Christmas Streak 12/88: BASICS!

<Solution 1>

Applying partial differentiation to both sides of both equations,

$\cases{\begin{aligned} & dy=\left(2t-\dfrac{1}{t^2}\right)dt \\\\ & dx=\left(1+\dfrac{4}{t^3}\right)dt \end{aligned}}$

Finding $\dfrac{dx}{dy},$ we get

$\frac{dx}{dy}=\frac{\left(1+\dfrac{4}{t^3}\right)dt}{\left(2t-\dfrac{1}{t^2}\right)dt}=\frac{1+\dfrac{4}{t^3}}{2t-\dfrac{1}{t^2}}$

Therefore, plugging in $t=1,$

$\left(\frac{dx}{dy}\right)_{t=1}=\frac{5}{1}=\boxed{5}.$

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<Solution 2>

Differentiating the two equations by $y$ and $t$ respectively,

$\cases{\begin{aligned} & 1=2t\cdot\dfrac{dt}{dy}-\dfrac{1}{t^2}\cdot\dfrac{dt}{dy} \\\\ & \dfrac{dx}{dt}=1+\dfrac{4}{t^3} \end{aligned}}$ $\cases{\begin{aligned} & \dfrac{1}{2t-\dfrac{1}{t^2}}=\dfrac{dt}{dy} \\\\ & \dfrac{dx}{dt}=1+\dfrac{4}{t^3} \end{aligned}}$

Then, according to the chain rule , we see that

$\begin{aligned} \frac{dx}{dy} &= \frac{dt}{dy}\cdot\frac{dx}{dt} \\\\ & =\dfrac{1}{2t-\dfrac{1}{t^2}}\cdot \left(1+\dfrac{4}{t^3}\right) \\\\ & = \dfrac{1+\dfrac{4}{t^3}}{2t-\dfrac{1}{t^2}} \\\\ \end{aligned}$

Therefore, plugging in $t=1,$

$\left(\frac{dx}{dy}\right)_{t=1}=\frac{5}{1}=\boxed{5}.$