Problem 24 - Calculus Volume 1

Question

For the following exercises, draw and label diagrams to help solve the related-rates problems. A triangle has a height that is increasing at a rate of 2 \(\mathrm{cm} /\) sec and its area is increasing at a rate of 4 \(\mathrm{cm}^{2} / \mathrm{sec}\) . Find the rate at which the base of the triangle is changing when the height of the triangle is 4 \(\mathrm{cm}\) and the area is 20 \(\mathrm{cm}^{2} .\)

Step-by-Step Solution

Answer

The base is decreasing at 3 cm/sec.

Step 1: Understand the Problem Context

We have a triangle with a height increasing at 2 cm/sec, and an area increasing at 4 cm²/sec. We need to find how fast the base, \(b\), is changing. Known values: \(h = 4\) cm, \(A = 20\) cm², \(\frac{dh}{dt} = 2\) cm/sec, \(\frac{dA}{dt} = 4\) cm²/sec.

Step 2: Write Down the Formula

The area \(A\) of a triangle is \(A = \frac{1}{2} b h\), where \(b\) is the base and \(h\) is the height.

Step 3: Differentiate the Area Formula with Respect to Time

Use the chain rule to differentiate the area equation with respect to time \(t\): \[\frac{dA}{dt} = \frac{1}{2} \left( b \frac{dh}{dt} + h \frac{db}{dt} \right)\]

Step 4: Substitute Known Rates and Values

Plug in the given values and rates into the differentiated equation: \[4 = \frac{1}{2} (b \cdot 2 + 4 \cdot \frac{db}{dt})\]Solve for \(b\) first using \(A = \frac{1}{2} \times b \times h = 20\). Hence, \(b \times 4 = 40\), so \(b = 10\) cm.

Step 5: Solve for the Rate of Change of Base \( \frac{db}{dt} \)

Substitute \(b = 10\) cm into the differentiated equation:\[4 = \frac{1}{2} (10 \times 2 + 4 \times \frac{db}{dt})\]This simplifies to \(4 = \frac{1}{2} (20 + 4 \frac{db}{dt})\). Simplifying gives:\(4 = 10 + 2 \frac{db}{dt}\), so \(\frac{db}{dt} = -3\) cm/sec.

Key Concepts

Understanding the Chain Rule

The chain rule is an essential concept in calculus utilized for differentiating composite functions. When functions are nested within each other, the chain rule allows us to differentiate without unraveling each function.
To differentiate a composite function effectively:

• Identify the outer function and inner function.
• Differentiate the outer function, keeping the inner function unchanged.
• Multiply by the derivative of the inner function.

In the context of this problem, we need to apply the chain rule to differentiate the area equation of the triangle with respect to time. By understanding how each component of the triangle—like base and height—changes over time, we are able to solve for unknown rates.

Calculating the Area of a Triangle

Understanding the formula for the area of a triangle is crucial for solving related rates problems involving triangles. The area (\(A\)) of a triangle is given by the formula:

• \(A = \frac{1}{2} \times b \times h\)

where \(b\) is the base and \(h\) is the height of the triangle.
This formula is used as the primary equation to relate changes in the triangle's dimensions to changes in its area. In problems where the area's rate of change is given, alongside the rates of change of base and height, this formula helps us set up the necessary relationship to find unknown rates.

The Process of Differentiation

Differentiation is the mathematical process of finding the rate at which a function is changing at any given point. In the problem, we differentiated the area formula of the triangle with respect to time to establish a relationship between the various rates of change.
To use differentiation effectively:

• Start with the function you want to differentiate. Here, it's the area equation \(A = \frac{1}{2} bh\).
• Apply the chain rule to break down the function's components that are changing over time.

In this case, we differentiated to find \(\frac{dA}{dt}\) in terms of \(b\), \(h\), \(\frac{db}{dt}\), and \(\frac{dh}{dt}\). This forms a new equation that you can use to substitute known values and solve for unknown rates of change.

Understanding Rate of Change

Rate of change refers to how a particular quantity alters over time. It's a fundamental aspect of calculus, especially in real-world applications like physics, economics, and geometry-related problems like this exercise.
In the exercise, we were given:

• The rate at which the height is increasing: \(\frac{dh}{dt} = 2 \text{ cm/sec}\)
• The rate at which the area is increasing: \(\frac{dA}{dt} = 4 \text{ cm}^2/\text{sec}\)

Utilizing these rates, we applied the differentiated area formula to derive the rate of change for the base \(\frac{db}{dt}\). This approach ties together various rates of change to solve for an unknown rate efficiently, highlighting the interdependence of dimensions in related rates problems.