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
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Author:James Stewart
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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 =
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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