Understanding Parametric Differentiation

This question asks us to find the derivative \(\frac{dx}{dy}\) given two parametric equations for \(x\) and \(y\) in terms of a parameter \(t\). The equations are:

• \(x = a\left(t + \frac{1}{t}\right)\)
• \(y = a\left(t - \frac{1}{t}\right)\)

To find \(\frac{dx}{dy}\), we can use two primary methods: directly differentiating \(x\) and \(y\) with respect to \(t\) and then applying the chain rule, or by first finding a relationship between \(x\) and \(y\) and then implicitly differentiating.

Method 1: Parametric Differentiation using Chain Rule

First, let's find the derivatives of \(x\) and \(y\) with respect to \(t\):

For \(x = a\left(t + \frac{1}{t}\right)\):

• Recall that \(\frac{d}{dt}\left(\frac{1}{t}\right) = \frac{d}{dt}\left(t^{-1}\right) = -1 \cdot t^{-2} = -\frac{1}{t^2}\).
• So, \(\frac{dx}{dt} = a\left(\frac{d}{dt}(t) + \frac{d}{dt}\left(\frac{1}{t}\right)\right)\)
• \(\frac{dx}{dt} = a\left(1 - \frac{1}{t^2}\right)\)
• \(\frac{dx}{dt} = a\left(\frac{t^2 - 1}{t^2}\right)\)

Next, for \(y = a\left(t - \frac{1}{t}\right)\):

• \(\frac{dy}{dt} = a\left(\frac{d}{dt}(t) - \frac{d}{dt}\left(\frac{1}{t}\right)\right)\)
• \(\frac{dy}{dt} = a\left(1 - \left(-\frac{1}{t^2}\right)\right)\)
• \(\frac{dy}{dt} = a\left(1 + \frac{1}{t^2}\right)\)
• \(\frac{dy}{dt} = a\left(\frac{t^2 + 1}{t^2}\right)\)

Now, we use the chain rule formula: \(\frac{dx}{dy} = \frac{dx/dt}{dy/dt}\)

• \(\frac{dx}{dy} = \frac{a\left(\frac{t^2 - 1}{t^2}\right)}{a\left(\frac{t^2 + 1}{t^2}\right)}\)
• \(\frac{dx}{dy} = \frac{t^2 - 1}{t^2 + 1}\)

While this is a correct derivative in terms of \(t\), the options are in terms of \(x\) and \(y\). We would need to express \(\frac{t^2 - 1}{t^2 + 1}\) in terms of \(x\) and \(y\), which can be complex. Let's explore a more direct method.

Method 2: Relating x and y directly using Algebraic Manipulation

We have the equations:

• (1) \(x = a\left(t + \frac{1}{t}\right)\)
• (2) \(y = a\left(t - \frac{1}{t}\right)\)

Let's square both equations:

• From (1): \(x^2 = \left[a\left(t + \frac{1}{t}\right)\right]^2\)
• \(x^2 = a^2\left(t^2 + 2 \cdot t \cdot \frac{1}{t} + \frac{1}{t^2}\right)\)
• \(x^2 = a^2\left(t^2 + 2 + \frac{1}{t^2}\right)\)

• From (2): \(y^2 = \left[a\left(t - \frac{1}{t}\right)\right]^2\)
• \(y^2 = a^2\left(t^2 - 2 \cdot t \cdot \frac{1}{t} + \frac{1}{t^2}\right)\)
• \(y^2 = a^2\left(t^2 - 2 + \frac{1}{t^2}\right)\)

Now, let's subtract the expression for \(y^2\) from the expression for \(x^2\):

• \(x^2 - y^2 = a^2\left(t^2 + 2 + \frac{1}{t^2}\right) - a^2\left(t^2 - 2 + \frac{1}{t^2}\right)\)
• \(x^2 - y^2 = a^2\left[\left(t^2 + 2 + \frac{1}{t^2}\right) - \left(t^2 - 2 + \frac{1}{t^2}\right)\right]\)
• \(x^2 - y^2 = a^2\left[t^2 + 2 + \frac{1}{t^2} - t^2 + 2 - \frac{1}{t^2}\right]\)
• \(x^2 - y^2 = a^2(4)\)
• \(x^2 - y^2 = 4a^2\)

This equation, \(x^2 - y^2 = 4a^2\), relates \(x\) and \(y\) directly. Since \(4a^2\) is a constant, we can now differentiate this implicit equation with respect to \(y\).

Implicit Differentiation to find \(\frac{dx}{dy}\)

Differentiate \(x^2 - y^2 = 4a^2\) with respect to \(y\):

• \(\frac{d}{dy}(x^2) - \frac{d}{dy}(y^2) = \frac{d}{dy}(4a^2)\)
• Using the chain rule for \(\frac{d}{dy}(x^2)\), we get \(2x \frac{dx}{dy}\).
• For \(\frac{d}{dy}(y^2)\), we get \(2y\).
• For \(\frac{d}{dy}(4a^2)\), since \(4a^2\) is a constant, its derivative is \(0\).

So, the equation becomes:

• \(2x \frac{dx}{dy} - 2y = 0\)
• Add \(2y\) to both sides: \(2x \frac{dx}{dy} = 2y\)
• Divide both sides by \(2x\): \(\frac{dx}{dy} = \frac{2y}{2x}\)
• \(\frac{dx}{dy} = \frac{y}{x}\)

Final Conclusion

By finding the direct relationship between \(x\) and \(y\) as \(x^2 - y^2 = 4a^2\) and then using implicit differentiation, we found that \(\frac{dx}{dy} = \frac{y}{x}\). This matches one of the provided options.