Answer

Verified

436.2k+ views

Hint: Use the quotient rule of differentiation in the given question to get the desired result.

Here, starting with \[\dfrac{dx}{dy}=\dfrac{1}{\dfrac{dy}{dx}}\], we have to prove that \[\dfrac{{{d}^{2}}y}{d{{y}^{2}}}=-\dfrac{\dfrac{{{d}^{2}}y}{d{{x}^{2}}}}{{{\left( \dfrac{dy}{dx} \right)}^{3}}}\].

Also, we have to prove \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}.\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=-\dfrac{2a}{{{y}^{3}}}\]for parabola \[{{y}^{2}}=4ax\].

Taking, \[\dfrac{dx}{dy}=\dfrac{1}{\dfrac{dy}{dx}}\]

Now, we will differentiate both sides with respect to \[y\].

Also, we know that quotient rule says that,

\[\dfrac{d}{dy}\left( \dfrac{f}{g} \right)=\dfrac{g\left( \dfrac{dt}{dy} \right)-f\left( \dfrac{dg}{dy} \right)}{{{g}^{2}}}\]

In \[\dfrac{dx}{dy}=\dfrac{1}{\dfrac{dy}{dx}}\]

\[f=1\]and \[g=\dfrac{dy}{dx}\]

Now, differentiating both sides with respect to \[y\].

We get, \[\dfrac{d}{dy}\left( \dfrac{dx}{dy} \right)=\dfrac{\left( \dfrac{dy}{dx} \right)\dfrac{d}{dy}\left( 1 \right)-\left( 1 \right)\left[ \dfrac{d}{dy}\left( \dfrac{dy}{dx} \right) \right]}{{{\left( \dfrac{dy}{dx} \right)}^{2}}}\]

As we know that \[\dfrac{d}{dy}\left( \text{constant} \right)=0\]

Therefore, we get

\[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{0-\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right)}{{{\left( \dfrac{dy}{dx} \right)}^{2}}}\]

Now we will multiply by \[\dfrac{dy}{dx}\] on both the numerator and denominator of \[RHS\].

We get \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right).\dfrac{dy}{dx}}{{{\left( \dfrac{dy}{dx} \right)}^{2}}.\dfrac{dy}{dx}}\]

We know that \[\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right).\dfrac{dy}{dx}=\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]

Therefore, we get \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-\dfrac{{{d}^{2}}y}{d{{x}^{2}}}}{{{\left( \dfrac{dy}{dx} \right)}^{2}}}\][Hence Proved]

which is our required result.

Now, we have to prove that \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}.\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=-\dfrac{2a}{{{y}^{3}}}\] for parabola \[{{y}^{2}}=4ax\].

Now we take parabola, \[{{y}^{2}}=4ax\].

So, we differentiate the above equation with respect to \[x\].

Also, we know that \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\]

Therefore, \[2y\dfrac{dy}{dx}=4a\]

\[\dfrac{dy}{dx}=\dfrac{2a}{y}....\left( i \right)\]

Again differentiating both sides with respect to \[x\],

We get, \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\left( 2a \right){{y}^{-1-1}}.\dfrac{dy}{dx}\]

Now, we put the value of \[\dfrac{dy}{dx}\]from equation \[\left( i \right)\].

We get \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-2a}{{{y}^{2}}}.\dfrac{2a}{y}\]

Therefore, \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-4{{a}^{2}}}{{{y}^{3}}}....\left( ii \right)\]

Now, from previous results, we know that

\[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{\dfrac{-{{d}^{2}}y}{d{{x}^{2}}}}{{{\left( \dfrac{dy}{dx} \right)}^{2}}}\]

By putting values from equation \[\left( i \right)\]and \[\left( ii \right)\]

We get, \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-\left[ \dfrac{-4{{a}^{2}}}{{{y}^{3}}} \right]}{{{\left[ \dfrac{2a}{y} \right]}^{3}}}\]

\[\Rightarrow \dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{\dfrac{+4{{a}^{2}}}{{{y}^{3}}}}{\dfrac{8{{a}^{3}}}{{{y}^{3}}}}\]

By cancelling the like terms,

We get, \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{+1}{2a}....\left( iii \right)\]

Now we will multiply the equation \[\left( ii \right)\]and \[\left( iii \right)\].

We get, \[\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)\left( \dfrac{{{d}^{2}}x}{d{{y}^{2}}} \right)=\left( \dfrac{-4{{a}^{2}}}{{{y}^{3}}} \right).\dfrac{1}{2a}\]

By cancelling the like terms,

We get, \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}.\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-2a}{{{y}^{3}}}\]which is the required result.

Note: In the term \[\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right)\], some students cancel \[dy\] from numerator and denominator considering them to be like terms but that is wrong and \[\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right)=\dfrac{\dfrac{{{d}^{2}}y}{d{{x}^{2}}}}{\dfrac{dy}{dx}}\].

