Question

If \[x = a\sec \theta ,y = b\tan \theta \], then prove that \[\dfrac{{{\partial ^2}y}}{{\partial {x^2}}} = - \dfrac{{{b^4}}}{{{a^2}{y^3}}}\].

Answer 1

Hint: We differentiate the values of $x$ and $y$ with respect to \[\theta \] which gives us the values of first derivatives, and then we divide the derivative of $y$ with respect to \[\theta \] by the derivative of $x$ with respect to \[\theta \] which gives us the derivative of $y$ with respect of $x$. Then we again differentiate this obtained value with respect to $x$ but we have to apply the chain rule of differentiation as the value contains \[\theta \].

Differentiation of some functions used in the solution are:

\[\dfrac{d}{{dx}}\sec x = \sec x\tan x\]

\[\dfrac{d}{{dx}}\tan x = {\sec ^2}x\]

\[\dfrac{d}{{dx}}\cos ecx = - \csc x \cot x\]

Complete step by step solution:

We have the values of $x$ and $y$ as \[x = a\sec \theta ,y = b\tan \theta \].

First we differentiate \[x = a\sec \theta \] with respect to \[\theta \].

\[ \Rightarrow \dfrac{{dx}}{{d\theta }} = \dfrac{d}{{d\theta }}(a\sec \theta )\]

Since $a$ is a constant value, it comes out of the differentiation

\[ \Rightarrow \dfrac{{dx}}{{d\theta }} = a\dfrac{d}{{d\theta }}(\sec \theta )\]

Since, we know \[\dfrac{d}{{dx}}\sec x = \sec x\tan x \]

\[ \Rightarrow \dfrac{{dx}}{{d\theta }} = a(\sec \theta \tan \theta )\]

Opening the brackets we get

\[ \Rightarrow \dfrac{{dx}}{{d\theta }} = a\sec \theta \tan \theta \] … (1)

Now we differentiate \[y = b\tan \theta \] with respect to \[\theta \].

\[ \Rightarrow \dfrac{{dy}}{{d\theta }} = \dfrac{d}{{d\theta }}(b\tan \theta )\]

Since $b$ is a constant value, it comes out of the differentiation

\[ \Rightarrow \dfrac{{dy}}{{d\theta }} = b\dfrac{d}{{d\theta }}(\tan \theta )\]

Since, we know \[\dfrac{d}{{dx}}\tan x = {\sec ^2}x\]

\[ \Rightarrow \dfrac{{dy}}{{d\theta }} = b ({\sec ^2}\theta )\]

Opening the brackets we get

\[ \Rightarrow \dfrac{{dy}}{{d\theta }} = b {\sec ^2}\theta \] … (2)

Now we divide equation (2) by equation (1)

\[ \Rightarrow \dfrac{{\dfrac{{dy}}{{d\theta }}}}{{\dfrac{{dx}}{{d\theta }}}} = \dfrac{{b{{\sec }^2}\theta }}{{a\sec \theta \tan \theta }}\]

We can write the LHS and RHS of the equation in simpler form as

\[ \Rightarrow \dfrac{{dy}}{{d\theta }} \times \dfrac{{d\theta }}{{dx}} = \dfrac{{b\sec \theta \times \sec \theta }}{{a\sec \theta \tan \theta }}\]

Cancel out same terms from numerator and denominator on each side of the equation.

\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{b\sec \theta }}{{a\tan \theta }}\]

Now, we know \[\sec \theta = \dfrac{1}{{\cos \theta }},\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]

Therefore, substituting the values in RHS of the equation we get

\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{b\dfrac{1}{{\cos \theta }}}}{{a\dfrac{{\sin \theta }}{{\cos \theta }}}}\]

We can write RHS of the equation in a simpler form as

\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{b}{a} \times \dfrac{1}{{\cos \theta }} \times \dfrac{{\cos \theta }}{{\sin \theta }}\]

Cancel out the same terms from numerator and denominator

\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{b}{{a\sin \theta }}\]

We know the value \[\dfrac{1}{{\sin x}} = \cos ecx\]

\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{b\cos ec\theta }}{a}\] … (3)

Now we differentiate equation (3) with respect to $x$.

