If we measure time in hours the differential equation in part (b) changes. What is the new differential equation? Use the letter s t denote time in hours.. (c) dT ds (d) = dF ds If we measure time in hours and we also measure temperature in degrees Fahrenheit, the differential equation in part (c) changes even more. What is the new differential equation? Use the letter F to denote temperature in degrees Fahrenheit. -307+28.5 cos (2T-s) = X
If we measure time in hours the differential equation in part (b) changes. What is the new differential equation? Use the letter s t denote time in hours.. (c) dT ds (d) = dF ds If we measure time in hours and we also measure temperature in degrees Fahrenheit, the differential equation in part (c) changes even more. What is the new differential equation? Use the letter F to denote temperature in degrees Fahrenheit. -307+28.5 cos (2T-s) = X
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
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![### Newton's Law of Cooling and Differential Equations
**Newton’s law of cooling** states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. If we measure temperature in degrees Celsius and time in minutes, the constant of proportionality \( k \) equals 0.5. Suppose the ambient temperature \( T_A(t) \) is equal to a constant 57 degrees Celsius. Write the differential equation that describes the time evolution of the temperature \( T \) of the object.
#### Solution:
(a)
\[ \frac{dT}{dt} = -0.5T + 28.5 \]
This equation describes the rate of change of the temperature \( T \) of the object with respect to time \( t \).
Suppose the ambient temperature \( T_A(t) = 57 \cos\left( \frac{\pi}{30} t \right) \) degrees Celsius (time measured in minutes). Write the differential equation that describes the time evolution of temperature \( T \) of the object.
(b)
\[ \frac{dT}{dt} = -0.5T + 28.5 \cos\left( \frac{\pi}{30} t \right) \]
If we measure time in hours, the differential equation in part (b) changes. What is the new differential equation? Use the letter \( s \) to denote time in hours.
(c) Incorrect Solution:
\[ \frac{dT}{ds} = -30T + 28.5 \cos\left( 2\pi \cdot s \right) \]
This equation is incorrect because it does not correctly transform the time units from minutes to hours.
If we measure time in hours and we also measure temperature in degrees Fahrenheit, the differential equation in part (c) changes even more. What is the new differential equation? Use the letter \( F \) to denote temperature in degrees Fahrenheit.
(d)
\[ \frac{dF}{ds} = \]
The conversion factors and exact new form of the equation would need to be derived from transformations that account for both the time unit change and the temperature unit change.
### Explanation of Graphs or Diagrams
There are no diagrams or graphs in this problem, but the question involves calculating the differential equations based on Newton’s law of cooling under varying conditions. These illustrate the changes in temperature over time, considering different ambient temperatures](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fae98b63a-c1e0-40a0-8ef8-c296ebcafc68%2F0ea9c173-a3a9-4213-ae49-5d137bae0e0b%2Fegjk1yb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Newton's Law of Cooling and Differential Equations
**Newton’s law of cooling** states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. If we measure temperature in degrees Celsius and time in minutes, the constant of proportionality \( k \) equals 0.5. Suppose the ambient temperature \( T_A(t) \) is equal to a constant 57 degrees Celsius. Write the differential equation that describes the time evolution of the temperature \( T \) of the object.
#### Solution:
(a)
\[ \frac{dT}{dt} = -0.5T + 28.5 \]
This equation describes the rate of change of the temperature \( T \) of the object with respect to time \( t \).
Suppose the ambient temperature \( T_A(t) = 57 \cos\left( \frac{\pi}{30} t \right) \) degrees Celsius (time measured in minutes). Write the differential equation that describes the time evolution of temperature \( T \) of the object.
(b)
\[ \frac{dT}{dt} = -0.5T + 28.5 \cos\left( \frac{\pi}{30} t \right) \]
If we measure time in hours, the differential equation in part (b) changes. What is the new differential equation? Use the letter \( s \) to denote time in hours.
(c) Incorrect Solution:
\[ \frac{dT}{ds} = -30T + 28.5 \cos\left( 2\pi \cdot s \right) \]
This equation is incorrect because it does not correctly transform the time units from minutes to hours.
If we measure time in hours and we also measure temperature in degrees Fahrenheit, the differential equation in part (c) changes even more. What is the new differential equation? Use the letter \( F \) to denote temperature in degrees Fahrenheit.
(d)
\[ \frac{dF}{ds} = \]
The conversion factors and exact new form of the equation would need to be derived from transformations that account for both the time unit change and the temperature unit change.
### Explanation of Graphs or Diagrams
There are no diagrams or graphs in this problem, but the question involves calculating the differential equations based on Newton’s law of cooling under varying conditions. These illustrate the changes in temperature over time, considering different ambient temperatures
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