Problem 22 - Calculus Volume 1

Question

Draw and label diagrams to help solve the related-rates problems. The base of a triangle is shrinking at a rate of \(1 \mathrm{~cm} / \mathrm{min}\) and the height of the triangle is increasing at a rate of \(5 \mathrm{~cm} /\) min. Find the rate at which the area of the triangle changes when the height is \(22 \mathrm{~cm}\) and the base is \(10 \mathrm{~cm}\).

Step-by-Step Solution

Answer

The area of the triangle increases at a rate of 14 cm²/min.

Step 1: Understand the Problem

We have a triangle with a base and a height that change over time. We need to find the rate at which the area changes.

Step 2: Diagram the Scenario

Sketch a triangle and label the base as \(b\) and the height as \(h\). Note that \( \frac{db}{dt} = -1 \) cm/min and \( \frac{dh}{dt} = 5 \) cm/min.

Step 3: Write the Area Formula

The area \( A \) of a triangle is given by \( A = \frac{1}{2} b h \).

Step 4: Differentiate the Area Formula

Apply implicit differentiation with respect to time \( t \) to the area formula: \( \frac{dA}{dt} = \frac{1}{2} \left( b \frac{dh}{dt} + h \frac{db}{dt} \right) \).

Step 5: Substitute Known Values

Plug in \( b = 10 \) cm, \( h = 22 \) cm, \( \frac{db}{dt} = -1 \) cm/min, and \( \frac{dh}{dt} = 5 \) cm/min into the differentiated equation: \( \frac{dA}{dt} = \frac{1}{2} \left( 10 \times 5 + 22 \times (-1) \right) \).

Step 6: Calculate the Rate of Change of Area

Simplify the expression: \( \frac{dA}{dt} = \frac{1}{2} (50 - 22) = \frac{1}{2} (28) = 14 \) cm²/min.

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique we use when dealing with equations that are not explicitly solved for one variable. In the context of related rates, we often deal with functions related to time. Here, implicit differentiation helps us find how one quantity changes with respect to time by incorporating the rates at which other quantities change.

• To apply implicit differentiation, consider a formula like the area of a triangle. We'll differentiate everything with respect to time.
• For example, if the area of a triangle is given by \( A = \frac{1}{2} b h \), where \( b \) (base) and \( h \) (height) are functions of time, implicit differentiation lets us find \( \frac{dA}{dt} \), the rate of change of the area.

This technique relies on the chain rule. The chain rule helps us differentiate composite functions by breaking them into simpler parts. With implicit differentiation, we systematically apply it to each variable and its time rate.

Changing Dimensions

In this problem, the triangle's base and height are changing, illustrating changing dimensions. This concept involves the rates at which dimensions like length, width, or height of a geometric figure change over time.

• For the triangle, the base shrinks at \(-1\) cm/min, and the height grows at \(5\) cm/min.
• These changes impact the shape, and hence the area of the triangle every minute.

Understanding changing dimensions is crucial as it helps predict how specific aspects like area or volume morph over time. With this knowledge, we can apply related rates to measure how quickly these changes happen at any given moment.

Triangle Area

Calculating the area of a triangle when its dimensions are changing is an essential part of solving related rates problems. The formula used is quite simple: \( A = \frac{1}{2} b h \), where

• \( b \) is the length of the base and
• \( h \) is the height of the triangle.

The area of the triangle can change even if one of these quantities remains constant. Therefore, in problems involving changing dimensions, the formula allows us to understand how both base and height contribute to the changing area.

Rate of Change

The rate of change is a fundamental concept in calculus, particularly in related rates. It tells us how quickly a quantity, like area, volume, or length, is changing with respect to time. In this triangle problem:

• The calculation \( \frac{dA}{dt} = \frac{1}{2} (b \cdot \frac{dh}{dt} + h \cdot \frac{db}{dt}) \) finds how fast the area changes.
• Substitute known rates and dimensions: \( b = 10 \), \( h = 22 \), \( \frac{db}{dt} = -1 \), and \( \frac{dh}{dt} = 5 \).

Finally, the result, \( \frac{dA}{dt} = 14 \) cm²/min, reflects how the shrinking base and growing height together affect the area per minute. This illustrates that rates of change take both growth and reductions into account, providing a complete picture of dynamic changes.