Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

The vertex angle \(\theta\) opposite the base of an isosceles triangle with equal sides of length 100 centimeters is increasing at \(\frac{1}{10}\) radian per minute. How fast is the area of the triangle increasing when the vertex angle measures \(\pi / 6\) radians? Hint: \(A=\frac{1}{2} a b \sin \theta\).

Step-by-Step Solution

Answer

The area increases at \( 250 \sqrt{3} \) cm\(^2\) per minute.

Step 1: Identify the Formula for the Area of the Triangle

The area \( A \) of an isosceles triangle with sides of length \( a = b = 100 \) cm and the angle \( \theta \) between them can be found using the formula for the area of a triangle: \[ A = \frac{1}{2} \cdot a \cdot b \cdot \sin \theta \] Therefore, \[ A = \frac{1}{2} \cdot 100 \cdot 100 \cdot \sin \theta = 5000 \cdot \sin \theta \]

Step 2: Differentiate the Area Function with Respect to Time

To find how fast the area is increasing over time, differentiate the area function \( A = 5000 \cdot \sin \theta \) with respect to time \( t \). Using the chain rule, \[ \frac{dA}{dt} = 5000 \cdot \cos \theta \cdot \frac{d\theta}{dt} \] Here, \( \frac{d\theta}{dt} = \frac{1}{10} \) radians per minute, as given in the problem.

Step 3: Evaluate the Derivative at the Given Angle

Substitute \( \theta = \frac{\pi}{6} \) and \( \frac{d\theta}{dt} = \frac{1}{10} \) into the derivative: \[ \frac{dA}{dt} = 5000 \cdot \cos \left( \frac{\pi}{6} \right) \cdot \frac{1}{10} \] We know that \( \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} \). Therefore, \[ \frac{dA}{dt} = 5000 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{10} \] \[ \frac{dA}{dt} = 250 \sqrt{3} \]

Step 4: Simplify the Result

The rate at which the area of the triangle is increasing is \( 250 \sqrt{3} \) cm\(^2\) per minute.This is the final result after calculating the differentiated area formula and substituting \( \theta \) and \( \frac{d\theta}{dt} \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that deals with the rate at which a function changes as its input changes. In simple terms, it tells us how quickly or slowly something is changing at a particular point. When you differentiate a function, you find its derivative, which gives you this rate of change.
In the context of our problem, we have a function that describes the area of a triangle in terms of the vertex angle \(\theta\). When we differentiate this function with respect to time, we're looking for how the area of the triangle changes over time as \(\theta\) changes.

• The formula for the area of the isosceles triangle is given by \(A = \frac{1}{2} \cdot 100 \cdot 100 \cdot \sin \theta\).
• We differentiate this with respect to \(\theta\) to get \(\frac{dA}{d\theta} = 5000 \cdot \cos \theta\).
• Using the chain rule, we can find \(\frac{dA}{dt}\), the rate of change of the area over time: \(\frac{dA}{dt} = 5000 \cdot \cos \theta \cdot \frac{d\theta}{dt}\).

In essence, differentiation helps us understand how the area of the triangle is affected by changes in the vertex angle over time.

Related Rates

Related rates problems involve finding the rate at which one quantity changes by relating it to other quantities whose rates of change are known. Usually, they involve derivatives with respect to time.
In our problem, we are examining how the area of an isosceles triangle changes as the vertex angle \(\theta\) changes over time. We know that the angle \(\theta\) is increasing at a rate of \(\frac{1}{10}\) radians per minute. We want to find how quickly the area \(A\) changes.

• We first express the area of the triangle as a function of \(\theta\): \(A = 5000 \cdot \sin \theta\).
• Then, we differentiate with respect to time to find \(\frac{dA}{dt}\).
• The chain rule allows us to differentiate \(A\) in terms of \(\theta\), and then multiply by \(\frac{d\theta}{dt}\) to relate the rates.

By substituting known values, we can solve for \(\frac{dA}{dt}\), the rate at which the area is expanding. Related rates problems are crucial in dynamic contexts where multiple variables change with respect to time.

Trigonometry

Trigonometry deals with the relationships between the sides and angles of triangles. It's particularly useful when working with angles in calculus problems.
In this exercise, trigonometry is used to determine the area of the triangle through a trigonometric function: the sine of the angle \(\theta\). We use the formula \(A = \frac{1}{2} ab \sin \theta\) to express the area.

• The sine function helps us capture the effect of the angle \(\theta\) on the triangle's dimensions.
• For \(\theta = \frac{\pi}{6}\), the cosine value \(\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\) is an essential part of calculating the rate of change of the area.
• Trigonometric identities like \(\cos \theta\) provide crucial information when we're differentiating functions involving angles.

Thus, trigonometry allows us to effectively link angles with side lengths and calculate how changes in angles impact spatial properties, like the area of a triangle.