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Question: Find the coordinates of the center of curvature of any point \((x,\;y)\) on the parabola \(x = a{t^2...

Find the coordinates of the center of curvature of any point (x,  y)(x,\;y) on the parabola x=at2,  y=2atx = a{t^2},\;y = 2at

Explanation

Solution

The equation of parabola is already written in parametric form, so to find the coordinates of the center of curvature of any point (x,  y)(x,\;y) on the parabola, first find derivative and double derivative of yy with respect to xx and then use the following formula to find the coordinates of the center of curvature directly
x‾=x−y1(1+y12)y2 y‾=y+(1+y12)y2  \overline x = x - \dfrac{{{y_1}\left( {1 + y_1^2} \right)}}{{{y_2}}} \\\ \overline y = y + \dfrac{{\left( {1 + y_1^2} \right)}}{{{y_2}}} \\\
Where (x‾,  y‾),  y1  and  y2\left( {\overline x ,\;\overline y } \right),\;{y_1}\;{\text{and}}\;{y_2} are coordinate of center of curvature, first derivative and second derivative respectively.

Formula used:
Formula to find coordinates of center of curvature of a parabola:
x‾=x−y1(1+y12)y2 y‾=y+(1+y12)y2  \overline x = x - \dfrac{{{y_1}\left( {1 + y_1^2} \right)}}{{{y_2}}} \\\ \overline y = y + \dfrac{{\left( {1 + y_1^2} \right)}}{{{y_2}}} \\\
Where (x‾,  y‾),  y1  and  y2\left( {\overline x ,\;\overline y } \right),\;{y_1}\;{\text{and}}\;{y_2} are coordinate of center of curvature, first derivative and second derivative respectively.

Complete step by step solution:
In order to find the coordinates of the center of curvature of any point (x,  y)(x,\;y) on the parabola x=at2,  y=2atx = a{t^2},\;y = 2at, we will first find first and second derivative of yy with respect to xx as follows
y1=dydx=dydtdxdt=d(2at)dtd(at2)dt=2a2at=1t{y_1} = \dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}} = \dfrac{{\dfrac{{d(2at)}}{{dt}}}}{{\dfrac{{d\left( {a{t^2}} \right)}}{{dt}}}} = \dfrac{{2a}}{{2at}} = \dfrac{1}{t}
Now finding second derivative,
y2=dy1dx=dy1dtdxdt=d(1t)dtd(at2)dt=−1t212at=−12at3{y_2} = \dfrac{{d{y_1}}}{{dx}} = \dfrac{{\dfrac{{d{y_1}}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}} = \dfrac{{\dfrac{{d\left( {\dfrac{1}{t}} \right)}}{{dt}}}}{{\dfrac{{d\left( {a{t^2}} \right)}}{{dt}}}} = - \dfrac{1}{{{t^2}}}\dfrac{1}{{2at}} = - \dfrac{1}{{2a{t^3}}}
Now we know that coordinates of center of curvature is given by following formula
x‾=x−y1(1+y12)y2 y‾=y+(1+y12)y2  \overline x = x - \dfrac{{{y_1}\left( {1 + y_1^2} \right)}}{{{y_2}}} \\\ \overline y = y + \dfrac{{\left( {1 + y_1^2} \right)}}{{{y_2}}} \\\
Putting required values in the above formula, in order to get the desired coordinates of the center of curvature, we will get
x‾=at2−1t(1+1t2)(−12at3)=at2+2at3t(1+1t2)=at2+2at2+2a=3at2+2a y‾=2at+(1+1t2)(−12at3)=2at−2at3(1+1t2)=2at−2at3−2at=−2at3  \overline x = a{t^2} - \dfrac{{\dfrac{1}{t}\left( {1 + \dfrac{1}{{{t^2}}}} \right)}}{{\left( { - \dfrac{1}{{2a{t^3}}}} \right)}} = a{t^2} + \dfrac{{2a{t^3}}}{t}\left( {1 + \dfrac{1}{{{t^2}}}} \right) = a{t^2} + 2a{t^2} + 2a = 3a{t^2} + 2a \\\ \overline y = 2at + \dfrac{{\left( {1 + \dfrac{1}{{{t^2}}}} \right)}}{{\left( { - \dfrac{1}{{2a{t^3}}}} \right)}} = 2at - 2a{t^3}\left( {1 + \dfrac{1}{{{t^2}}}} \right) = 2at - 2a{t^3} - 2at = - 2a{t^3} \\\
Therefore, (3at2+2a,  −2at3)\left( {3a{t^2} + 2a,\; - 2a{t^3}} \right) is the required coordinate of the center of curvature of the given parabola.

Note: The given parabola is already expressed in the parametric form that means relating the present variables, indirectly with the help of a third variable. So, if you tackle this type of question where you have to find the coordinates of the center of curvature of a parabola then if the equation of parabola is written in parametric form then use the above formula directly and if not then first express it in parametric form first in order to use the above formula.