How do you differentiate the following parametric equation: $x (t) = \sqrt{t - 2}, y (t) = t^{2} - 2 t^{3} e^{t}$ $x(t)=sqrt(t-2), y(t)= t^2-2t^3e^(t)$ ?

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How do you differentiate the following parametric equation: $x (t) = \sqrt{t - 2}, y (t) = t^{2} - 2 t^{3} e^{t}$ $x(t)=sqrt(t-2), y(t)= t^2-2t^3e^(t)$ ?

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Answer 1 · 2018-04-21T01:02:50

First, differentiate each individual function as you have been all year.

using the chain rule,

$x (t) = {(t - 2)}^{\frac{1}{2}}$ $x(t)=(t-2)^(1/2)$

$\Rightarrow \frac{d x}{d t} = \frac{1}{2} {(t - 2)}^{- \frac{1}{2}} \cdot \frac{d}{d t} (t - 2)$ $" "" "=>dx/dt=1/2(t-2)^(-1/2)*d/dt(t-2)$

$\Rightarrow \frac{d x}{d t} = \frac{1}{2} {(t - 2)}^{- \frac{1}{2}} \cdot 1$ $" "" "color(white)(=>dx/dt)=1/2(t-2)^(-1/2)*1$

$\Rightarrow \frac{d x}{d t} = \frac{1}{2 \sqrt{t - 2}}$ $" "" "color(white)(=>dx/dt)=color(red)(1/(2sqrt(t-2))$

and, by the product rule,

$y (t) = t^{2} - 2 t^{3} e^{t}$ $y(t)=t^2-2t^3e^t$

$\Rightarrow \frac{d y}{d t} = 2 t - [(\frac{d}{d t} 2 t^{3}) e^{t} - 2 t^{3} (\frac{d}{d t} e^{t})]$ $" "" "=>dy/dt=2t-[(d/dt2t^3)e^t-2t^3(d/dte^t)]$

$\Rightarrow \frac{d y}{d t} = 2 t - [6 t^{2} e^{t} - 2 t^{3} e^{t}]$ $" "" "color(white)(=>dy/dt)=2t-[6t^2e^t-2t^3e^t]$

$\Rightarrow \frac{d y}{d t} = 2 t (1 - 3 t e^{t} + t^{2} e^{t})$ $" "" "color(white)(=>dy/dt)=color(red)(2t(1-3te^t+t^2e^t)$

if we want to know $\frac{d y}{d x}$ $dy/dx$ , we see that:

$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{2 t (1 - 3 t e^{t} + t^{2} e^{t})}{\frac{1}{2 \sqrt{t - 2}}} = 4 t \sqrt{t - 2} (1 - 3 t e^{t} + t^{2} e^{t})$ $dy/dx=(dy/dt)/(dx/dt)=(2t(1-3te^t+t^2e^t))/(1/(2sqrt(t-2)))=color(blue)(4tsqrt(t-2)(1-3te^t+t^2e^t)$

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How do you differentiate the following parametric equation: $x (t) = \sqrt{t - 2}, y (t) = t^{2} - 2 t^{3} e^{t}$ $x(t)=sqrt(t-2), y(t)= t^2-2t^3e^(t)$ ?

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How do you differentiate the following parametric equation: $x (t) = \sqrt{t - 2}, y (t) = t^{2} - 2 t^{3} e^{t}$ $x(t)=sqrt(t-2), y(t)= t^2-2t^3e^(t)$ ?