Problem 4
Question
Both \(x\) and \(y\) denote functions of \(t\) that are related by the given equation. Use this equation and the given derivative information to find the specified derivative. $$ \begin{array}{l}{\text { Equation: } x^{2}+y^{2}=2 x+4 y} \\ {\text { (a) Given that } d x / d t=-5 \text { , find } d y / d t \text { when }} \\ {(x, y)=(3,1) .} \\ {\text { (b) Given that } d y / d t=6, \text { find } d x / d t \text { when }} \\ {\quad(x, y)=(1+\sqrt{2}, 2+\sqrt{3})}\end{array} $$
Step-by-Step Solution
Verified Answer
(a) \( \frac{dy}{dt} = -10 \). (b) \( \frac{dx}{dt} = -6\sqrt{6} \).
1Step 1: Differentiate each term with respect to t
We start with the given equation: \( x^2 + y^2 = 2x + 4y \). To find the derivative of this equation with respect to \( t \), apply the chain rule. The derivative of each term is: \( 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2 \frac{dx}{dt} + 4 \frac{dy}{dt} \).
2Step 2: Rearrange the derivative equation
Rewriting the differentiated equation gives us: \( 2x \frac{dx}{dt} + 2y \frac{dy}{dt} - 2 \frac{dx}{dt} - 4 \frac{dy}{dt} = 0 \) or \( (2x - 2) \frac{dx}{dt} + (2y - 4) \frac{dy}{dt} = 0 \).
3Step 3: Solve for \( \frac{dy}{dt} \) when \( \frac{dx}{dt} = -5 \)
Substitute \( x = 3 \), \( y = 1 \), and \( \frac{dx}{dt} = -5 \) into the rearranged equation: \( (2 \times 3 - 2)(-5) + (2 \times 1 - 4) \frac{dy}{dt} = 0 \). Simplify to find \( -20 + (-2) \frac{dy}{dt} = 0 \). Solve for \( \frac{dy}{dt} \): \( \frac{dy}{dt} = -10 \).
4Step 4: Solve for \( \frac{dx}{dt} \) with given \( \frac{dy}{dt} = 6 \)
Substitute \( x = 1+\sqrt{2} \), \( y = 2+\sqrt{3} \), and \( \frac{dy}{dt} = 6 \) into the equation: \( (2(1+\sqrt{2}) - 2) \frac{dx}{dt} + (2(2+\sqrt{3}) - 4) (6) = 0 \). Simplify to get \( 2\sqrt{2} \frac{dx}{dt} + 4\sqrt{3} \times 6 = 0 \). Solve for \( \frac{dx}{dt} \): \( \frac{dx}{dt} = -12\sqrt{3}/\sqrt{2} \). Simplify to get \( \frac{dx}{dt} = -6\sqrt{6} \).
Key Concepts
Understanding the Chain RuleThe Role of Differential EquationsGrasping Implicit Differentiation
Understanding the Chain Rule
The chain rule is a fundamental concept in calculus used to differentiate the composite functions, especially when a variable is itself a function of another variable. This rule is particularly useful for addressing problems involving parametric equations, where both dependent and independent variables depend on a common parameter, like time.
Imagine you have two functions, such as \(y(x(t))\) and \(x(t)\), where both are functions of \(t\). In this context, the chain rule provides a way to express the derivative of \(y\) with respect to \(t\), based on each individual derivative: \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\).
In our exercise, we differentiate each term of the equation \(x^2 + y^2 = 2x + 4y\) with respect to \(t\) using the chain rule. We then apply it to both \(x\) and \(y\), capturing how these variables change over time. This imposes the rule: \(2x \frac{dx}{dt} + 2y \frac{dy}{dt} - 2 \frac{dx}{dt} - 4 \frac{dy}{dt} = 0\). By rearranging and solving this equation for either \(\frac{dx}{dt}\) or \(\frac{dy}{dt}\), we can find the rate of change of one variable given the rate of change of the other.
Imagine you have two functions, such as \(y(x(t))\) and \(x(t)\), where both are functions of \(t\). In this context, the chain rule provides a way to express the derivative of \(y\) with respect to \(t\), based on each individual derivative: \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\).
