Question Statement

Given the parametric equations: $x=t+\frac{1}{t},y=t+1$

Find $\frac{dy}{dx}$ in terms of $t$.

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Background and Explanation

This problem involves parametric differentiation, where both $x$ and $y$ are expressed in terms of a parameter $t$. To find $\frac{dy}{dx}$, we calculate the derivatives of $x$ and $y$ with respect to $t$ separately, then apply the chain rule: $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$.

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Solution

First, rewrite $x$ using negative exponents to prepare for differentiation: $x=t+\frac{1}{t}=t+t^{-1}$

Differentiate $x$ with respect to $t$:

$\frac{dx}{dt}=\frac{d}{dt}(t)+\frac{d}{dt}\left(t^{-1}\right)$

Applying the power rule: $\frac{dx}{dt}=1+(-1)t^{-1-1}=1-t^{-2}$

Converting back to fractional form: $\frac{dx}{dt}=1-\frac{1}{t^2}$

Combining over a common denominator: $\frac{dx}{dt}=\frac{t^2-1}{t^2}$

Differentiate $y$ with respect to $t$:

Given $y=t+1$:

$\frac{dy}{dt}=\frac{d}{dt}(t)+\frac{d}{dt}(1)=1+0=1$

Apply the Chain Rule:

To find $\frac{dy}{dx}$, we use: $\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}$

Since $\frac{dx}{dt}=\frac{t^2-1}{t^2}$, the reciprocal is $\frac{dt}{dx}=\frac{t^2}{t^2-1}$.

Substituting the values: $\frac{dy}{dx}=1\times \frac{t^2}{t^2-1}$

Therefore: $\frac{dy}{dx}=\frac{t^2}{t^2-1}$

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Key Formulas or Methods Used

• Power Rule: $\frac{d}{dt}(t^n)=n{t}^{n-1}$ (used to differentiate $t^{-1}$)
• Chain Rule for Parametric Equations: $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{dy}{dt}\times \frac{dt}{dx}$
• Negative Exponent Rule: $t^{-n}=\frac{1}{t^n}$ and $\frac{1}{t^n}=t^{-n}$

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Summary of Steps

1. Rewrite $x=t+\frac{1}{t}$ as $x=t+t^{-1}$ to apply the power rule
2. Differentiate $x$ with respect to $t$: $\frac{dx}{dt}=1-t^{-2}=\frac{t^2-1}{t^2}$
3. Differentiate $y=t+1$ with respect to $t$: $\frac{dy}{dt}=1$
4. Apply the chain rule: $\frac{dy}{dx}=\frac{dy}{dt}÷\frac{dx}{dt}=1\times \frac{t^2}{t^2-1}=\frac{t^2}{t^2-1}$