Question

Find the coordinates of the centre of curvature for any point $(x,y)$ on the parabola $x=a{{t}^{2}}\ ,\ y=2at$ .

Answer 1

Hint : We have given the equations in the parametric form where ‘t’ is the parameter. First we need to differentiate with respect to ‘t’ and then later we do the second derivative with respect to ‘t’ . Now, we solve for the coordinates of the center of the curvature by applying the formula and putting all the values in the formula, we will get our required solution.
Formula used:
If $\left( \alpha ,\beta \right)$ are the coordinates of the center of the curvature of the curve $y=f\left( x \right)$ at $\left( {{x}_{1}},{{y}_{1}} \right)$
Then,
$\alpha ={{x}_{1}}-\dfrac{\dfrac{dy}{dx}\left( 1+{{\left( \dfrac{dy}{dx} \right)}^{2}} \right)}{\dfrac{{{d}^{2}}y}{d{{x}^{2}}}}$
And,
$\beta ={{y}_{1}}-\dfrac{\left( 1+{{\left( \dfrac{dy}{dx} \right)}^{2}} \right)}{\dfrac{{{d}^{2}}y}{d{{x}^{2}}}}$

Complete step-by-step answer :
We have given that,
$\Rightarrow x=a{{t}^{2}}\ ,\ y=2at$ ---- (in parametric form)
First derivative of the above equations with respect to ‘t’, we get
$\Rightarrow y'=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{2a}{2at}$
Simplifying the above, we get
$\Rightarrow y'=\dfrac{1}{t}$
Second derivative with respect to ‘t’, we get
$\Rightarrow y''=\dfrac{d}{dx}y'=\dfrac{d}{dt}\left( \dfrac{1}{t} \right)\dfrac{dt}{dx}$
$\Rightarrow y''=-\dfrac{1}{{{t}^{2}}}\times \dfrac{1}{2at}$
Simplifying the above derivative, we get
$\Rightarrow y''=-\dfrac{1}{2a{{t}^{3}}}$
Now, solving for the coordinates of the centre of the curvature.
If $\left( \alpha ,\beta \right)$ are the coordinates of the center of the curvature of the curve $y=f\left( x \right)$ at $\left( {{x}_{1}},{{y}_{1}} \right)$
Then,
$\alpha ={{x}_{1}}-\dfrac{\dfrac{dy}{dx}\left( 1+{{\left( \dfrac{dy}{dx} \right)}^{2}} \right)}{\dfrac{{{d}^{2}}y}{d{{x}^{2}}}}$
And,
$\beta ={{y}_{1}}-\dfrac{\left( 1+{{\left( \dfrac{dy}{dx} \right)}^{2}} \right)}{\dfrac{{{d}^{2}}y}{d{{x}^{2}}}}$
Therefore,
Let the coordinates of the center of the curvature at any point (x, y) be $\left( {{x}_{1}},{{y}_{1}} \right)$ .
Thus,
$\Rightarrow {{x}_{1}}=x-\dfrac{y'\left( 1+{{\left( y' \right)}^{2}} \right)}{y''}$
As we know that,
Here $x=a{{t}^{2}}$
Now putting all the values, we get
$\Rightarrow {{x}_{1}}=a{{t}^{2}}-\dfrac{\dfrac{1}{t}\times \left( 1+{{\left( \dfrac{1}{t} \right)}^{2}} \right)}{-\dfrac{1}{2a{{t}^{3}}}}$
Simplifying the above, we get
$\Rightarrow {{x}_{1}}=a{{t}^{2}}+\dfrac{2a{{t}^{3}}}{t}\times \left( 1+\dfrac{1}{{{t}^{2}}} \right)$
Simplifying the above result, we get
$\Rightarrow {{x}_{1}}=3a{{t}^{2}}+2a$
Now,
$\Rightarrow {{y}_{1}}=y+\dfrac{\left( 1+{{\left( y' \right)}^{2}} \right)}{y''}$
Putting all the values from the above, we get
$\Rightarrow {{y}_{1}}=2at+\dfrac{\left( 1+{{\left( \dfrac{1}{t} \right)}^{2}} \right)}{-\dfrac{1}{2a{{t}^{3}}}}=2at+\dfrac{\left( 1+\dfrac{1}{{{t}^{2}}} \right)}{-\dfrac{1}{2a{{t}^{3}}}}$
Simplifying the above, we get
$\Rightarrow {{y}_{1}}=-2a{{t}^{3}}$
Therefore,
$\Rightarrow \left( {{x}_{1}},{{y}_{1}} \right)=\left( 3a{{t}^{2}}+2a,-2a{{t}^{3}} \right)$
Hence, the coordinate of the centre of curvature at any point $(x,y)$ on the parabola are $\left( 3a{{t}^{2}}+2a,-2a{{t}^{3}} \right)$ .
So, the correct answer is “$\left( 3a{{t}^{2}}+2a,-2a{{t}^{3}} \right)$”.

Note : While solving these types of problems, students should know about the differentiation. We should also keep in mind all the properties of the derivative to start solving the question as it is the first step while solving the question. Students also know the basic concepts and the formulae for finding the coordinates of the center of the curvature at any point (x, y).