Problem 108 - Calculus Early Transcendentals

Question

Assume a curve is given by the parametric equations \(x=f(t)\) and \(y=g(t),\) where \(f\) and \(g\) are twice differentiable. Use the Chain Rule to show that $$y^{\prime \prime}(x)=\frac{f^{\prime}(t) g^{\prime \prime}(t)-g^{\prime}(t) f^{\prime \prime}(t)}{\left(f^{\prime}(t)\right)^{3}}.$$

Step-by-Step Solution

Answer

To find the second derivative of a curve defined by parametric equations \(x=f(t)\) and \(y=g(t)\), follow these steps: 1. Compute the first derivatives of \(g(t)\) and \(f(t)\), denoted \(g^{\prime}(t)\) and \(f^{\prime}(t)\). 2. Use the Chain Rule to find \(y^{\prime}(x) = \frac{g^{\prime}(t)}{f^{\prime}(t)}\). 3. Compute the second derivatives of \(g(t)\) and \(f(t)\), denoted \(g^{\prime\prime}(t)\) and \(f^{\prime\prime}(t)\). 4. Differentiate \(y^{\prime}(x)\) with respect to \(x\) using the Chain Rule and Quotient Rule, then simplify the resulting expression to obtain \(y^{\prime\prime}(x) = \frac{f^{\prime}(t) g^{\prime\prime}(t) - g^{\prime}(t) f^{\prime\prime}(t)}{\left(f^{\prime}(t)\right)^{3}}\).

Step 1: Find \(g'(t)\) and \(f'(t)\)

Differentiate \(g(t)\) and \(f(t)\) with respect to \(t\) to obtain their first derivatives: $$g^{\prime}(t) = \frac{d}{dt}g(t) \quad \text{and} \quad f^{\prime}(t) = \frac{d}{dt}f(t).$$

Step 2: Find \(y'(x)\) using the Chain Rule

By the Chain Rule, we have: $$y^{\prime}(x) = \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{g^{\prime}(t)}{f^{\prime}(t)}.$$

Step 3: Differentiate \(f'(t)\) and \(g'(t)\) again

Find the second derivatives of \(g(t)\) and \(f(t)\) with respect to \(t\): $$g^{\prime\prime}(t) = \frac{d^2}{dt^2}g(t) \quad \text{and} \quad f^{\prime\prime}(t) = \frac{d^2}{dt^2}f(t).$$

Step 4: Find \(y''(x)\) using the Chain Rule again

Now, differentiate \(y^{\prime}(x) = \frac{g^{\prime}(t)}{f^{\prime}(t)}\) with respect to \(x\): $$y^{\prime\prime}(x) = \frac{d}{dx}\left(\frac{g^{\prime}(t)}{f^{\prime}(t)}\right).$$ Using the chain rule and quotient rule, we obtain: $$y^{\prime\prime}(x) = \frac{d}{dt}\left(\frac{g^{\prime}(t)}{f^{\prime}(t)}\right) \cdot \frac{dt}{dx} = \frac{f^{\prime}(t)\cdot g^{\prime\prime}(t) - g^{\prime}(t) \cdot f^{\prime\prime}(t)}{(f^{\prime}(t))^2} \cdot \frac{1}{f^{\prime}(t)}.$$ Finally, simplify the expression: $$y^{\prime\prime}(x) = \frac{f^{\prime}(t) g^{\prime\prime}(t) - g^{\prime}(t) f^{\prime\prime}(t)}{\left(f^{\prime}(t)\right)^{3}}.$$

Key Concepts

Understanding the Chain Rule

The Chain Rule is a fundamental concept in calculus used to differentiate composite functions. It allows us to find the derivative of a function that is composed of two or more functions. In the context of parametric equations, where functions are expressed in terms of a parameter, the Chain Rule becomes incredibly helpful.

For a given function defined as a composition, like \(y = g(t)\), and another function \(x = f(t)\), we apply the Chain Rule to express the rate of change of \(y\) with respect to \(x\) instead of \(t\). This is accomplished by:

• First finding the derivative of \(y\) with respect to \(t\), represented by \(g'(t)\).
• Then, finding the derivative of \(x\) with respect to \(t\), denoted as \(f'(t)\).
• Finally, expressing \(\frac{dy}{dx}\) as \(\frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}\).

This process allows us to translate how changes in \(t\) affect \(y\) in terms of changes in \(x\), which is essential for working with parametric curves.

Calculating the Second Derivative

Once we have the first derivative, we often need the second derivative to understand the curvature or acceleration of the function described by the parametric equations. The second derivative, \(y''(x)\), tells us how the rate of change of \(y'\) varies with respect to \(x\).

To find \(y''(x)\) from a parametric perspective, we need to differentiate \(y'(x)\) again:

• We start with \(y'(x) = \frac{g'(t)}{f'(t)}\).
• Applying both the Chain Rule and Quotient Rule, we differentiate \(\frac{g'(t)}{f'(t)}\) with respect to \(t\), and multiply by \(\frac{dt}{dx}\).
• The result becomes \(\frac{f'(t)g''(t) - g'(t)f''(t)}{(f'(t))^2} \, \cdot \frac{1}{f'(t)}\).

This simplifies to the second derivative \(y''(x) = \frac{f'(t)g''(t) - g'(t)f''(t)}{(f'(t))^3}\). This gives a full picture of how \(y\)'s rate of change is evolving as \(x\) changes.

Differentiation Techniques in Parametric Equations

Differentiation techniques are vital for handling parametric equations. These techniques involve using derivatives to explore concepts like slope, velocity, and curvature in parametrized forms.

Key techniques include:

• **First Derivative**: Calculate \(y'(x)\) by using \(g'(t)\) and \(f'(t)\), giving us the slope of the tangent to the curve at any point.
• **Second Derivative**: This involves differentiating \(y'(x) = \frac{g'(t)}{f'(t)}\) again to measure the curve's concavity or acceleration. It is found using formulas derived from the Chain Rule and Quotient Rule.
• **Quotient Rule**: When differentiating \(\frac{u}{v}\), we use \(\frac{vu' - uv'}{v^2}\), which is critical in finding second derivatives of parametric equations.

Mastering these techniques is essential for understanding the behavior and geometry of curves defined parametrically, ensuring a comprehensive grasp of how changes in a parameter affect all aspects of the curve.