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If $x=\phi(t), y=\psi(t)$, then $\frac{d^{2} y}{d x^{2}}$ is equal to
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Verified Answer
The correct answer is:
$\frac{\phi^{\prime} \psi^{\prime \prime}-\psi^{\prime} \phi^{\prime \prime}}{\left(\phi^{\prime}\right)^{3}}$
We have, $x=\phi(t), y=\psi(t)$
$$
\begin{array}{l}
\therefore \quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{\psi^{\prime}}{\phi^{\prime}} \\
\Rightarrow \quad \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{\psi^{\prime}}{\phi^{\prime}}\right)=\frac{d}{d t}\left(\frac{\psi^{\prime}}{\phi^{\prime}}\right) \frac{d t}{d x} \\
=\frac{\phi^{\prime} \psi^{\prime \prime}-\psi^{\prime} \phi^{\prime \prime}}{\left(\phi^{\prime}\right)^{2}} \cdot \frac{1}{\phi^{\prime}}=\frac{\phi^{\prime} \psi^{\prime \prime}-\psi^{\prime} \phi^{\prime \prime}}{\left(\phi^{\prime}\right)^{3}}
\end{array}
$$
$$
\begin{array}{l}
\therefore \quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{\psi^{\prime}}{\phi^{\prime}} \\
\Rightarrow \quad \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{\psi^{\prime}}{\phi^{\prime}}\right)=\frac{d}{d t}\left(\frac{\psi^{\prime}}{\phi^{\prime}}\right) \frac{d t}{d x} \\
=\frac{\phi^{\prime} \psi^{\prime \prime}-\psi^{\prime} \phi^{\prime \prime}}{\left(\phi^{\prime}\right)^{2}} \cdot \frac{1}{\phi^{\prime}}=\frac{\phi^{\prime} \psi^{\prime \prime}-\psi^{\prime} \phi^{\prime \prime}}{\left(\phi^{\prime}\right)^{3}}
\end{array}
$$
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