Problem 13 - Calculus Single Variable

Question

Variables \(x\) and \(y,\) which depend on \(t,\) are related by a given equation. A point \(P_{0}\) on the graph of that equation is also given, as is one of the following two values: $$v_{0}=\left.\frac{d x}{d t}\right|_{P_{0}} \quad \text { or } \quad s_{0}=\left.\frac{d y}{d t}\right|_{P_{0}}$$ Find the other. $$ x^{2}+x y=2, \quad P_{0}=(-2,1), \quad s_{0}=3 $$

Step-by-Step Solution

Answer

The value of \( \frac{dx}{dt} \bigg|_{P_0} \) is \(-2\).

Step 1: Differentiate the Equation Implicitly

We are given the equation \( x^2 + xy = 2 \). Since both \( x \) and \( y \) are functions of \( t \), differentiate each term with respect to \( t \) using implicit differentiation. Recall that when differentiating \( y \), we will use \( \frac{dy}{dt} \), and for \( x \), we will use \( \frac{dx}{dt} \).\[ \frac{d}{dt}(x^2) + \frac{d}{dt}(xy) = \frac{d}{dt}(2) \]The derivative of \( x^2 \) is \( 2x \frac{dx}{dt} \). The derivative of \( xy \) using the product rule is \( x \frac{dy}{dt} + y \frac{dx}{dt} \). The derivative of a constant is zero, so the equation becomes:\[ 2x \frac{dx}{dt} + x \frac{dy}{dt} + y \frac{dx}{dt} = 0 \]

Step 2: Substitute Known Values into the Derivative Equation

Now substitute the known values into the derivative equation \(2x \frac{dx}{dt} + x \frac{dy}{dt} + y \frac{dx}{dt} = 0\). These known values are \(P_0 = (-2, 1)\) and \(s_0 = \frac{dy}{dt}\bigg|_{P_0} = 3\). Substitute \(x = -2\), \(y = 1\), and \(\frac{dy}{dt} = 3\) into the equation:\[ 2(-2) \frac{dx}{dt} + (-2)(3) + 1 \frac{dx}{dt} = 0 \]

Step 3: Simplify and Solve for \(\frac{dx}{dt}\)

Simplify the equation \(-4 \frac{dx}{dt} - 6 + \frac{dx}{dt} = 0\). Combine like terms:\[ -3 \frac{dx}{dt} - 6 = 0 \]Now solve for \(\frac{dx}{dt}\) by adding 6 to both sides and then dividing by -3:\[ -3 \frac{dx}{dt} = 6 \]Divide both sides by \(-3\):\[ \frac{dx}{dt} = -2 \]

Key Concepts

Implicit Differentiation Steps

Implicit differentiation is a powerful technique used when dealing with equations involving multiple variables that are related implicitly. Here, you do not solve for one variable in terms of the other before differentiating. Instead, you differentiate each term with respect to a given variable and make use of the chain rule to incorporate derivatives of other variables. This method is ideal for functions that are not easily separable or explicit, meaning you cannot write them as one variable in terms of the other directly.

To apply implicit differentiation, follow these straightforward steps:

• Identify the equation in which the variables are implicitly related.
• Differentiating every term in the equation with respect to the variable in concern (for instance, time \( t \) in our example).
• Use the chain rule for derivatives of dependent variables, introducing terms like \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \).
• Simplify the resulting expression by combining like terms.

Implicit differentiation greatly simplifies finding derivatives when variables intertwine, as seen with the equation \(x^2 + xy = 2\), where differentiating explicitly is cumbersome.

Product Rule in Differentiation

The product rule is essential when differentiating expressions where two functions are multiplied together. The rule states that the derivative of a product of two functions, say \( u(t) \) and \( v(t) \), is given by:

• \( \frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t) \)

In our exercise, when differentiating the term \( xy \), both \( x \) and \( y \) are functions of \( t \). Thus, we apply the product rule:

• The derivative of \( x \) is \( \frac{dx}{dt} \), and the derivative of \( y \) is \( \frac{dy}{dt} \).
• The product differentiation becomes \( x \cdot \frac{dy}{dt} + y \cdot \frac{dx}{dt} \).

Understanding this rule ensures you'll confidently tackle expressions involving products, leading to accurate simplification and solution finding.

Chain Rule in Differentiation

The chain rule is a fundamental differentiation technique used primarily when a function is composed of other functions. It allows you to compute the derivative of a composite function by differentiating the outer function and multiplying it by the derivative of the inner function. This technique is especially valuable in implicit differentiation as it relates to variables changing with respect to time or another parameter.

In implicit differentiation, we frequently encounter terms where both variables depend on another variable, like time \( t \). For example, with \( x^2 \), when differentiating with respect to \( t \), we apply the chain rule:

• The outer function is \( x^2 \), whose derivative is \( 2x \).
• Since \( x \) depends on \( t \), its derivative is \( \frac{dx}{dt} \).
• Thus, the derivative of \( x^2 \) becomes \( 2x \frac{dx}{dt} \).

Through the chain rule, functions within functions become manageable, simplifying the differentiation of complex expressions and ensuring precision in your calculations.