Problem 15 - Calculus Concepts

Question

Dog Weight For the first 9 months of life, the average weight \(w,\) in pounds, of a certain breed of dog increases at a rate that is inversely proportional to time, \(t,\) in months. A 1 -month-old puppy weighs 6 pounds, and a 9 -monthold puppy weighs 80 pounds. a. Write a differential equation describing the rate of change of the weight of the puppy. b. Give the particular solution for this differential equation on the basis of the information given. c. Estimate the weight of the puppy at 3 months and at 6 months. d. Why does this differential equation describe weight gain for only 8 months instead of for the life span of the dog?

Step-by-Step Solution

Answer

Differential equation: \( \frac{dw}{dt} = \frac{k}{t} \). Weight equation: \( w = \frac{74}{\ln(9)} \ln(t) + 6 \). Use this to find weight at given months. Only valid for 9 months due to growth changes.

Step 1: Define the Differential Equation

Since the rate of change of weight \( w \) is inversely proportional to time \( t \), we can write this as \( \frac{dw}{dt} = \frac{k}{t} \), where \( k \) is a constant. This is the form of the differential equation we need.

Step 2: Solve the Differential Equation

To find the solution, integrate both sides of the equation. \[\int dw = \int \frac{k}{t} \, dt\]This gives us \( w = k \ln(t) + C \), where \( C \) is an integration constant.

Step 3: Apply Initial Conditions

We know that at \( t = 1 \), \( w = 6 \). Plugging these values into the equation \( w = k \ln(t) + C \), we get:\[6 = k \ln(1) + C\]Which simplifies to \( 6 = C \) since \( \ln(1) = 0 \).

Step 4: Determine the Value of k

Use the condition at \( t = 9 \), where \( w = 80 \):\[80 = k \ln(9) + 6\]Solving for \( k \), we find:\[k = \frac{80 - 6}{\ln(9)}\]

Step 5: Write the Particular Solution

Plugging in \( C = 6 \) and \( k = \frac{74}{\ln(9)} \) back into the general solution, we get:\[w = \frac{74}{\ln(9)} \ln(t) + 6\]

Step 6: Estimate the Weight at 3 Months

Substitute \( t = 3 \) into the equation:\[w = \frac{74}{\ln(9)} \ln(3) + 6\]

Step 7: Estimate the Weight at 6 Months

Substitute \( t = 6 \) into the equation:\[w = \frac{74}{\ln(9)} \ln(6) + 6\]

Step 8: Explain Time Limitation of Model

This differential equation only applies up to 9 months because the inverse relationship model does not reflect how growth rates naturally slow and eventually cease as the dog approaches its full adult size.

Key Concepts

Inverse Proportionality

In mathematics, inverse proportionality describes a relationship where one variable increases while the other decreases. In the context of this exercise, the weight ( w ) of the puppy grows at a rate inversely related to time ( t ).

• Imagine two quantities: weight and time.
• If weight increases proportionally to a decrease in time, each unit of time has a smaller effect.

This is expressed as:\[\frac{dw}{dt} = \frac{k}{t}\] where k is a constant. It's like saying, "As the clock ticks, the rate of gaining weight gradually decreases." Over time, this affects how rapidly or slowly the pup's weight changes.

Integration

Integration is a powerful mathematical tool used to find a function when its derivative is known. When we integrate, we essentially reverse the process of differentiation to find the original function.

• The differential equation given was: \( \frac{dw}{dt} = \frac{k}{t} \).
• To solve it, we integrate: \( \int dw = \int \frac{k}{t} \, dt \).
• This yields: \( w = k \ln(t) + C \).

Here C is the integration constant, which we'll determine using provided values. The function we find describes how weight changes over time.

Initial Conditions

Initial conditions help us find specific values for constants in general solutions of differential equations. They anchor the equation to known real-world values at certain points.

• At one month, the weight ( w ) is 6 pounds. At this point, \( t = 1 \).
• By substituting these values into our weight equation, \( w = k \ln(t) + C \), we get \( 6 = C \) since \( \ln(1) = 0 \).
• Another condition is at nine months, where the puppy weighs 80 pounds. Joining this info gives more insight into k : \( 80 = k \ln(9) + 6 \).

Solving for k adjusts our general solution to reflect these precise conditions, allowing a more accurate description of the dog's growth at varying times.

Weight Estimation

Weight estimation involves using an established model to predict outcomes at different times. With the differential equation, we can project the puppy's weight for any given month.

• Substituting t = 3 provides an estimated weight at three months: \( w = \frac{74}{\ln(9)} \ln(3) + 6 \).
• Similarly, for six months: \( w = \frac{74}{\ln(9)} \ln(6) + 6 \).

This step transforms theoretical concepts into practical predictions. However, this equation eventually loses applicability beyond nine months since it hinges on an oversimplified pattern of growth—dictated solely by time and not accounting for natural maturity limits.