Problem 4 - Calculus of a Single Variable

Question

Using Related Rates In Exercises \(1-4,\) assume that \(x\) and \(y\) are both differentiable functions of \(t\) and find the required values of \(d y / d t\) and \(d x / d t .\) $$ \begin{aligned} x^{2}+y^{2}=25 & \text { (a) } \frac{d y}{d t} \text { when } x=3, y=4 \quad \frac{d x}{d t}=8 \\ & \text { (b) } \frac{d x}{d t} \text { when } x=4, y=3 \quad \frac{d y}{d t}=-2 \end{aligned} $$

Step-by-Step Solution

Answer

The value of \(dy/dt\) under the conditions of part (a) is -6, and the value of \(dx/dt\) under the conditions of part (b) is 1.5.

Step 1: Differentiate the Given Equation with Respect to Time

First, the given equation \(x^{2}+y^{2}=25\) is differentiated implicitly with respect to \(t\) using the chain rule. Doing so provides \(2x \cdot dx/dt + 2y \cdot dy/dt = 0\). This equation can be simplified to \(x dx/dt + y dy/dt = 0\).

Step 2: Solve for \(dy/dt\) under the Conditions of Part (a)

Substitute \(x=3, y=4, dx/dt =8\) into the derived equation to get \(3(8) + 4(dy/dt) = 0\). Solving for \(dy/dt\) gives \(dy/dt = -6\).

Step 3: Solve for \(dx/dt\) under the Conditions of Part (b)

Substitute \(x=4, y=3, dy/dt =-2\) into the derived equation to get \(4(dx/dt) + 3(-2) = 0\). Solving for \(dx/dt\) gives \(dx/dt = 1.5\).

Key Concepts

Implicit Differentiation

Implicit differentiation is a powerful technique in calculus used to find the derivative of an equation that cannot be easily solved for one variable in terms of the other. In related rates problems, where multiple variables change with time, it is often necessary to differentiate an equation with respect to time without explicitly solving for one variable.

For instance, consider an equation like the one in our exercise, \(x^{2}+y^{2}=25\). This equation does not give \(y\) explicitly as a function of \(x\), or vice versa; they are implicitly related through their relationship to another variable, time \(t\). When differentiating implicitly, we use \(\frac{dy}{dt}\) or \(\frac{dx}{dt}\) to denote the derivatives of \(y\) and \(x\) with respect to time, acknowledging that both \(x\) and \(y\) are functions of \(t\).

Applying implicit differentiation to our circle equation step by step, we take the derivative of each term with respect to \(t\), using the chain rule when needed. The \(x^{2}\) term, when differentiated, becomes \(2x \(\frac{dx}{dt}\)\), and the \(y^{2}\) term becomes \(2y \(\frac{dy}{dt}\)\). Implicit differentiation helps us to solve for one rate by knowing the other, exemplified when solving for \(\frac{dy}{dt}\) given \(\frac{dx}{dt}\), or vice versa.

Chain Rule Calculus

The chain rule in calculus is a formula to compute the derivative of a composite function. When we deal with related rates, we are often interested in how one rate affects another rate; the chain rule provides the connection between these rates. In the context of our exercise, where \(x\) and \(y\) are functions of \(t\), we use the chain rule to differentiate each term of \(x^{2}+y^{2}=25\) with respect to time \(t\).

The procedure reveals how the rate of change of the function's outer layer (squared terms) depends on the rate of change of its inner layer (\(x\) and \(y\) with respect to \(t\)). Specifically, the derivative of \( x^{2} \) with respect to time is \( 2x \(\frac{dx}{dt}\)\), and similarly, for \( y^{2} \) we get \( 2y \(\frac{dy}{dt}\)\) - exactly due to the chain rule application, where \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) are respectively the rates at which \( x \) and \( y \) change over time.

This essential calculus principle not only helps in finding new relationships between changing quantities but also ensures we accurately reflect the dependent nature of rates in differentiable functions.

Differential Equations

Differential equations are equations that involve one or more functions and their derivatives. They are central to modeling situations where change is continuous and variables are interdependent, such as in related rates problems in calculus.

In our textbook problem, the equation \(x^{2}+y^{2}=25\) is effectively a differential equation once we understand that \(x\) and \(y\) are functions of time \(t\), and we are looking for their rates of change. We have intuitively used a very simple kind of a differential equation where the relationship between the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) is directly found through implicit differentiation.

By substituting known values of \(x\), \(y\), \(\frac{dx}{dt}\), and/or \(\frac{dy}{dt}\), we are solving for the unknown rates, which is the essence of solving a differential equation. Differential equations can take many different forms and can require various complex methods to solve, but in the context of related rates problems like this one, the differential equation lends itself to a straightforward application of algebra and calculus techniques to find the solution.