A point is moving along the graph of a given function such that dx/dt is 3 centimeters per second. Find dy/dt for the given values of x. y = 3x2 + 2 (a) x = -1 dy/dt = cm/sec (b) x = 0 dy/dt = | cm/sec (c) x = 2 dy/dt = | cm/sec
A point is moving along the graph of a given function such that dx/dt is 3 centimeters per second. Find dy/dt for the given values of x. y = 3x2 + 2 (a) x = -1 dy/dt = cm/sec (b) x = 0 dy/dt = | cm/sec (c) x = 2 dy/dt = | cm/sec
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
A point is moving along the graph of a given function such that \( \frac{dx}{dt} \) is 3 centimeters per second. Find \( \frac{dy}{dt} \) for the given values of \( x \).
The function given is:
\[ y = 3x^2 + 2 \]
For the following values of \( x \):
**(a) \( x = -1 \)**
\[ \frac{dy}{dt} = \_\_\_\_\_\_\_\_ \text{ cm/sec} \]
**(b) \( x = 0 \)**
\[ \frac{dy}{dt} = \_\_\_\_\_\_\_\_ \text{ cm/sec} \]
**(c) \( x = 2 \)**
\[ \frac{dy}{dt} = \_\_\_\_\_\_\_\_ \text{ cm/sec} \]
#### Step-by-Step Explanation
1. **Differentiate the Given Function:**
To find \( \frac{dy}{dt} \), we need to apply the chain rule to differentiate \( y = 3x^2 + 2 \) with respect to \( t \):
\[ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \]
2. **Find \( \frac{dy}{dx} \):**
\[ \frac{dy}{dx} = \frac{d}{dx}(3x^2 + 2) = 6x \]
3. **Calculate \( \frac{dy}{dt} \) Using \( \frac{dy}{dx} \) and \( \frac{dx}{dt} \):**
Since we know \( \frac{dx}{dt} = 3 \) cm/sec, we can substitute this and the value of \( \frac{dy}{dx} \):
\[ \frac{dy}{dt} = 6x \cdot 3 \]
4. **Apply the Specific Values of \( x \):**
**(a) When \( x = -1 \)**
\[ \frac{dy}{dt} = 6(-1) \cdot 3 = -18 \text{ cm/sec} \]
**(b) When \( x =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F08c083ee-5dc4-4392-bfaa-a5f4117ce6af%2F0315df1d-a0f5-4fea-9d2f-ed0e0f8d0fac%2F4uefyxn.png&w=3840&q=75)
Transcribed Image Text:### Related Rates Problem
#### Problem Statement
A point is moving along the graph of a given function such that \( \frac{dx}{dt} \) is 3 centimeters per second. Find \( \frac{dy}{dt} \) for the given values of \( x \).
The function given is:
\[ y = 3x^2 + 2 \]
For the following values of \( x \):
**(a) \( x = -1 \)**
\[ \frac{dy}{dt} = \_\_\_\_\_\_\_\_ \text{ cm/sec} \]
**(b) \( x = 0 \)**
\[ \frac{dy}{dt} = \_\_\_\_\_\_\_\_ \text{ cm/sec} \]
**(c) \( x = 2 \)**
\[ \frac{dy}{dt} = \_\_\_\_\_\_\_\_ \text{ cm/sec} \]
#### Step-by-Step Explanation
1. **Differentiate the Given Function:**
To find \( \frac{dy}{dt} \), we need to apply the chain rule to differentiate \( y = 3x^2 + 2 \) with respect to \( t \):
\[ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \]
2. **Find \( \frac{dy}{dx} \):**
\[ \frac{dy}{dx} = \frac{d}{dx}(3x^2 + 2) = 6x \]
3. **Calculate \( \frac{dy}{dt} \) Using \( \frac{dy}{dx} \) and \( \frac{dx}{dt} \):**
Since we know \( \frac{dx}{dt} = 3 \) cm/sec, we can substitute this and the value of \( \frac{dy}{dx} \):
\[ \frac{dy}{dt} = 6x \cdot 3 \]
4. **Apply the Specific Values of \( x \):**
**(a) When \( x = -1 \)**
\[ \frac{dy}{dt} = 6(-1) \cdot 3 = -18 \text{ cm/sec} \]
**(b) When \( x =
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