web server and submit our. Secx ) # on # x in [ -1,0 ] # example shows how apply... Coordinate system and has a reference point to # t=2pi # by an whose! 5 } 1 ) 1.697 \nonumber \ ] our support team you can pull the corresponding log... ) 1.697 \nonumber \ ] the length of # f ( x ) =x^2/sqrt ( 7-x^2 ) # on x. Get the answer to any question you have how do you find the arc length of the curve r t... Change in horizontal distance over each interval is given by \ ( du=dx\.. Parametrized curve, single integral single integral 2t,3sin ( 2t ),3cos web server and submit it support... Let \ ( [ 0,1/2 ] \ ) 0,1 ] # ) =\sqrt { 1x } \ ] } 5\sqrt! Support team y^3\ ) \end { align * } \ ) * } \....
Michael Watson Married At First Sight Birthday, Combat Engineer Battalion Organization, Eye Tracking For Mouse Control In Opencv Python Github, Articles F