Problem 27 - Calculus

Question

A growing box has length \(t,\) width \(1 /(1+t)\), and height \(\cos t\) (a) What is the rate of change of the volume? (b) What is the rate of change of the surface area?

Step-by-Step Solution

Answer

Compute \( \frac{dV}{dt} \) and \( \frac{dS}{dt} \) using derivatives of volume and surface area formulas.

Step 1: Write the Expression for Volume

The volume \( V \) of a box is calculated by multiplying its length, width, and height. Given: \( L = t \), \( W = \frac{1}{1+t} \), and \( H = \cos t \). Thus, the volume is \( V = t \cdot \frac{1}{1+t} \cdot \cos t \).

Step 2: Differentiate Volume with Respect to t

To find the rate of change of volume, we differentiate \( V = t \frac{1}{1+t} \cos t \) with respect to \( t \). Use the product rule and chain rule to find \( \frac{dV}{dt} \).

Step 3: Simplification of Volume's Derivative

Apply the product rule to \( t \cdot \frac{1}{1+t} \cdot \cos t \). Let \( u = t \), \( v = \frac{1}{1+t} \), \( w = \cos t \). Then, \( \frac{dV}{dt} = \left( u \frac{dv}{dt} \right)w + uv \left( \frac{dw}{dt} \right) + \frac{du}{dt}vw \), and compute each derivative.

Step 4: Calculate the Surface Area

The surface area \( S \) of a box is computed by the formula \( S = 2(LW + LH + WH) \). Substitute \( L = t \), \( W = \frac{1}{1+t} \), \( H = \cos t \) into this formula.

Step 5: Differentiate Surface Area with Respect to t

To find the rate of change of the surface area, differentiate \( S = 2(t \cdot \frac{1}{1+t} + t \cdot \cos t + \frac{1}{1+t} \cdot \cos t) \) with respect to \( t \) using the product rule and chain rule.

Step 6: Simplification of Surface Area's Derivative

Apply the derivative rules to each term in the surface area expression, focusing on the interactions of \( t \), \( \frac{1}{1+t} \), and \( \cos t \). Sum the resulting derivatives to find \( \frac{dS}{dt} \).

Key Concepts

Derivative

In calculus, a derivative represents the rate at which a function is changing at any given point. It is sometimes called the slope of the function at a certain point and gives us a way to understand dynamics in processes. For instance, in our exercise, the derivative helps in finding how quickly the volume or surface area of the box changes with respect to time.

• To find a derivative, you differentiate the function with respect to the variable of interest, which is usually time or space in practical problems.
• Differentiation follows certain rules and techniques, such as the power rule, product rule, or chain rule, each suited to specific types of functions and compositions.

In this exercise, we use derivatives to monitor the changes in volume and surface area over time, which are crucial in understanding the box's spatial dynamics.

Product Rule

The product rule is essential when differentiating expressions where two or more functions are multiplied together. It ensures that each part of the product is differentiated correctly and added accordingly.Imagine multiplying function \( u \) by function \( v \):

• The derivative of the product \( uv \) is given by the formula: \( \frac{d}{dt}(uv) = u \frac{dv}{dt} + v \frac{du}{dt} \).

This formula indicates that you differentiate each function separately, multiply, and sum the results.In the context of this exercise, the product rule is applied to components like \( t \cdot \frac{1}{1+t} \cdot \cos(t) \), where each function is treated separately, following the rule to account for each piece's contribution to the change.

Chain Rule

The chain rule is a fundamental technique for differentiating composite functions. It allows you to differentiate functions inside other functions.Consider a composite function \( f(g(t)) \), where \( g \) is a function of \( t \), and \( f \) is a function of \( g(t) \):

• The chain rule states that \( \frac{df}{dt} = \frac{df}{dg} \cdot \frac{dg}{dt} \).

This approach effectively "chains" together the derivatives of the inner and the outer functions.In our exercise, the chain rule becomes crucial for functions like \( \cos t \) or \( \frac{1}{1+t} \), as it permits the breaking down of more complex functions into manageable parts for differentiation.

Rate of Change

The concept of the rate of change captures how a quantity changes over time, which is vital for modeling dynamic systems. In this context, it answers questions like how the volume or surface area of the box evolves over a given duration.The rate of change is specifically calculated by finding the derivative of the function representing the physical quantity you're interested in:

• For volume, this involves differentiating the volume formula \( V(t) = t \cdot \frac{1}{1+t} \cdot \cos t \).
• For the surface area, you differentiate its respective formula.

The rates of change provide insights into how quickly aspects of the system, such as volume or surface area, are increasing or decreasing, offering a deeper understanding of the parameters affecting these changes.