Question

If we have the condition as \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] and \[k=\dfrac{1}{a}\], then k is equal to:
A. \[\dfrac{y''}{\sqrt{1+y'}}\]
B. \[\dfrac{\left| y'' \right|}{\sqrt{{{\left( 1+y{{'}^{2}} \right)}^{3}}}}\]
C. \[\dfrac{2y''}{\sqrt{1+y'}}\]
D. \[\dfrac{y''}{2\sqrt{{{\left( 1+y{{'}^{2}} \right)}^{3}}}}\]

Answer 1

Hint: First of all we will have to know about y’ and y’’. y’ is equal to differentiation of y with respect to x and y’’ is equal to double differentiation of y or derivative of y’. We will find the values of y’ and y’’ and then we will check the options one by one.

Complete step-by-step answer:
We have been given \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] and \[k=\dfrac{1}{a}\].
Now we will find the values of y’ and y’’ then will check the options one by one.
y’ is nothing but derivative of y with respect to x and y’’ is the second derivative of y.
We have \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\].
On differentiation the equation with respect to ‘x’ we get as follows:
\[\begin{align}
  & \dfrac{d}{dx}\left( {{x}^{2}}+{{y}^{2}} \right)=\dfrac{d}{dx}\left( {{a}^{2}} \right) \\
 & \dfrac{d}{dx}\left( {{x}^{2}} \right)+\dfrac{d}{dx}\left( {{y}^{2}} \right)=0 \\
\end{align}\]
Since the derivative of a constant term is zero.
\[\begin{align}
  & \Rightarrow 2x+2y\dfrac{dy}{dx}=0 \\
 & \Rightarrow 2y\dfrac{dy}{dx}=-2x \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{-2x}{2y} \\
 & \Rightarrow y'=\dfrac{-x}{y}......(1) \\
\end{align}\]
Again differentiating the equating (1) with respect to ‘x’ we get as follows:
Since we know that the quotient rule of derivative in which we have to find the derivative of \[F(x)\] divided by \[g(x)\] is given by as follows:
\[\dfrac{d}{dx}\left[ \dfrac{F\left( x \right)}{g\left( x \right)} \right]=\dfrac{g\left( x \right)F'\left( x \right)-F\left( x \right)g'\left( x \right)}{{{\left[ g\left( x \right) \right]}^{2}}}\]
\[\begin{align}
  & \dfrac{d}{dx}\left( y' \right)=\dfrac{d}{dx}\left( \dfrac{-x}{y} \right) \\
 & \Rightarrow y''=\dfrac{y\dfrac{d}{dx}\left( -x \right)-\left( -x \right)\dfrac{dy}{dx}}{{{y}^{2}}} \\
\end{align}\]
Substituting the value of \[\dfrac{dy}{dx}\] from equation (1) we get as follows:
\[y''=\dfrac{-y+x\left( \dfrac{-x}{y} \right)}{{{y}^{2}}}=\dfrac{-y-\dfrac{{{x}^{2}}}{y}}{{{y}^{2}}}=\dfrac{\dfrac{-{{y}^{2}}-{{x}^{2}}}{y}}{{{y}^{2}}}=\dfrac{-\left( {{x}^{2}}+{{y}^{2}} \right)}{{{y}^{3}}}\]
Since we have given \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\]
\[y''=\dfrac{{{-a}^{2}}}{{{y}^{3}}}\]

Now we will check the options one by one as follows:
A. \[\dfrac{y''}{\sqrt{1+y'}}\]
Substituting the values y’ and y’’ we get as follows:
\[\Rightarrow \dfrac{-\left( {{a}^{2}} \right)}{{{y}^{3}}\sqrt{1+\left( \dfrac{-x}{y} \right)}}=\dfrac{-{{a}^{2}}}{{{y}^{3}}\sqrt{\dfrac{y-x}{y}}}\ne k\]
Hence, this is not the correct option.

