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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Question
![**Problem Statement:**
Find the equation of the tangent line to \( y = e^{-6t} \) at \( t = 0 \).
**Solution Approach:**
To solve this problem, follow these steps:
1. **Find the derivative \( \frac{dy}{dt} \) of \( y = e^{-6t} \):**
- The derivative of \( y = e^{-6t} \) with respect to \( t \) is \( \frac{dy}{dt} = -6e^{-6t} \).
2. **Evaluate the derivative at \( t = 0 \) to find the slope of the tangent line:**
- Substituting \( t = 0 \) into the derivative gives \( \frac{dy}{dt}\bigg|_{t=0} = -6e^{0} = -6 \).
3. **Find the point on the curve at \( t = 0 \):**
- Substituting \( t = 0 \) into the original function \( y = e^{-6t} \) gives \( y = e^{0} = 1 \).
- So, the point is \( (0, 1) \).
4. **Use the point-slope form of the equation of a line:**
- The point-slope form is \( y - y_1 = m(t - t_1) \), where \( m \) is the slope and \( (t_1, y_1) \) is the point on the line.
- Substitute \( m = -6 \), \( t_1 = 0 \), and \( y_1 = 1 \) to get:
\[
y - 1 = -6(t - 0)
\]
\[
y = -6t + 1
\]
**Conclusion:**
The equation of the tangent line to \( y = e^{-6t} \) at \( t = 0 \) is \( y = -6t + 1 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff1ac4a5c-5c91-4743-9545-fc03c7695e62%2Fcc7ccb54-2996-4bdf-b78e-988972307f1b%2Fn1usatm_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the equation of the tangent line to \( y = e^{-6t} \) at \( t = 0 \).
**Solution Approach:**
To solve this problem, follow these steps:
1. **Find the derivative \( \frac{dy}{dt} \) of \( y = e^{-6t} \):**
- The derivative of \( y = e^{-6t} \) with respect to \( t \) is \( \frac{dy}{dt} = -6e^{-6t} \).
2. **Evaluate the derivative at \( t = 0 \) to find the slope of the tangent line:**
- Substituting \( t = 0 \) into the derivative gives \( \frac{dy}{dt}\bigg|_{t=0} = -6e^{0} = -6 \).
3. **Find the point on the curve at \( t = 0 \):**
- Substituting \( t = 0 \) into the original function \( y = e^{-6t} \) gives \( y = e^{0} = 1 \).
- So, the point is \( (0, 1) \).
4. **Use the point-slope form of the equation of a line:**
- The point-slope form is \( y - y_1 = m(t - t_1) \), where \( m \) is the slope and \( (t_1, y_1) \) is the point on the line.
- Substitute \( m = -6 \), \( t_1 = 0 \), and \( y_1 = 1 \) to get:
\[
y - 1 = -6(t - 0)
\]
\[
y = -6t + 1
\]
**Conclusion:**
The equation of the tangent line to \( y = e^{-6t} \) at \( t = 0 \) is \( y = -6t + 1 \).
Expert Solution
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