Here, starting with \[\dfrac{dx}{dy}=\dfrac{1}{\dfrac{dy}{dx}}\], we have to prove that \[\dfrac{{{d}^{2}}y}{d{{y}^{2}}}=-\dfrac{\dfrac{{{d}^{2}}y}{d{{x}^{2}}}}{{{\left( \dfrac{dy}{dx} \right)}^{3}}}\].

Also, we have to prove \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}.\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=-\dfrac{2a}{{{y}^{3}}}\]for parabola \[{{y}^{2}}=4ax\].

Taking, \[\dfrac{dx}{dy}=\dfrac{1}{\dfrac{dy}{dx}}\]

Now, we will differentiate both sides with respect to \[y\].

Also, we know that quotient rule says that,

\[\dfrac{d}{dy}\left( \dfrac{f}{g} \right)=\dfrac{g\left( \dfrac{dt}{dy} \right)-f\left( \dfrac{dg}{dy} \right)}{{{g}^{2}}}\]

In \[\dfrac{dx}{dy}=\dfrac{1}{\dfrac{dy}{dx}}\]

\[f=1\]and \[g=\dfrac{dy}{dx}\]

Now, differentiating both sides with respect to \[y\].

We get, \[\dfrac{d}{dy}\left( \dfrac{dx}{dy} \right)=\dfrac{\left( \dfrac{dy}{dx} \right)\dfrac{d}{dy}\left( 1 \right)-\left( 1 \right)\left[ \dfrac{d}{dy}\left( \dfrac{dy}{dx} \right) \right]}{{{\left( \dfrac{dy}{dx} \right)}^{2}}}\]

As we know that \[\dfrac{d}{dy}\left( \text{constant} \right)=0\]

Therefore, we get

\[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{0-\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right)}{{{\left( \dfrac{dy}{dx} \right)}^{2}}}\]

Now we will multiply by \[\dfrac{dy}{dx}\] on both the numerator and denominator of \[RHS\].

We get \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right).\dfrac{dy}{dx}}{{{\left( \dfrac{dy}{dx} \right)}^{2}}.\dfrac{dy}{dx}}\]

We know that \[\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right).\dfrac{dy}{dx}=\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]

Therefore, we get \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-\dfrac{{{d}^{2}}y}{d{{x}^{2}}}}{{{\left( \dfrac{dy}{dx} \right)}^{2}}}\][Hence Proved]

which is our required result.

Now, we have to prove that \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}.\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=-\dfrac{2a}{{{y}^{3}}}\] for parabola \[{{y}^{2}}=4ax\].

Now we take parabola, \[{{y}^{2}}=4ax\].

So, we differentiate the above equation with respect to \[x\].

Also, we know that \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\]

Therefore, \[2y\dfrac{dy}{dx}=4a\]

\[\dfrac{dy}{dx}=\dfrac{2a}{y}....\left( i \right)\]

Again differentiating both sides with respect to \[x\],

We get, \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\left( 2a \right){{y}^{-1-1}}.\dfrac{dy}{dx}\]

Now, we put the value of \[\dfrac{dy}{dx}\]from equation \[\left( i \right)\].

We get \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-2a}{{{y}^{2}}}.\dfrac{2a}{y}\]

Therefore, \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-4{{a}^{2}}}{{{y}^{3}}}....\left( ii \right)\]

Now, from previous results, we know that

\[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{\dfrac{-{{d}^{2}}y}{d{{x}^{2}}}}{{{\left( \dfrac{dy}{dx} \right)}^{2}}}\]

By putting values from equation \[\left( i \right)\]and \[\left( ii \right)\]

We get, \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-\left[ \dfrac{-4{{a}^{2}}}{{{y}^{3}}} \right]}{{{\left[ \dfrac{2a}{y} \right]}^{3}}}\]

\[\Rightarrow \dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{\dfrac{+4{{a}^{2}}}{{{y}^{3}}}}{\dfrac{8{{a}^{3}}}{{{y}^{3}}}}\]

By cancelling the like terms,

We get, \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{+1}{2a}....\left( iii \right)\]

Now we will multiply the equation \[\left( ii \right)\]and \[\left( iii \right)\].

We get, \[\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)\left( \dfrac{{{d}^{2}}x}{d{{y}^{2}}} \right)=\left( \dfrac{-4{{a}^{2}}}{{{y}^{3}}} \right).\dfrac{1}{2a}\]

By cancelling the like terms,

We get, \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}.\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-2a}{{{y}^{3}}}\]which is the required result.

Note: In the term \[\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right)\], some students cancel \[dy\] from numerator and denominator considering them to be like terms but that is wrong and \[\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right)=\dfrac{\dfrac{{{d}^{2}}y}{d{{x}^{2}}}}{\dfrac{dy}{dx}}\].

Recently Updated Pages

Basicity of sulphurous acid and sulphuric acid are

Assertion The resistivity of a semiconductor increases class 13 physics CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

What is the stopping potential when the metal with class 12 physics JEE_Main

The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main

Using the following information to help you answer class 12 chemistry CBSE

Trending doubts

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Draw a diagram showing the external features of fish class 11 biology CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

What is BLO What is the full form of BLO class 8 social science CBSE