Chapter 24: Problem 11
$$\text { Solve the problems in related rates.}$$ The electric resistance \(R\) (in \(\Omega\) ) of a certain resistor as a function of the temperature \(T\) (in \(^{\circ} \mathrm{C}\) ) is \(R=4.000+0.003 T^{2} .\) If the temperature is increasing at the rate of \(0.100^{\circ} \mathrm{C} / \mathrm{s}\), find how fast the resistance changes when \(T=150^{\circ} \mathrm{C}\)
Short Answer
Step by step solution
Identify Given Information
Differentiate the Resistance Function
Calculate \( \frac{dR}{dT} \)
Substitute Given Values
Calculate \( \frac{dR}{dt} \)
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
For example, in our problem, resistance \( R \) is a function of temperature \( T \), and temperature \( T \) in turn varies with time \( t \). Thus, to find how quickly \( R \) changes with time, we apply the chain rule:
- First, calculate \( \frac{dR}{dT} \), the rate at which resistance changes with temperature.
- Then, multiply it by \( \frac{dT}{dt} \), which is how quickly temperature changes over time.
The chain rule transforms complex real-world relationships into manageable calculations, making it a crucial tool in physics and engineering.
Differentiation
To identify how resistance changes with temperature, we start by differentiating the function given: \( R = 4.000 + 0.003T^2 \).
Here, let's differentiate step by step:
- The derivative of the constant 4.000 is zero since constants do not change.
- The derivative of \( 0.003T^2 \) with respect to \( T \) is \( 0.006T \) (applying the power rule).
Electric Resistance
The given relation \( R = 4.000 + 0.003T^2 \) captures how resistance \( R \) responds to changes in temperature \( T \) in this specific setup. At a glance, this equation implies:
- As temperature \( T \) rises, \( R \) may increase since the \( 0.003T^2 \) term grows larger.
- The constants define the resistance at a baseline state (when no temperature effect is present).
Temperature Change
This rise means every second, the temperature boosts by \( 0.1^{\circ} \mathrm{C} \), affecting the resistance through the effect captured in the derivative \( \frac{dR}{dT} \).
To calculate the impact:
- We know that the temperature change is steady, providing a constant \( \frac{dT}{dt} \).
- The increase impacts \( R \) as you multiply \( 0.006T \) by this constant \( 0.1 \).