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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![The problem presented is to find the derivative of \( y \) with respect to \( t \) evaluated at \( t = 3 \), where \( y = (t^2 + 1)(t + 9) \).
### Steps to Solve:
1. **Apply the Product Rule:**
The function \( y \) is the product of two functions: \( u = t^2 + 1 \) and \( v = t + 9 \).
The product rule states that the derivative of a product \( uv \) is given by:
\[
\frac{dy}{dt} = u'v + uv'
\]
2. **Differentiate Each Function:**
- \( u = t^2 + 1 \)
\( u' = \frac{du}{dt} = 2t \)
- \( v = t + 9 \)
\( v' = \frac{dv}{dt} = 1 \)
3. **Substitute and Simplify:**
Substitute \( u \), \( u' \), \( v \), and \( v' \) into the product rule:
\[
\frac{dy}{dt} = (2t)(t + 9) + (t^2 + 1)(1)
\]
\[
\frac{dy}{dt} = 2t(t + 9) + (t^2 + 1)
\]
Simplify:
\[
2t(t + 9) = 2t^2 + 18t
\]
\[
\frac{dy}{dt} = 2t^2 + 18t + t^2 + 1 = 3t^2 + 18t + 1
\]
4. **Evaluate at \( t = 3 \):**
\[
\frac{dy}{dt}\bigg|_{t=3} = 3(3)^2 + 18(3) + 1
\]
\[
= 3(9) + 54 + 1
\]
\[
= 27 + 54 + 1 = 82
\]
The value of \( \frac{dy}{dt} \) at \( t = 3 \) is](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F569db0f8-be64-498b-98dc-01e49b6c2a48%2F7b214301-8d46-4da5-86c4-3fa18b30fde8%2Fz3mvrj9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The problem presented is to find the derivative of \( y \) with respect to \( t \) evaluated at \( t = 3 \), where \( y = (t^2 + 1)(t + 9) \).
### Steps to Solve:
1. **Apply the Product Rule:**
The function \( y \) is the product of two functions: \( u = t^2 + 1 \) and \( v = t + 9 \).
The product rule states that the derivative of a product \( uv \) is given by:
\[
\frac{dy}{dt} = u'v + uv'
\]
2. **Differentiate Each Function:**
- \( u = t^2 + 1 \)
\( u' = \frac{du}{dt} = 2t \)
- \( v = t + 9 \)
\( v' = \frac{dv}{dt} = 1 \)
3. **Substitute and Simplify:**
Substitute \( u \), \( u' \), \( v \), and \( v' \) into the product rule:
\[
\frac{dy}{dt} = (2t)(t + 9) + (t^2 + 1)(1)
\]
\[
\frac{dy}{dt} = 2t(t + 9) + (t^2 + 1)
\]
Simplify:
\[
2t(t + 9) = 2t^2 + 18t
\]
\[
\frac{dy}{dt} = 2t^2 + 18t + t^2 + 1 = 3t^2 + 18t + 1
\]
4. **Evaluate at \( t = 3 \):**
\[
\frac{dy}{dt}\bigg|_{t=3} = 3(3)^2 + 18(3) + 1
\]
\[
= 3(9) + 54 + 1
\]
\[
= 27 + 54 + 1 = 82
\]
The value of \( \frac{dy}{dt} \) at \( t = 3 \) is
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