Chapter 10 - Section 10.2 - Calculus with Parametric Curves - 10.2 Exercises - Page 679: 2

$\frac{dx}{dt}=1-\frac{1}{t}$ $\frac{dy}{dt}=2t+2t^{-3}$ $\frac{dy}{dx}=\frac{2t^4+2}{t^3-t^2}$

Find $dx/dt$: $\frac{dx}{dt}=1-\frac{1}{t}$ Find $dy/dt$: $\frac{dy}{dt}=2t+2t^{-3}$ Find $dy/dx$: $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2t+2t^{-3}}{1-\frac{1}{t}}=\frac{2t+2t^{-3}}{1-\frac{1}{t}}\times \frac{t^3}{t^3}=\frac{2t^4+2}{t^3-t^2}$