Problem 43 - Calculus - AP Edition

Question

Find \(d y / d x\) if $$2 y^{3} t+t^{3} y=1 \quad \text { and } \quad \frac{d t}{d x}=\frac{1}{\cos t}$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{-2y^3 \frac{1}{\cos t} - 3t^2 \frac{y}{\cos t}}{6y^2t + t^3}\)

Step 1: Implicit Differentiate Both Sides

Given the equation \(2 y^3 t + t^3 y = 1\), we'll differentiate both sides with respect to \(x\). This requires using the product rule and chain rule, as the equation involves both \(y\) and \(t\), which are functions of \(x\). The left side becomes:\[\frac{d}{dx}(2y^3t) + \frac{d}{dx}(t^3y)\]

Step 2: Differentiate the First Term

For \(2y^3t\), apply the product rule: \[\frac{d}{dx}(2y^3t) = 2 \left( \frac{d}{dx}(y^3) \cdot t + y^3 \cdot \frac{d}{dx}(t) \right)\]Since \(y^3\) differentiates to \(3y^2\cdot \frac{dy}{dx} \) and \( \frac{dt}{dx} = \frac{1}{\cos t}\), we get:\[6y^2t \frac{dy}{dx} + 2y^3 \frac{1}{\cos t}\]

Step 3: Differentiate the Second Term

For \(t^3y\), apply the product rule:\[\frac{d}{dx}(t^3y) = \frac{d}{dx}(t^3) \cdot y + t^3 \cdot \frac{dy}{dx}\]Since \(d(t^3)/dx = 3t^2\cdot \frac{1}{\cos t}\), it becomes:\[3t^2 \frac{y}{\cos t} + t^3 \frac{dy}{dx}\]

Step 4: Combine and Simplify

Putting it together, we have:\[6y^2t \frac{dy}{dx} + 2y^3 \frac{1}{\cos t} + 3t^2 \frac{y}{\cos t} + t^3 \frac{dy}{dx} = 0\]Combine terms involving \( \frac{dy}{dx} \):\[(6y^2t + t^3)\frac{dy}{dx} = -2y^3 \frac{1}{\cos t} - 3t^2 \frac{y}{\cos t}\]

Step 5: Solve for \( \frac{dy}{dx} \)

To isolate \( \frac{dy}{dx} \), divide both sides of the equation by \((6y^2t + t^3)\):\[\frac{dy}{dx} = \frac{-2y^3 \frac{1}{\cos t} - 3t^2 \frac{y}{\cos t}}{6y^2t + t^3}\]

Key Concepts

Product Rule

The product rule is a fundamental concept in calculus used to differentiate expressions where two functions are multiplied together. When you have a product of two functions, let's say \(u(x)\) and \(v(x)\), the derivative is given by:

• \(\frac{d}{dx}(u(x)v(x)) = u'(x)v(x) + u(x)v'(x)\)

This rule is particularly useful in implicit differentiation when dealing with equations that involve products of variables dependent on \(x\).
This exercise requires applying the product rule twice: one for the term \(2y^3t\) and another for \(t^3y\). For \(2y^3t\), the functions are \(2y^3\) and \(t\), and for \(t^3y\), they are \(t^3\) and \(y\).
Understanding when and how to apply the product rule helps simplify and solve equations more effectively, which is crucial for correctly differentiating complex expressions in calculus.

Chain Rule

The chain rule is another key concept in calculus that facilitates finding the derivative of composite functions. When a function is nested inside another, the chain rule allows us to differentiate it systematically.
The chain rule states:

• If \(y = f(g(x))\), then \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\)

In this exercise, the chain rule is applied while differentiating \(y^3\) with respect to \(x\) because \(y\) itself is a function of \(x\).
Thus, the derivative \(\frac{d}{dx}y^3\) includes the term \(3y^2 \cdot \frac{dy}{dx}\).
Employing the chain rule in this context connects all the dependent functions correctly, ensuring every part of the function is considered during differentiation.

Differentiation Steps

Differentiation is performed using specific steps to ensure accuracy and completeness in the process. Here is a structured breakdown:

• **Identify the equation:** Start by writing down the given equation \(2y^3 t + t^3 y = 1\).
• **Apply differentiation rules:** Decide which rules (product rule, chain rule) apply to each term of the equation.
• **Differentiate each term separately:** This involves applying the product rule to both \(2y^3t\) and \(t^3y\), and then using the chain rule where appropriate.
• **Combine and simplify terms:** After differentiating, combine the derivatives properly to reach a simpler form.
• **Isolate the desired derivative:** Solve for \(\frac{dy}{dx}\) by isolating it on one side of the equation.

Incorporating these steps will help in systematically tackling similar problems, arriving at the solution efficiently without missing crucial parts of the differentiation process.