Understanding Parametric Differentiation

The problem asks us to find the second derivative, denoted as \(\rm \dfrac{d^2y}{dx^2}\), for a curve defined parametrically by the equations:

• \(x = a \cos t\)
• \(y = b \sin t\)

We need to use the rules of parametric differentiation to solve this. The key is to first find the first derivative \(\rm \dfrac{dy}{dx}\) and then differentiate that result with respect to \(x\) again.

Calculating the First Derivative \(\rm \dfrac{dy}{dx}\)

To find \(\rm \dfrac{dy}{dx}\) for parametric equations, we use the formula:

\(\rm \dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}\)

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

• Derivative of \(x\) w.r.t. \(t\): \(\rm \dfrac{dx}{dt} = \dfrac{d}{dt}(a \cos t) = -a \sin t\)
• Derivative of \(y\) w.r.t. \(t\): \(\rm \dfrac{dy}{dt} = \dfrac{d}{dt}(b \sin t) = b \cos t\)

Now, substitute these into the formula for \(\rm \dfrac{dy}{dx}\):

\(\rm \dfrac{dy}{dx} = \dfrac{b \cos t}{-a \sin t} = -\dfrac{b}{a} \cot t\)

Calculating the Second Derivative \(\rm \dfrac{d^2y}{dx^2}\)

To find the second derivative, \(\rm \dfrac{d^2y}{dx^2}\), we differentiate \(\rm \dfrac{dy}{dx}\) with respect to \(x\). We use the chain rule:

\(\rm \dfrac{d^2y}{dx^2} = \dfrac{d}{dx} \left( \dfrac{dy}{dx} \right) = \dfrac{d}{dt} \left( \dfrac{dy}{dx} \right) \cdot \dfrac{dt}{dx}\)

We already found \(\rm \dfrac{dy}{dx} = -\dfrac{b}{a} \cot t\). Let's differentiate this with respect to \(t\):

\(\rm \dfrac{d}{dt} \left( \dfrac{dy}{dx} \right) = \dfrac{d}{dt} \left( -\dfrac{b}{a} \cot t \right) = -\dfrac{b}{a} (-\csc^2 t) = \dfrac{b}{a} \csc^2 t\)

We also need \(\rm \dfrac{dt}{dx}\). Since \(\rm \dfrac{dx}{dt} = -a \sin t\), we have:

\(\rm \dfrac{dt}{dx} = \dfrac{1}{dx/dt} = \dfrac{1}{-a \sin t}\)

Now, substitute these back into the formula for \(\rm \dfrac{d^2y}{dx^2}\):

\(\rm \dfrac{d^2y}{dx^2} = \left( \dfrac{b}{a} \csc^2 t \right) \cdot \left( \dfrac{1}{-a \sin t} \right)\)

\(\rm \dfrac{d^2y}{dx^2} = -\dfrac{b}{a^2} \dfrac{\csc^2 t}{\sin t}\)

Using the identity \(\rm \csc t = \dfrac{1}{\sin t}\), we get:

\(\rm \dfrac{d^2y}{dx^2} = -\dfrac{b}{a^2} \dfrac{(1/\sin^2 t)}{\sin t} = -\dfrac{b}{a^2} \dfrac{1}{\sin^3 t}\)

Expressing \(\rm \dfrac{d^2y}{dx^2}\) in terms of \(y\)

The options are given in terms of \(x\) and \(y\). We have the original parametric equation \(y = b \sin t\). From this, we can express \(\sin t\) as:

\(\rm \sin t = \dfrac{y}{b}\)

Substitute this expression for \(\sin t\) into our result for \(\rm \dfrac{d^2y}{dx^2}\):

\(\rm \dfrac{d^2y}{dx^2} = -\dfrac{b}{a^2} \dfrac{1}{(\frac{y}{b})^3}\)

\(\rm \dfrac{d^2y}{dx^2} = -\dfrac{b}{a^2} \dfrac{1}{y^3 / b^3}\)

\(\rm \dfrac{d^2y}{dx^2} = -\dfrac{b}{a^2} \cdot \dfrac{b^3}{y^3}\)

\(\rm \dfrac{d^2y}{dx^2} = -\dfrac{b^4}{a^2 y^3}\)

Conclusion

Comparing our result with the given options, we find that the correct expression for \(\rm \dfrac{d^2y}{dx^2}\) is \(-\rm \dfrac{b^4}{a^2y^3}\).