In our exercise, we differentiate each term of the equation \(x^2 + y^2 = 2x + 4y\) with respect to \(t\) using the chain rule. We then apply it to both \(x\) and \(y\), capturing how these variables change over time. This imposes the rule: \(2x \frac{dx}{dt} + 2y \frac{dy}{dt} - 2 \frac{dx}{dt} - 4 \frac{dy}{dt} = 0\). By rearranging and solving this equation for either \(\frac{dx}{dt}\) or \(\frac{dy}{dt}\), we can find the rate of change of one variable given the rate of change of the other.
The Role of Differential Equations
Differential equations play a crucial role in modeling and understanding systems where the rate of change of a function is a point of interest. These equations involve derivatives, linking a function to its rates of change. In parametric functions, differential equations allow the describing of relationships between variables that change over time.
In the given exercise, the task is to address how changes in \(x\) and \(y\) are interrelated especially as functions of \(t\). Here, the relationship becomes a differential equation: \( (2x - 2) \frac{dx}{dt} + (2y - 4) \frac{dy}{dt} = 0 \). By understanding how these terms interact, we can solve for required derivatives, which in turn illustrate the dynamic behavior of these functions at given points.
Differential equations help us see how parametric representations of the relationship between \(x\) and \(y\) govern their combined motion. It provides insights into the dependency and impacts of change rate on each variable through time.
In the given exercise, the task is to address how changes in \(x\) and \(y\) are interrelated especially as functions of \(t\). Here, the relationship becomes a differential equation: \( (2x - 2) \frac{dx}{dt} + (2y - 4) \frac{dy}{dt} = 0 \). By understanding how these terms interact, we can solve for required derivatives, which in turn illustrate the dynamic behavior of these functions at given points.
Differential equations help us see how parametric representations of the relationship between \(x\) and \(y\) govern their combined motion. It provides insights into the dependency and impacts of change rate on each variable through time.
Grasping Implicit Differentiation
Implicit differentiation is a technique used to find derivatives of functions that are not explicitly solved for one variable in terms of another. Instead of rearranging a function explicitly, you differentiate both sides of an equation with respect to one independent variable, often using the chain rule.
In the equation \(x^2 + y^2 = 2x + 4y\), both \(x\) and \(y\) are expressed implicitly with respect to \(t\). Hence, implicit differentiation is applied. Here, we differentiate each element of the equation with respect to \(t\), while treating \(x\) and \(y\) as functions of \(t\). This results in equations connecting \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
Using implicit differentiation in parametric forms highlights how changes in interconnected variables are linked without explicit expressions. It allows simpler ways to solve the derivatives for equations involving multiple dependent variables, which might be tedious to reformat otherwise. This method respects the innate complexity of the relationship between \(x\) and \(y\) in scenarios like this exercise, allowing us to derive accurate derivatives to solve the problem constraints.
In the equation \(x^2 + y^2 = 2x + 4y\), both \(x\) and \(y\) are expressed implicitly with respect to \(t\). Hence, implicit differentiation is applied. Here, we differentiate each element of the equation with respect to \(t\), while treating \(x\) and \(y\) as functions of \(t\). This results in equations connecting \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
Using implicit differentiation in parametric forms highlights how changes in interconnected variables are linked without explicit expressions. It allows simpler ways to solve the derivatives for equations involving multiple dependent variables, which might be tedious to reformat otherwise. This method respects the innate complexity of the relationship between \(x\) and \(y\) in scenarios like this exercise, allowing us to derive accurate derivatives to solve the problem constraints.
Other exercises in this chapter
Problem 4
Find \(d y / d x\) $$ y=\ln (2+\sqrt{x}) $$
View solution Problem 4
Determine whether the statement is true or false. Explain your answer. For any polynomial \(p(x), \lim _{x \rightarrow+\infty} \frac{p(x)}{e^{x}}=0\).
View solution Problem 4
Find \(\left(f^{-1}\right)^{\prime}(x)\) using Formula (2), and check your answer by differentiating \(f^{-1}\) directly. $$ f(x)=\ln (2 x+1) $$
View solution Problem 4
Find \(d y / d x\) by implicit differentiation. \(x^{3}+y^{3}=3 x y^{2}\)
View solution