1. The position of a particle moving in the $x y$-plane is given by the parametric equations $x(t)=\frac{6 t}{t+1}$ and $y(t)=\frac{-8}{t^{2}+4}$. What is the slope of the line tangent to the path of the particle at the point where $t=2$ ?
(A) $\frac{1}{2}$
(B) $\frac{2}{3}$
(C) $\frac{3}{4}$
(D) $\frac{4}{3}$

Question

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Differentiate $x(t)=\frac{6t}{t+1}$ with respect to $t$: ‖ ‖ $ \frac{d}{dt}\left(\frac{6t}{t+1}\right) = \frac{6(t+1) - 6t}{(t+1)^2} = \frac{6}{(t+1)^2} $ ‖ ‖ step 2 ⋮ Differentiate $y(t)=\frac{-8}{t^2+4}$ with respect to $t$: ‖ ‖ $ \frac{d}{dt}\left(\frac{-8}{t^2+4}\right) = \frac{16t}{(t^2+4)^2} $ ‖ ‖ step 3 ⋮ Evaluate $\frac{dx}{dt}$ and $\frac{dy}{dt}$ at $t=2$: ‖ ‖ $ \frac{dx}{dt}\bigg|_{t=2} = \frac{6}{(2+1)^2} = \frac{6}{9} = \frac{2}{3} $ ‖ ‖ $ \frac{dy}{dt}\bigg|_{t=2} = \frac{16 \cdot 2}{(2^2+4)^2} = \frac{32}{(4+4)^2} = \frac{32}{64} = \frac{1}{2} $ ‖ ‖ step 4 ⋮ Find the slope of the tangent line using $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$: ‖ ‖ $ \frac{dy}{dx}\bigg|_{t=2} = \frac{\frac{1}{2}}{\frac{2}{3}} = \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4} $ ‖ ∻Answer∻ ⚹ $\frac{3}{4}$ ⚹ ∻Key Concept∻ ⚹Slope of the tangent line⚹ ∻Explanation∻ ⚹The slope of the tangent line to the path of the particle at a given point is found by taking the derivative of the parametric equations and evaluating the ratio $\frac{dy}{dx}$ at the specified value of $t$.⚹◊What is the formula for the slope of a tangent line to a parametric curve given by $x(t)$ and $y(t)$ at a specific point $(x(t_0), y(t_0))$?⍭ Generate me a similar question◊

Sia