A beautiful ice sculpture is melting, which its volume changing at a rate of V'(t)= 60/(sqrt(t)(sqrt(t)-10)) ft^3/hr, for 0 <t<90.  If after 1 hour the volume of the sculpture is 290ft^3, find its volume after 25 hours. If the sculpture has melted completely, put 0. (round 3 decimal places)


We are given the rate of change of the volume of an ice sculpture, \( V'(t) = \frac{60}{\sqrt{t}(\sqrt{t}-10)} \) ft³/hr, for \( 0 < t < 90 \), and need to determine the volume of the sculpture after 25 hours.

### Step-by-Step Solution:

1. **Initial Setup:**
   - \( V'(t) = \frac{60}{\sqrt{t}(\sqrt{t} - 10)} \)
   - \( V(1) = 290 \) ft³

We need to find \( V(25) \), the volume after 25 hours.

2. **Volume Function:**
   To find the volume, we need to integrate the rate of change \( V'(t) \):
   \[
   V(t) = \int V'(t) \, dt
   \]
   Given \( V(1) = 290 \), we can find the constant of integration after performing the integration.

3. **Integration and Substitution:**
   I'll first solve the integral and then apply the boundary condition \( V(1) = 290 \) to find the volume at \( t = 25 \).

Let's compute it.The volume of the ice sculpture after 25 hours is approximately \( 219.466 \) ft³.

Would you like further details or have any questions? Here are 5 related questions you might find interesting:

1. How does the rate of change of volume affect the melting process over time?
2. What happens to the volume function as \( t \to 90 \)?
3. Can the sculpture's melting be described by a different type of equation?
4. How would the volume change if the initial volume were different?
5. What if the melting rate increased exponentially over time?

**Tip:** Integration is a powerful tool for determining quantities like volume when given a rate of change (derivative).

A beautiful ice sculpture is melting, with its volume changing at a rate of V'(t)= 60/(sqrt(t)(sqrt(t)-10)) ft^3/hr, for 0 < t < 90. If after 1 hour the volume of the sculpture is 290ft^3, find its volume after 25 hours. If the sculpture has melted completely, put 0. (round 3 decimal places)

Discover how to calculate the volume of a melting ice sculpture after 25 hours using the rate of change function V'(t) = 60/sqrt(t)(sqrt(t)-10). Learn step-by-step how integration helps determine volume from a rate of change, with boundary conditions V(1) = 290 ft^3. Suitable for advanced high school or undergraduate calculus students.

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