Problem 29 - The Calculus with Analytic Geometry

Question

At a given instant, the length of one leg of a right triangle is \(10 \mathrm{ft}\) and it is increasing at the rate of \(1 \mathrm{ft} / \mathrm{min}\) and the length of the other leg of the right triangle is \(12 \mathrm{ft}\) and it is decreasing at the rate of \(2 \mathrm{ft} / \mathrm{min}\). Find the rate of change of the measure of the acute angle opposite the leg of length \(12 \mathrm{ft}\) at the given instant.

Step-by-Step Solution

Answer

The rate of change of the angle is \(\frac{2}{9} \text{ rad/min}\).

Step 1 - Identify Known Values

Let the lengths of the legs be denoted by: - The leg increasing in length is denoted by \(a\) with \(a = 10 \text{ ft}\) and \( \frac{da}{dt} = 1 \text{ ft/min} \). - The leg decreasing in length is denoted by \(b\) with \(b = 12 \text{ ft}\) and \( \frac{db}{dt} = -2 \text{ ft/min} \) (negative because it is decreasing).

Step 2 - Use Pythagorean Theorem

For a right triangle, the hypotenuse \(c\) can be defined as: \[ c^2 = a^2 + b^2 \]. Differentiating both sides with respect to time \(t\), we get: \[ 2c \frac{dc}{dt} = 2a \frac{da}{dt} + 2b \frac{db}{dt} \]. Simplifying and substituting the values for \(a, b, \frac{da}{dt}, \text{ and } \frac{db}{dt}\), we get: \[ c \frac{dc}{dt} = a \frac{da}{dt} + b \frac{db}{dt} \].

Step 3 - Calculate Hypotenuse and Derivative

First, calculate the hypotenuse using: \[ c = \sqrt{10^2 + 12^2} = 15 \text{ ft} \]. Now substitute back to find \( \frac{dc}{dt} \): \[ 15 \frac{dc}{dt} = 10 \times 1 + 12 \times (-2) \]. Therefore, \[ 15 \frac{dc}{dt} = 10 - 24 = -14 \]. Simplify to find: \[ \frac{dc}{dt} = -\frac{14}{15} \text{ ft/min} \].

Step 4 - Use Trigonometric Relation

Next, use \( \tan(\theta) = \frac{a}{b} \) where \( \theta \) is the acute angle opposite leg \(b\). Differentiate implicitly: \[ \frac{d}{dt} \tan(\theta) = \frac{d}{dt} \frac{a}{b} \]. Simplifying, \[ \frac{1}{\tan^2(\theta)} \frac{d\theta}{dt} \times \frac{d}{dt}(\tan(\theta)) = \frac{b \frac{da}{dt}}{b^2} - \frac{a \frac{db}{dt}}{b^2} \]. We know \( \tan(\theta) = \frac{10}{12} = \frac{5}{6} \).

Step 5 - Calculate Rate of Change of Angle

To find the rate of change \(\frac{d\theta}{dt}\), \[ \tan^2(\theta) + 1 = \frac{d\theta}{dt} \times \frac{b \frac{da}{dt} - a \frac{db}{dt}}{b^2} \], substitute values: \[ \frac{d\theta}{dt} = \frac{12 \times 1 - 10 \times (-2)}{10^2 + 12^2} \]. Simplify to: \[ \frac{d\theta}{dt} = \frac{12 + 20}{144} = \frac{32}{144} = \frac{2}{9} \text{ rad/min} \].

Key Concepts

Pythagorean theorem

The Pythagorean theorem is fundamental in solving problems involving right triangles. It states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. Mathematically, this can be represented as: \[c^2 = a^2 + b^2\]where \(c\) is the hypotenuse, and \(a\) and \(b\) are the other two sides. This theorem helps us determine the length of one side if we know the lengths of the other two. In our problem, we first used the Pythagorean theorem to find the hypotenuse when \(a = 10 \text{ ft}\) and \(b = 12 \text{ ft}\). We calculated it as: \[c = \sqrt{10^2 + 12^2} = 15 \text{ ft} \].Understanding this concept is key to solving many problems involving right triangles and helps set the base for further calculations like differentiation.

differentiation

Differentiation in calculus is used to determine how a function changes as its input changes. In the context of this exercise, we are interested in how the lengths of the sides of the triangle change over time. We use the rates of change given (\(\frac{da}{dt}\) and \(\frac{db}{dt}\)) to find the rate of change of the hypotenuse (\(\frac{dc}{dt}\)). We start by differentiating the Pythagorean theorem implicitly with respect to time \(t\), which gives us: \[2c \frac{dc}{dt} = 2a \frac{da}{dt} + 2b \frac{db}{dt}\].By substituting the known values of \(a\), \(b\), \(\frac{da}{dt}\), and \(\frac{db}{dt}\), we find: \[15 \frac{dc}{dt} = 10 \times 1 + 12 \times (-2) = -14\],thus, \[\frac{dc}{dt} = -\frac{14}{15} \text{ ft/min}\].Differentiation makes it possible to analyze and solve problems involving rates of change and motion.

implicit differentiation

Implicit differentiation is a technique used to differentiate equations when the variables are not easily separated. This is particularly useful in our problem because the equation involves both \(a\) and \(b\). To find how the angle \(\theta\) changes over time, we use the trigonometric relationship \( \tan(\theta) = \frac{a}{b} \).Differentiating both sides concerning time \(t\), we get: \[ \frac{d}{dt} \tan(\theta) = \frac{d}{dt} \left( \frac{a}{b} \right) \],which simplifies to: \[ \frac{1}{\cos^2 (\theta)} \frac{d \theta}{dt} = \frac{b \frac{da}{dt} - a \frac{db}{dt}}{b^2} \].Putting the known values and solving for \( \frac{d \theta}{dt} \), we find:\[ \frac{d \theta}{dt} = \frac{12 \times 1 - 10 \times (-2)}{10^2 + 12^2} = \frac{32}{144} = \frac{2}{9} \text{ rad/min} \].Implicit differentiation lets us find derivatives in scenarios where direct separation isn't feasible.

trigonometric identities

Trigonometric identities are equations that involve trigonometric functions and are true for any angle. In this exercise, we use the identity involving the tangent function: \[\tan(\theta) = \frac{a}{b}\],where \(\theta\) is the angle opposite side \(b\). Using this identity allows for the angle's rate of change to be connected to the rates of change in the sides of the triangle.Upon differentiating implicitly with respect to time, knowing: \[ \tan(\theta) = \frac{10}{12} = \frac{5}{6}\],and differentiating, we employ the trigonometric identity: \[\frac{1}{ \cos^2 (\theta) } = 1 + \tan^2 (\theta)\].By linking these trigonometric identities, we find the rate of change of the angle over time: \[ \frac{d \theta}{dt} = \frac{12 \times 1 - 10 \times (-2)}{b^2 + a^2} \].Understanding and utilizing trigonometric identities effectively bridges the gap between geometric shapes and calculus-based problems.