\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {\dfrac{{b\cos ec\theta }}{a}} \right)\]

Since the constant values can be taken out from the differentiation, we write

\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{b}{a}[\dfrac{d}{{dx}}\left( {\cos ec\theta } \right)]\]

Since we know \[\dfrac{d}{{dx}}\cos ecx = - \cos ecx\cot x\]

\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{b}{a}[ - \cos ec\theta .\cot \theta \times \dfrac{{d\theta }}{{dx}}]\] … (4)

Now we know from equation (1)

\[ \Rightarrow \dfrac{{dx}}{{d\theta }} = a\sec \theta \tan \theta \]

Taking reciprocal on both sides of the equation we get

\[ \Rightarrow \dfrac{{d\theta }}{{dx}} = \dfrac{1}{{a\sec \theta \tan \theta }}\]

Substitute this value in equation (4)

\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{b}{a}[ - \cos ec\theta .\cot \theta \times \dfrac{1}{{a\sec \theta \tan \theta }}]\]

We take out the negative sign with the constants and write all the terms in simplest form of Sine and cosine as $\tan x = \dfrac{{\sin x}}{{\cos x}},\cot x = \dfrac{{\cos x}}{{\sin x}},\sec x = \dfrac{1}{{\cos x}},\cos ecx = \dfrac{1}{{\sin x}} $

\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{b}{{a \times a}}[\dfrac{1}{{\sin \theta }} \times \dfrac{{\cos \theta }}{{\sin \theta }} \times \dfrac{1}{{\dfrac{1}{{\cos \theta }} \times \dfrac{{\sin \theta }}{{\cos \theta }}}}]\]

Write RHS of the equation in simpler form

\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{b}{{{a^2}}}[\dfrac{1}{{\sin \theta }} \times \dfrac{{\cos \theta }}{{\sin \theta }} \times \dfrac{{\cos \theta \times \cos \theta }}{{\sin \theta }}]\]

Collecting the same terms as we can write \[\underbrace {p \times p \times ..... \times p}_n = {p^n}\]

\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{b}{{{a^2}}}\left( {\dfrac{{{{\cos }^3}\theta }}{{{{\sin }^3}\theta }}} \right)\]

\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{b}{{{a^2}}}{\left( {\dfrac{{\cos \theta }}{{\sin \theta }}} \right)^3}\]

Using \[\dfrac{{\cos x}}{{\sin x}} = \cot x\]

\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{b}{{{a^2}}}{\left( {\cot \theta } \right)^3}\]

We know $\cot x = \dfrac{1}{{\tan x}}$

\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{b}{{{a^2}}}{\left( {\dfrac{1}{{\tan \theta }}} \right)^3}\] … (5)

We are given the value of \[y = b\tan \theta \]

Dividing both sides by b

\[\dfrac{y}{b} = \dfrac{{b\tan \theta }}{b}\]

Cancel same terms from numerator and denominator.

\[\dfrac{y}{b} = \tan \theta \]

Take reciprocal on both sides

\[\dfrac{b}{y} = \dfrac{1}{{\tan \theta }}\]

Substitute the value of \[\dfrac{b}{y} = \dfrac{1}{{\tan \theta }}\]in equation (5)

\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{b}{{{a^2}}}{\left( {\dfrac{b}{y}} \right)^3}\]

Since we know the powers are added when the base is same

\[   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{{{b^{1 + 3}}}}{{{a^2}{y^3}}} \]

\[   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{{{b^4}}}{{{a^2}{y^3}}}  \]

Hence Proved.

Note:

Students many times skip the step of differentiating \[\theta \] with respect to $x$ while we are calculating the second derivative which is wrong. Also, keep in mind always take reciprocal on both sides of the equation, and always convert complex trigonometric functions in sine and cosine for easy calculations.