The rate at which ice is melting in a pond is given by R (t) = V1+ 2', where Ris measured in cubic feet per minute, and t is the time in minutes. What amount of ice has melted in the first 5 minutes? O 1.931 ft3 O 12.970 ft3 O 14.530 ft3 5.745 ft3
The rate at which ice is melting in a pond is given by R (t) = V1+ 2', where Ris measured in cubic feet per minute, and t is the time in minutes. What amount of ice has melted in the first 5 minutes? O 1.931 ft3 O 12.970 ft3 O 14.530 ft3 5.745 ft3
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Understanding the Rate at Which Ice Melts in a Pond**
**Problem Statement:**
The rate at which ice is melting in a pond is given by \( R(t) = \sqrt{1 + 2t} \), where \( R \) is measured in cubic feet per minute, and \( t \) is the time in minutes. What amount of ice has melted in the first 5 minutes?
**Options:**
- O 1.931 ft³
- O 12.970 ft³
- O 14.530 ft³
- O 5.745 ft³
**Solution:**
To determine the total amount of ice that has melted in the first 5 minutes, we need to integrate the rate function, \( R(t) \), from \( t = 0 \) to \( t = 5 \). This can be done using definite integration.
The integral we need to evaluate is:
\[ \int_{0}^{5} \sqrt{1 + 2t} \, dt \]
This integral represents the total amount of ice melted within the given time period. Once we solve this integral, we can compare the answer with the given options to identify the correct one.
1. Set up the integral:
\[ \int_{0}^{5} \sqrt{1 + 2t} \, dt \]
2. Perform the integration:
First, we'll use a substitution to simplify the integral. Let:
\[ u = 1 + 2t \implies du = 2 \, dt \]
\[ dt = \frac{du}{2} \]
Adjust the limits of integration for \( u \):
- When \( t = 0 \), \( u = 1 \)
- When \( t = 5 \), \( u = 11 \)
The integrand and differential become:
\[ \int_{1}^{11} \sqrt{u} \cdot \frac{1}{2} \, du = \frac{1}{2} \int_{1}^{11} \sqrt{u} \, du \]
The integral of \( \sqrt{u} \) is:
\[ \int \sqrt{u} \, du = \int u^{1/2} \, du = \frac{2}{3} u^{3/2} \]
Evaluate from \( u = 1 \) to \( u = 11 \):
\[](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4143f45a-5c96-4449-af2f-abda697f3e91%2F14967d97-3ede-4562-ab37-942ccbca3687%2Ffnz5mnl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Understanding the Rate at Which Ice Melts in a Pond**
**Problem Statement:**
The rate at which ice is melting in a pond is given by \( R(t) = \sqrt{1 + 2t} \), where \( R \) is measured in cubic feet per minute, and \( t \) is the time in minutes. What amount of ice has melted in the first 5 minutes?
**Options:**
- O 1.931 ft³
- O 12.970 ft³
- O 14.530 ft³
- O 5.745 ft³
**Solution:**
To determine the total amount of ice that has melted in the first 5 minutes, we need to integrate the rate function, \( R(t) \), from \( t = 0 \) to \( t = 5 \). This can be done using definite integration.
The integral we need to evaluate is:
\[ \int_{0}^{5} \sqrt{1 + 2t} \, dt \]
This integral represents the total amount of ice melted within the given time period. Once we solve this integral, we can compare the answer with the given options to identify the correct one.
1. Set up the integral:
\[ \int_{0}^{5} \sqrt{1 + 2t} \, dt \]
2. Perform the integration:
First, we'll use a substitution to simplify the integral. Let:
\[ u = 1 + 2t \implies du = 2 \, dt \]
\[ dt = \frac{du}{2} \]
Adjust the limits of integration for \( u \):
- When \( t = 0 \), \( u = 1 \)
- When \( t = 5 \), \( u = 11 \)
The integrand and differential become:
\[ \int_{1}^{11} \sqrt{u} \cdot \frac{1}{2} \, du = \frac{1}{2} \int_{1}^{11} \sqrt{u} \, du \]
The integral of \( \sqrt{u} \) is:
\[ \int \sqrt{u} \, du = \int u^{1/2} \, du = \frac{2}{3} u^{3/2} \]
Evaluate from \( u = 1 \) to \( u = 11 \):
\[
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