Complete each related rate problem. a) The edges of a square will grow at the rate of 3cm per minute, starting from length 4cm at time 0. At what rate is the area of the square changing after 3 minutes?
Complete each related rate problem. a) The edges of a square will grow at the rate of 3cm per minute, starting from length 4cm at time 0. At what rate is the area of the square changing after 3 minutes?
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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![**Related Rates Problem:**
**Problem Statement:**
Complete each related rate problem.
a) The edges of a square will grow at the rate of 3 cm per minute, starting from a length of 4 cm at time 0. At what rate is the area of the square changing after 3 minutes?
**Explanation for Students:**
This problem involves understanding how the rate of change in one dimension (the side length of the square) affects the rate of change in a related dimension (the area of the square).
Let's break it down step-by-step:
1. **Determine the initial conditions:**
- Initial side length of the square, \( s = 4 \) cm at time \( t = 0 \).
2. **Growth rate of the side length:**
- Given that the side length is growing at a rate of 3 cm per minute, this is expressed as \( \frac{ds}{dt} = 3 \) cm/min.
3. **Find the side length after 3 minutes:**
- After 3 minutes, the side length will be:
\[
s = 4 \, \text{cm} + (3 \, \text{cm/min}) \times (3 \, \text{min}) = 4 \, \text{cm} + 9 \, \text{cm} = 13 \, \text{cm}
\]
4. **Express the area of the square in terms of its side length:**
- The area \( A \) of a square is given by \( A = s^2 \).
5. **Differentiate the area with respect to time:**
- Using the chain rule, the rate of change of the area with respect to time is:
\[
\frac{dA}{dt} = 2s \cdot \frac{ds}{dt}
\]
6. **Substitute the side length and its rate of change at 3 minutes:**
- At \( t = 3 \) minutes, the side length \( s = 13 \) cm and \( \frac{ds}{dt} = 3 \) cm/min:
\[
\frac{dA}{dt} = 2 \cdot 13 \, \text{cm} \cdot 3 \, \text{cm/min} = 78 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff65fb7b2-ba85-4e67-8fe1-01f97a564d1e%2Fff96843f-ba04-450e-9945-d34e37489006%2Fqurggf_processed.png&w=3840&q=75)
Transcribed Image Text:**Related Rates Problem:**
**Problem Statement:**
Complete each related rate problem.
a) The edges of a square will grow at the rate of 3 cm per minute, starting from a length of 4 cm at time 0. At what rate is the area of the square changing after 3 minutes?
**Explanation for Students:**
This problem involves understanding how the rate of change in one dimension (the side length of the square) affects the rate of change in a related dimension (the area of the square).
Let's break it down step-by-step:
1. **Determine the initial conditions:**
- Initial side length of the square, \( s = 4 \) cm at time \( t = 0 \).
2. **Growth rate of the side length:**
- Given that the side length is growing at a rate of 3 cm per minute, this is expressed as \( \frac{ds}{dt} = 3 \) cm/min.
3. **Find the side length after 3 minutes:**
- After 3 minutes, the side length will be:
\[
s = 4 \, \text{cm} + (3 \, \text{cm/min}) \times (3 \, \text{min}) = 4 \, \text{cm} + 9 \, \text{cm} = 13 \, \text{cm}
\]
4. **Express the area of the square in terms of its side length:**
- The area \( A \) of a square is given by \( A = s^2 \).
5. **Differentiate the area with respect to time:**
- Using the chain rule, the rate of change of the area with respect to time is:
\[
\frac{dA}{dt} = 2s \cdot \frac{ds}{dt}
\]
6. **Substitute the side length and its rate of change at 3 minutes:**
- At \( t = 3 \) minutes, the side length \( s = 13 \) cm and \( \frac{ds}{dt} = 3 \) cm/min:
\[
\frac{dA}{dt} = 2 \cdot 13 \, \text{cm} \cdot 3 \, \text{cm/min} = 78 \
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