B. \[\dfrac{\left| y'' \right|}{\sqrt{{{\left( 1+y{{'}^{2}} \right)}^{3}}}}\]
On substituting the values of y’ and y’’ we get as follows:
\[\Rightarrow \dfrac{\left| \dfrac{{{a}^{2}}}{{{y}^{3}}} \right|}{\sqrt{{{\left[ 1+{{\left( \dfrac{-x}{y} \right)}^{2}} \right]}^{3}}}}=\dfrac{{{a}^{2}}}{{{y}^{3}}\sqrt{{{\left[ 1+\dfrac{{{x}^{2}}}{{{y}^{2}}} \right]}^{3}}}}=\dfrac{{{a}^{2}}}{{{y}^{3}}\sqrt{{{\left( \dfrac{{{y}^{2}}+{{x}^{2}}}{{{y}^{2}}} \right)}^{3}}}}\]
Since we have been given \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\], so,
\[\Rightarrow \dfrac{\left| y'' \right|}{\sqrt{{{\left( 1+y{{'}^{2}} \right)}^{3}}}}=\dfrac{{{a}^{2}}}{{{y}^{3}}\sqrt{{{\left( \dfrac{{{a}^{2}}}{{{y}^{2}}} \right)}^{3}}}}=\dfrac{{{a}^{2}}}{{{y}^{3}}\sqrt{\dfrac{{{a}^{6}}}{{{y}^{6}}}}}=\dfrac{{{a}^{2}}\times \sqrt{{{y}^{6}}}}{{{y}^{3}}\times \sqrt{{{a}^{6}}}}=\dfrac{{{a}^{2}}\times {{y}^{3}}}{{{y}^{3}}\times {{a}^{3}}}=\dfrac{1}{a}=k\]
Hence, this is the correct option.

C. \[\dfrac{2y''}{\sqrt{\left( 1+y' \right)}}\]
On substituting the values of y’ and y’’, we get as follows:
\[\Rightarrow \dfrac{2\left( \dfrac{-{{a}^{2}}}{{{y}^{3}}} \right)}{\sqrt{1+\left( \dfrac{-x}{y} \right)}}=\dfrac{-2{{a}^{2}}}{{{y}^{3}}\sqrt{\dfrac{y-x}{y}}}\ne k\]
Hence, this is not the correct option.

D. \[\dfrac{y''}{2\sqrt{{{\left( 1+y{{'}^{2}} \right)}^{3}}}}\]
On substituting the values of y’ and y’’ we get as follows:
\[\Rightarrow \dfrac{\dfrac{-{{a}^{2}}}{{{y}^{3}}}}{2{{\sqrt{\left[ 1+{{\left( \dfrac{-x}{y} \right)}^{2}} \right]}}^{3}}}=\dfrac{-{{a}^{2}}}{2{{y}^{3}}\sqrt{{{\left( \dfrac{{{y}^{2}}+{{x}^{2}}}{{{y}^{2}}} \right)}^{3}}}}\]
Since we have \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\]
\[\Rightarrow \dfrac{y''}{2\sqrt{{{\left( 1+y{{'}^{2}} \right)}^{3}}}}=\dfrac{-{{a}^{2}}}{2{{y}^{3}}\sqrt{{{\left( \dfrac{{{a}^{2}}}{{{y}^{2}}} \right)}^{3}}}}=\dfrac{-{{a}^{2}}\times \sqrt{{{y}^{6}}}}{2{{y}^{3}}\times \sqrt{{{a}^{6}}}}=\dfrac{-{{a}^{2}}\times {{y}^{3}}}{2{{y}^{3}}\times {{a}^{3}}}=\dfrac{-1}{2a}\ne k\]
Hence, this option is not correct.

Therefore, the correct option of the above question is option B.

Note: Be careful while finding the derivative y’ and y’’ as there is a chance that you might make a sign mistake specially, while finding the y’’ and using the quotient rule of derivative. Also, be careful while checking the options and substituting the values of y’ and y’’. The best way to solve this type of question is by checking each option one by one as it includes y' and y'' terms, which cannot be found otherwise.