Question

The length \(L\) metres of a certain metal rod at temperature \(t^{\circ} \mathrm{C}\) is given by: \(L=1+\) \(0.00003 t+0.0000004 t^{2}\). Determine the rate of change of length, in \(\mathrm{mm} /{ }^{\circ} \mathrm{C}\), when the temperature is (a) \(100^{\circ} \mathrm{C}\) and (b) \(250^{\circ} \mathrm{C}\)

Short answer

The rates of change are 0.11 mm/°C at 100°C and 0.23 mm/°C at 250°C.

Step-by-step solution

Formula for Rate of Change

The rate of change of length with respect to temperature is the derivative of the length formula with respect to temperature. Differentiating the given function \(L = 1 + 0.00003t + 0.0000004t^2\) with respect to \(t\) yields: \(\frac{dL}{dt} = 0.00003 + 2 \times 0.0000004 \times t = 0.00003 + 0.0000008 t\).

Evaluate Derivative at 100°C

Insert \(t = 100\) into the derivative: \(\frac{dL}{dt} = 0.00003 + 0.0000008 \times 100 = 0.00003 + 0.00008 = 0.00011\).

Convert to Millimeters per Degree Celsius

Since the rate is given in meters/°C, convert \(0.00011\) meters/°C to millimeters/°C by multiplying by 1000: \(0.00011 \times 1000 = 0.11\).

Evaluate Derivative at 250°C

Insert \(t = 250\) into the derivative: \(\frac{dL}{dt} = 0.00003 + 0.0000008 \times 250 = 0.00003 + 0.0002 = 0.00023\).

Convert the 250°C Rate to Millimeters per Degree Celsius

Convert \(0.00023\) meters/°C to millimeters/°C by multiplying by 1000: \(0.00023 \times 1000 = 0.23\).

Key concepts

Understanding the Rate of Change

The rate of change is a fundamental concept that helps us understand how one quantity changes with respect to another. In this context, we're interested in how the length of a metal rod changes as the temperature changes. This can give us insight into the material's properties.

To find the rate of change, we use derivatives. When dealing with functions, the derivative gives the rate at which one variable changes as another variable changes. In our case, the length of the rod, denoted by \(L\), is a function of temperature \(t\). By differentiating this function, we can determine \(\frac{dL}{dt}\), the rate of change of length with respect to temperature.

• Derivatives help measure how fast one quantity changes in relation to another.
• The formula used to find this rate is obtained by differentiating \(L\) with respect to \(t\).

This particular rate of change will tell us how many millimeters the rod's length changes for each degree increase in temperature.

Temperature and Length Relationship

The length of materials often depends on temperature due to thermal expansion. When materials are heated, their molecules have more energy and move apart, leading to an increase in length.

In our example, the length \(L\) of the rod is given by a quadratic equation. The formula \(L = 1 + 0.00003t + 0.0000004t^2\) shows that length increases with temperature, but not linearly:

• The constant term \(1\) indicates the initial length of the rod at \(0^{\circ} \text{C}\).
• The linear term \(0.00003t\) represents the direct proportional increase of length with temperature.
• The quadratic term \(0.0000004t^2\) accounts for the additional changes as temperature rises and reflects the non-linear part of growth.

Understanding the temperature-length relationship helps in designing materials that can withstand temperature variations without undergoing significant changes in shape or size.

Evaluating the Derivative for Specific Temperatures

Let's evaluate the derivative \(\frac{dL}{dt}\) at specific temperatures: \(100^{\circ} \text{C}\) and \(250^{\circ} \text{C}\). This allows us to understand how much the rod's length changes at these conditions.

First, calculate the derivative expression: \(\frac{dL}{dt} = 0.00003 + 0.0000008t\). Then substitute the temperature values:
- At \(100^{\circ} \text{C}\): Substitute \(t = 100\) into the derivative to get \[\frac{dL}{dt} = 0.00003 + 0.0000008 \times 100 = 0.00011\, \text{meters/°C}\]. Convert to millimeters: \(0.00011 \times 1000 = 0.11\, \text{mm/°C}\).
- At \(250^{\circ} \text{C}\): Substitute \(t = 250\) into the derivative to get \[\frac{dL}{dt} = 0.00003 + 0.0000008 \times 250 = 0.00023\, \text{meters/°C}\]. Convert to millimeters: \(0.00023 \times 1000 = 0.23\, \text{mm/°C}\).

Evaluating the derivative at these temperatures shows us that the rate of expansion increases with temperature. This is crucial for applications requiring precise measurements and control of expansion.