Problem 8 - Calculus - AP Edition

Question

Let \(l\) be the length of a diagonal of a rectangle whose sides have lengths \(x\) and \(y,\) and assume that \(x\) and \(y\) vary with time. (a) How are \(d l / d t, d x / d t,\) and \(d y / d t\) related? (b) If \(x\) increases at a constant rate of \(\frac{1}{2} \mathrm{ft} / \mathrm{s}\) and \(y \mathrm{de}-\) creases at a constant rate of \(\frac{1}{4} \mathrm{ft} / \mathrm{s}\), how fast is the size of the diagonal changing when \(x=3\) ft and \(y=4\) ft? Is the diagonal increasing or decreasing at that instant?

Step-by-Step Solution

Answer

(a) \(\frac{dl}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{\sqrt{x^2+y^2}}\). (b) Diagonal increases at \(\frac{1}{10}\) ft/s.

Step 1: Understanding the Problem

We have a rectangle with sides of length \(x\) and \(y\). The length of the diagonal \(l\) can be found using the Pythagorean theorem: \(l = \sqrt{x^2 + y^2}\). We want to find the relationship between \(\frac{dl}{dt}, \frac{dx}{dt},\) and \(\frac{dy}{dt}\).

Step 2: Finding the Relationship Between Derivatives

Differentiate \(l = \sqrt{x^2 + y^2}\) with respect to \(t\) using implicit differentiation. Start by differentiating both sides:\[\frac{d}{dt}l = \frac{d}{dt}\sqrt{x^2 + y^2}.\]This gives:\[\frac{dl}{dt} = \frac{1}{2\sqrt{x^2 + y^2}} \cdot (2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt})\]Simplify to:\[\frac{dl}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{\sqrt{x^2 + y^2}}.\]

Step 3: Substituting Given Values for Part (b)

We know that \(\frac{dx}{dt} = \frac{1}{2}\) ft/s and \(\frac{dy}{dt} = -\frac{1}{4}\) ft/s. Substitute these, along with \(x = 3\) ft and \(y = 4\) ft, into the formula:\[\frac{dl}{dt} = \frac{3 \cdot \frac{1}{2} + 4 \cdot (-\frac{1}{4})}{\sqrt{3^2 + 4^2}}.\]Calculate \(l\): \(\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\) ft.

Step 4: Calculating the Rate of Change of the Diagonal

Now continue with:\[\frac{dl}{dt} = \frac{\frac{3}{2} - 1}{5} = \frac{\frac{1}{2}}{5} = \frac{1}{10}\text{ ft/s}.\]Since \(\frac{1}{10}\) is positive, the diagonal length is increasing.

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique in calculus used when dealing with equations that are not easily solved for one variable in terms of others. In our problem, we have the equation for the diagonal length of a rectangle: \( l = \sqrt{x^2 + y^2} \), where both \( x \) and \( y \) change over time. To find how the diagonal length \( l \) changes with respect to time \( t \), we need to differentiate with respect to \( t \). This involves differentiating both sides of the equation with respect to time, treating both \( x \) and \( y \) as functions of \( t \).
The main steps in implicit differentiation involve:

• Differentiating each term with respect to \( t \).
• Applying the chain rule, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. For example, when differentiating \( x^2 \), we have \( 2x \cdot \frac{dx}{dt} \).

If you follow these steps, you will generate an expression connecting \( \frac{dl}{dt} \), \( \frac{dx}{dt} \), and \( \frac{dy}{dt} \). This will let us compute the rate of change of the diagonal length given changes in the sides of the rectangle.

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in geometry, expressing a relation between the sides of a right triangle. When applied to our rectangle, it helps determine the length of the diagonal \( l \) when given the lengths of the sides \( x \) and \( y \). The theorem states that the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
Mathematically, for our rectangle's diagonal:

• The diagonal is the hypotenuse of a right triangle formed by the rectangle's sides as the other two sides.
• Therefore, the equation is \( l = \sqrt{x^2 + y^2} \).

When solving related rates problems, knowing this formula allows us to find \( l \) at any instant, which is crucial for calculating \( \frac{dl}{dt} \). Understanding this theorem ensures you can relate the sides of the rectangle to its diagonal effectively, a critical step before applying derivative calculations.

Derivative Calculation

Derivative calculation is a key process in the analysis of changing quantities. It tells us how fast one variable changes with respect to another. In our problem, we calculate how the length of the diagonal \( l \) changes over time. Derivatives consider minute changes in a function, and calculating \( \frac{dl}{dt} \) involves applying the principles of implicit differentiation we've discussed.
Once the implicit differentiation provides the expression for \( \frac{dl}{dt} \), the next task is to substitute known rates of \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) into that expression. Here, \( \frac{dx}{dt} = \frac{1}{2} \text{ ft/s} \) and \( \frac{dy}{dt} = -\frac{1}{4} \text{ ft/s} \) were given.

• Plug these values along with \( x = 3 \) ft and \( y = 4 \) ft into the derivative formula \( \frac{dl}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{\sqrt{x^2 + y^2}} \).
• Calculate each term separately and simplify the resulting expression to derive \( \frac{dl}{dt} \).

This results in \( \frac{dl}{dt} = \frac{1}{10} \text{ ft/s} \), indicating that the diagonal is increasing. By understanding derivatives in the context of rates, you're equipped to analyze dynamic systems where multiple variables change over time.