Problem 4 - Calculus an Applied Approach

Question

use the given values to find \(d y / d t\) and \(d x / d t\). $$ x^{2}+y^{2}=25 $$ $$ \begin{array}{ll}{\text { (a) } \frac{d y}{d t}} & {x=3, y=4, \frac{d x}{d t}=8} \\ {\text { (b) } \frac{d x}{d t}} & {x=4, y=3, \frac{d y}{d t}=-2}\end{array} $$

Step-by-Step Solution

Answer

The derived values are \(dy/dt = -6\) for part (a) and \(dx/dt = 3\) for part (b).

Step 1: Differentiate both sides of the equation with respect to t

Start by differentiating the equation \(x^{2}+y^{2}=25\) with respect to \(t\). Applying the chain rule to \(x^2\) and \(y^2\), we get \(2x\cdot dx/dt + 2y\cdot dy/dt = 0\).

Step 2: Solve for dy/dt in (a)

For part (a), we are given \(x=3\), \(y=4\), and \(dx/dt=8\). We substitute these values into the equation derived in Step 1. This gives us \(2(3)\cdot8 + 2(4)\cdot dy/dt = 0\). Solve this equation for \(dy/dt\).

Step 3: Solve for dx/dt in (b)

For part (b), we are given \(x=4\), \(y=3\), and \(dy/dt=-2\). Substitute these values into the equation from Step 1, yielding \(2(4)\cdot dx/dt + 2(3)\cdot -2 = 0\). Solve this equation for \(dx/dt\).

Key Concepts

Rate of Change

In mathematical terms, a "Rate of Change" refers to how a quantity changes in relation to another. It is often used when dealing with dynamic processes like motion or growth. In the given exercise, the rate of change is depicted as either \(\frac{dx}{dt}\) or \(\frac{dy}{dt}\). These notations represent how the variables \(x\) and \(y\) change with respect to time \(t\).

Understanding rate of change is crucial when solving problems involving moving objects or changing values. For instance, if a relationship between \(x\) and \(y\) is defined by an equation, knowing their rate of change with respect to \(t\) helps determine how altering one variable influences the other.

In the exercise, we find these rates using partial derivatives, which account for the change in variables over time. This is significant for accurately analyzing how different parts of a system interact and evolve over time.

Chain Rule

The "Chain Rule" is a fundamental principle in calculus used to differentiate composite functions. Essentially, it allows us to find the derivative of a function nested within another function. It’s essential when working with implicit differentiation, which is applied in the exercise where both variables \(x\) and \(y\) are functions of \(t\).

Consider the exercise: the equation \(x^2 + y^2 = 25\) implicates that both \(x\) and \(y\) change over time. The chain rule helps differentiate this equation with respect to \(t\). We get

• For \(x^2\), the derivative is \(2x\cdot \frac{dx}{dt}\).
• For \(y^2\), the derivative is \(2y\cdot \frac{dy}{dt}\).

The chain rule connects these derivatives back to \(t\), enabling us to find how \(x\) and \(y\) change as \(t\) changes. This is crucial for solving both parts (a) and (b) of the exercise as it helps us isolate the required rates of change.

Related Rates

"Related Rates" problems involve finding the rate at which one quantity changes concerning another, given a relationship between them. These types of problems are prevalent in physics and engineering and often use implicit differentiation to solve.

For the equation \(x^2 + y^2 = 25\), \(x\) and \(y\) are related and their rates are influenced by each other because they collectively maintain the fixed sum of their squares. We use implicit differentiation to derive a general formula linking \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). Then, by substituting known values, we find unknown rates.

In the exercise:

• Part (a) uses the derived formula to find \(\frac{dy}{dt}\) given \(x\), \(y\), and \(\frac{dx}{dt}\).
• Part (b) involves finding \(\frac{dx}{dt}\) given the values for \(x\), \(y\), and \(\frac{dy}{dt}\).

Related rates problems underscore how changes in one part of a system affect others, making them powerful tools for modeling real-world phenomena.