applied calculus

Question

The weight of a laboratory mouse between 3 and 11 weeks of age can be modeled as

w(t) = 11.4 + 7.33 ln t

 grams

where the age of the mouse is t + 2 weeks.

 

How rapidly is its weight changing? (Round your answer to three decimal places.)

(b) What is the average rate of change in the weight of the mouse between ages 3 and 7 weeks? (Round your answer to three decimal places.)

 

Answer 1

(a) To find how rapidly the weight of the mouse is changing, we need to calculate the derivative of the weight function \(w(t)\) with respect to time (\(t\)). In this case, \(t\) represents the age of the mouse in weeks.

The weight function is given as:

\[w(t) = 11.4 + 7.33 \ln(t)\]

Now, let's find \(w'(t)\), the derivative of \(w(t)\) with respect to \(t\):

\[w'(t) = \frac{d}{dt}\left(11.4 + 7.33 \ln(t)\right)\]

Using the chain rule for differentiation, the derivative of \(\ln(t)\) with respect to \(t\) is \(\frac{1}{t}\), so:

\[w'(t) = 7.33 \cdot \frac{1}{t}\]

Now, we can calculate the rate of change of weight at a specific age, say at \(t = 3\) weeks:

\[w'(3) = 7.33 \cdot \frac{1}{3} = 2.443\]

So, the rate at which the mouse's weight is changing at 3 weeks of age is approximately 2.443 grams per week.

(b) To find the average rate of change in the weight of the mouse between ages 3 and 7 weeks, we can use the formula for the average rate of change:

Average Rate of Change = \(\frac{\text{Change in Weight}}{\text{Change in Time}}\)

The change in weight is \(w(7) - w(3)\), and the change in time is \(7 - 3\).

First, calculate \(w(7)\) and \(w(3)\) using the given weight function:

\[w(7) = 11.4 + 7.33 \ln(7) \approx 29.147\]

\[w(3) = 11.4 + 7.33 \ln(3) \approx 21.378\]

Now, calculate the average rate of change:

Average Rate of Change = \(\frac{w(7) - w(3)}{7 - 3} = \frac{29.147 - 21.378}{4} \approx 1.942\)

So, the average rate of change in the weight of the mouse between ages 3 and 7 weeks is approximately 1.942 grams per week.