The value, in dollars, of a stock t hours into a day of trading is modeled by
V(t) = √1400t+7000. Find the rate at which the stock's value is increasing or
decreasing after 4 hours. (Round your answer to the nearest cent.)
The stock's value is
?
by
?
after 4 hour

To find the rate at which the stock's value is increasing or decreasing after 4 hours, we need to compute the derivative of the function $ V(t) = \sqrt{1400t + 7000} $ with respect to time $ t $. This will give us the rate of change of the stock's value.

### Step 1: Differentiate the function
The function is given as $ V(t) = \sqrt{1400t + 7000} $. Using the chain rule, we can differentiate it:

\[
V'(t) = \frac{d}{dt} \left( \sqrt{1400t + 7000} \right)
\]
The derivative of $ \sqrt{u} $ is $ \frac{1}{2\sqrt{u}} $, where $ u = 1400t + 7000 $, so applying this:

\[
V'(t) = \frac{1}{2\sqrt{1400t + 7000}} \cdot \frac{d}{dt} (1400t + 7000)
\]
Since the derivative of $ 1400t + 7000 $ with respect to $ t $ is $ 1400 $, we have:

\[
V'(t) = \frac{1}{2\sqrt{1400t + 7000}} \cdot 1400
\]
\[
V'(t) = \frac{1400}{2\sqrt{1400t + 7000}} = \frac{700}{\sqrt{1400t + 7000}}
\]

### Step 2: Calculate the rate at $ t = 4 $
Now, substitute $ t = 4 $ into the derivative:

\[
V'(4) = \frac{700}{\sqrt{1400(4) + 7000}} = \frac{700}{\sqrt{5600 + 7000}} = \frac{700}{\sqrt{12600}}
\]

Next, compute $ \sqrt{12600} $:

\[
V'(4) = \frac{700}{112.25} \approx 6.23
\]

Thus, the rate of change of the stock's value after 4 hours is approximately **6.23 dollars per hour**.

### Conclusion:
The stock's value is **increasing** by **$6.23 per hour** after 4 hours.

Would you like more details on any of the steps? Here are 5 related questions you might find interesting:

1. How would the rate change after 6 hours?
2. What is the stock's value at $ t = 4 $ hours?
3. How can we interpret the behavior of the stock based on this model?
4. How would the result change if the model was $ V(t) = \sqrt{1600t + 8000} $?
5. What is the second derivative of $ V(t) $, and what does it tell us about the acceleration of the stock's value?

**Tip:** Always differentiate carefully, paying attention to any chain rule or product rule situations!

The value, in dollars, of a stock t hours into a day of trading is modeled by V(t) = √1400t+7000. Find the rate at which the stock's value is increasing or decreasing after 4 hours. (Round your answer to the nearest cent.) The stock's value is ? by ? after 4 hours.

Learn how to find the rate at which a stock's value is increasing or decreasing after 4 hours using derivatives. This calculus problem involves differentiating the function V(t) = √1400t+7000 to find the rate of change at a specific time. Suitable for students in Grades 11-12.

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