Problem 16 - 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. $$ x^{2}+x \exp (y)+y=6, \quad P_{0}=(2,0), \quad s_{0}=-1 $$

Step-by-Step Solution

Answer

\( \frac{dx}{dt} = \frac{3}{5} \).

Step 1: Differentiate the Equation Implicitly

We need to find \( \frac{dx}{dt} \), given that \( s_0 = -1 \), which means \( \frac{dy}{dt} = -1 \). Begin by differentiating the entire equation \( x^2 + x e^y + y = 6 \) with respect to \( t \). Remember that both \( x \) and \( y \) are functions of \( t \), so apply the chain rule where necessary.

Step 2: Apply the Chain Rule

Differentiate term by term:- For \( x^2 \), the derivative is \( 2x \frac{dx}{dt} \).- For \( x e^y \), apply the product rule: \( \frac{d}{dt}(x e^y) = e^y \frac{dx}{dt} + x e^y \frac{dy}{dt} \).- For \( y \), the derivative is \( \frac{dy}{dt} \).Set the derivative of the constant 6 to zero. The differentiated equation is:\[ 2x \frac{dx}{dt} + e^y \frac{dx}{dt} + x e^y \frac{dy}{dt} + \frac{dy}{dt} = 0 \]

Step 3: Substitute Known Values at Point

We know \( P_0 = (2,0) \) and \( \frac{dy}{dt} = -1 \), so substitute \( x = 2 \), \( y = 0 \), and \( \frac{dy}{dt} = -1 \) into the differentiated equation:\[ 2(2) \frac{dx}{dt} + e^0 \frac{dx}{dt} + 2 e^0 (-1) - 1 = 0 \].

Step 4: Simplify and Solve for \( \frac{dx}{dt} \)

Simplify the equation:\[ 4 \frac{dx}{dt} + \frac{dx}{dt} - 2 - 1 = 0 \].Combine like terms:\[ 5 \frac{dx}{dt} = 3 \].Solve for \( \frac{dx}{dt} \):\[ \frac{dx}{dt} = \frac{3}{5} \].

Key Concepts

Chain Rule

The chain rule is a fundamental technique used in calculus to differentiate composite functions. It allows us to break down more complicated derivatives into simpler parts. This is especially handy when dealing with functions where one variable is a function of another. For instance, if we have a function like \( y = f(g(t)) \), the chain rule helps us differentiate it as follows:

• First, differentiate \( g(t) \) with respect to \( t \), giving us \( g'(t) \).
• Second, differentiate \( f(u) \) with respect to \( u \), where \( u = g(t) \), giving us \( f'(u) \).

Therefore, using the chain rule, the derivative \( \frac{dy}{dt} = f'(g(t)) \cdot g'(t) \).
This rule is also utilized when we perform implicit differentiation, especially effectively together with the product and sum rules as shown in exercises involving multiple terms like \( x^2 + xe^y + y = 6 \). By understanding and applying the chain rule correctly, we can tackle complex differentiation problems effectively.

Implicit Differentiation

Implicit differentiation is a powerful tool used when you have an equation with multiple variables that define one variable in terms of another, but not explicitly. It allows us to differentiate equations where the variable dependencies are not isolated, like in the equation \( x^2 + x \exp(y) + y = 6 \).
When using implicit differentiation, every term of the equation needs to be differentiated with respect to a common variable, typically \( t \), and the chain rule often comes in handy for the intermediate steps. Here each variable is treated as a function of \( t \), and its derivative is included in the differentiation process.

• For instance, the derivative of \( y \) with respect to \( t \) becomes \( \frac{dy}{dt} \).
• The equation's structure leads us to apply implicit differentiation, resulting in derivative equations that we can solve for unknown rates — such as \( \frac{dx}{dt} \) given \( \frac{dy}{dt} = -1 \) in this particular problem.

Implicit differentiation is especially useful in solving problems involving complex relationships between variables.

Related Rates

Related rates problems involve finding the rate at which one quantity changes in relation to another. These problems hinge on the understanding that rates of change are interconnected via a common variable, usually time \( t \).
In such problems, an equation relates the different variables involved, such as \( x^2 + x \exp(y) + y = 6 \), where \( x \) and \( y \) are both dependent on \( t \). By differentiating the entire equation with respect to \( t \), we can solve for unknown rates like \( \frac{dx}{dt} \) when \( \frac{dy}{dt} \) is given.

• This involves substituting known values at given points (e.g., \( P_0 = (2,0) \)) into the derived equation.
• Simplifying and solving the resulting expression gives us the rate of change of one of the variables.

Understanding related rates is vital in applications across physics and engineering, where dynamic systems depend on the interplay of different changing quantities.

Derivatives

The concept of derivatives is central to calculus, particularly in understanding how quantities change. In simplest terms, a derivative represents the slope or rate of change of a function as its variables change. It tells us how a function responds to changes in its input.
In the context of the equation \( x^2 + x \exp(y) + y = 6 \), derivatives allow us to track how changes in \( x \) and \( y \) influence each other. There are specific rules to finding derivatives:

• The power rule enables us to differentiate powers of a variable, like \( x^2 \) becoming \( 2x \frac{dx}{dt} \).
• For composite functions involving exponential terms such as \( x \exp(y) \), the product and chain rules come into play.

Derivatives are used widely in calculating rates — a crucial aspect of fields like physics, economics, and biology. Through derivatives, we gain detailed insights into the behavior and dynamics of various systems.