Question

If a curve is represented parametrically by the equations $x=f\left(t\right)$ and $y=g\left(t\right),$ then $\frac{\left(\frac{d^{2}y}{d{x}^2}\right)}{\left(\frac{d^{2}x}{d{y}^2}\right)}$ is equal to
$($where $f^{\prime }\left(t\right)\ne 0$ and $g^{\prime }\left(t\right)\ne 0)$

• A. $1$
• B. $\frac{g^{\prime }\left(t\right)}{f^{\prime }\left(t\right)}$
• C. $-{\left(\frac{g^{\prime }\left(t\right)}{f^{\prime }\left(t\right)}\right)}^2$
• D. $-{\left(\frac{g^{\prime }\left(t\right)}{f^{\prime }\left(t\right)}\right)}^3$

Answer 1

The correct option is D $-{\left(\frac{g^{\prime }\left(t\right)}{f^{\prime }\left(t\right)}\right)}^3$

We know that $\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$
Differentiating w.r.t. $y,$ we get
$\frac{d^{2}x}{d{y}^2}=\frac{d}{dy}\left(\frac{1}{\frac{dy}{dx}}\right)$
$\Rightarrow \frac{d^{2}x}{d{y}^2}=\frac{d}{dx}\left(\frac{1}{\frac{dy}{dx}}\right)\cdot \frac{dx}{dy}$
$\Rightarrow \frac{d^{2}x}{d{y}^2}=\frac{-1}{{\left(\frac{dy}{dx}\right)}^2}\cdot \left(\frac{d^{2}y}{d{x}^2}\right)\cdot \left(\frac{1}{\frac{dy}{dx}}\right)$
$\Rightarrow \frac{d^{2}x}{d{y}^2}=\frac{-\frac{d^{2}y}{d{x}^2}}{{\left(\frac{dy}{dx}\right)}^3}$
$\Rightarrow \frac{\frac{d^{2}y}{d{x}^2}}{\frac{d^{2}x}{d{y}^2}}=-{\left(\frac{dy}{dx}\right)}^3=-{\left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right)}^3$
$\Rightarrow \frac{\frac{d^{2}y}{d{x}^2}}{\frac{d^{2}x}{d{y}^2}}=-{\left(\frac{g^{\prime }\left(t\right)}{f^{\prime }\left(t\right)}\right)}^3$