Question

If $x=a{t}^3$ and $y=b{t}^2$, where $t$ is a parameter, then prove that $\frac{d^{3}y}{d{x}^3}=\frac{8.b}{27a^3.t^7}$

Answer 1

$x=a{t}^{3}y=b{t}^2$

$\frac{dx}{dt}=3a{t}^2\frac{dy}{dt}=2bt$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2bt}{3a{t}^2}=\frac{2b}{3a{t}^2}$

$\frac{d^{2}y}{d{x}^2}=\frac{2b}{3a}\times \frac{-1}{t^2}\frac{dt}{dx}$

$=\frac{-2ab}{3a}\times \frac{1}{t^2}\times \frac{1}{3a{t}^2}$

$=\frac{-2b}{9a^2{t}^4}$

$\frac{d^{3}y}{d{x}^3}=\frac{-2b}{9a^2}\times \frac{-4}{t^5}\times \frac{dt}{dx}$

$=\frac{8b}{9a^2}\times \frac{1}{3a{t}^2}$

$=\frac{8b}{27a^3{t}^7}$