If x=phi(t) and y=psi(t),then frac{d^{2} y}{d | vedclass

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(B) Given $x=\phi(t)$ and $y=\psi(t)$.First,we find the first derivative $\frac{dy}{dx}$ using the chain rule:$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \

Vedclass · Accepted Answer

$\frac{\phi^{\prime} \psi^{\prime \prime}-\psi^{\prime} \phi^{\prime \prime}}{\left(\phi^{\prime}\right)^{3}}$

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(B) Given $x=\phi(t)$ and $y=\psi(t)$.First,we find the first derivative $\frac{dy}{dx}$ using the chain rule:$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\psi^{\prime}(t)}{\phi^{\prime}(t)}$.Now,we differentiate $\frac{dy}{dx}$ with respect to $x$:$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\left(\frac{\psi^{\prime}}{\phi^{\prime}}\right) = \frac{d}{dt}\left(\frac{\psi^{\prime}}{\phi^{\prime}}\right) \cdot \frac{dt}{dx}$.Using the quotient rule for $\frac{d}{dt}\left(\frac{\psi^{\prime}}{\phi^{\prime}}\right)$:$= \frac{\phi^{\prime}\psi^{\prime\prime} - \psi^{\prime}\phi^{\prime\prime}}{(\phi^{\prime})^{2}} \cdot \frac{1}{\phi^{\prime}}$.$= \frac{\phi^{\prime}\psi^{\prime\prime} - \psi^{\prime}\phi^{\prime\prime}}{(\phi^{\prime})^{3}}$.

If $x=\phi(t)$ and $y=\psi(t)$,then $\frac{d^{2} y}{d x^{2}}$ is equal to

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