Suppose that the annual sales S (in dollars) of a company may be approximated empirically by the formula S = 20,000*sqrt(e^sqrt t), where t is the number of years beyond some fixed reference date. Use a logarithmic derivative to determine the percentage rate of growth of sales at t=9.

To find the percentage rate of growth of the sales \( S \) at \( t = 9 \), we'll use the logarithmic derivative approach. The sales function is given by:

\[
S(t) = 20,000 \cdot \sqrt{e^{\sqrt{t}}}
\]

### Step 1: Take the natural logarithm of both sides.

Taking the natural logarithm of \( S(t) \), we have:

\[
\ln(S(t)) = \ln\left(20,000 \cdot \sqrt{e^{\sqrt{t}}}\right)
\]

This simplifies to:

\[
\ln(S(t)) = \ln(20,000) + \ln\left(\sqrt{e^{\sqrt{t}}}\right)
\]

Using the logarithm property \( \ln(ab) = \ln(a) + \ln(b) \) and \( \ln(\sqrt{x}) = \frac{1}{2}\ln(x) \), we get:

\[
\ln(S(t)) = \ln(20,000) + \frac{1}{2} \ln(e^{\sqrt{t}})
\]

Since \( \ln(e^{\sqrt{t}}) = \sqrt{t} \):

\[
\ln(S(t)) = \ln(20,000) + \frac{1}{2} \cdot \sqrt{t}
\]

### Step 2: Differentiate both sides with respect to \( t \).

Now, differentiate \( \ln(S(t)) \) with respect to \( t \):

\[
\frac{d}{dt}[\ln(S(t))] = \frac{d}{dt}\left[\ln(20,000) + \frac{1}{2} \sqrt{t}\right]
\]

Since \( \ln(20,000) \) is a constant, its derivative is 0. Thus:

\[
\frac{d}{dt}[\ln(S(t))] = \frac{1}{2} \cdot \frac{d}{dt}[\sqrt{t}]
\]

The derivative of \( \sqrt{t} \) with respect to \( t \) is:

\[
\frac{d}{dt}[\sqrt{t}] = \frac{1}{2\sqrt{t}}
\]

So:

\[
\frac{d}{dt}[\ln(S(t))] = \frac{1}{2} \cdot \frac{1}{2\sqrt{t}} = \frac{1}{4\sqrt{t}}
\]

### Step 3: Evaluate at \( t = 9 \).

Substituting \( t = 9 \) into the derivative:

\[
\frac{d}{dt}[\ln(S(t))] \Big|_{t=9} = \frac{1}{4\sqrt{9}} = \frac{1}{4 \cdot 3} = \frac{1}{12}
\]

### Step 4: Interpret the result.

The logarithmic derivative \( \frac{d}{dt}[\ln(S(t))] \) gives the relative rate of change of \( S(t) \), which is the percentage rate of growth. So, the percentage rate of growth of sales at \( t = 9 \) is:

\[
\frac{1}{12} \times 100\% = \frac{100\%}{12} \approx 8.33\%
\]

Therefore, the percentage rate of growth of sales at \( t = 9 \) is approximately \( 8.33\% \).

---

Would you like more details or have any questions? Here are some related questions:

1. What is the significance of the logarithmic derivative in understanding growth rates?
2. How does the rate of growth change over time for this sales function?
3. What would the percentage rate of growth be at \( t = 16 \)?
4. How do constant factors like 20,000 affect the rate of growth in such functions?
5. What is the difference between logarithmic derivatives and regular derivatives in the context of growth rates?

**Tip:** When dealing with growth rates, logarithmic derivatives are powerful tools as they directly give the relative rate of change or percentage growth rate, simplifying the analysis significantly.

Explore how to determine the percentage rate of growth of a company's sales using logarithmic derivatives, with an example where the sales function is S(t) = 20,000 * sqrt(e^sqrt(t)). Learn step-by-step differentiation techniques and the application of logarithmic properties. Ideal for students studying advanced calculus.

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