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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Can you please help me with that calculus question? PLEASEEE
![---
**Question 4: Tangent Line to a Parametric Curve**
**Objective:**
Find the equation of the tangent line to the curve at the specified value of the parameter.
**Given:**
\[ x = -3t^2 + 6t + 1 \]
\[ y = 5t^3 - 4t^2 \]
**Specified Parameter Value:**
\[ t = -2 \]
---
To solve this problem, we need to:
1. **Calculate the values of \( x \) and \( y \) at \( t = -2 \).**
2. **Determine the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).**
3. **Evaluate these derivatives at \( t = -2 \) to find the slope of the tangent line.**
4. **Use the point-slope form of the line equation to get the equation of the tangent line.**
---
**Step 1: Compute \( x \) and \( y \) at \( t = -2 \).**
\[ x = -3(-2)^2 + 6(-2) + 1 \]
\[ x = -3(4) - 12 + 1 \]
\[ x = -12 - 12 + 1 \]
\[ x = -23 \]
\[ y = 5(-2)^3 - 4(-2)^2 \]
\[ y = 5(-8) - 4(4) \]
\[ y = -40 - 16 \]
\[ y = -56 \]
So, the coordinates are \( (-23, -56) \).
---
**Step 2: Find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).**
\[ \frac{dx}{dt} = \frac{d(-3t^2 + 6t + 1)}{dt} \]
\[ \frac{dx}{dt} = -6t + 6 \]
\[ \frac{dy}{dt} = \frac{d(5t^3 - 4t^2)}{dt} \]
\[ \frac{dy}{dt} = 15t^2 - 8t \]
---
**Step 3: Evaluate derivatives at \( t = -2 \).**
\[ \frac{dx}{dt} \Big|_{t = -](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fefaa76e1-646a-4849-8550-5f6170986306%2Ff4376763-3a22-4d1f-ad0c-da0724de1ee4%2F4txg4qd6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:---
**Question 4: Tangent Line to a Parametric Curve**
**Objective:**
Find the equation of the tangent line to the curve at the specified value of the parameter.
**Given:**
\[ x = -3t^2 + 6t + 1 \]
\[ y = 5t^3 - 4t^2 \]
**Specified Parameter Value:**
\[ t = -2 \]
---
To solve this problem, we need to:
1. **Calculate the values of \( x \) and \( y \) at \( t = -2 \).**
2. **Determine the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).**
3. **Evaluate these derivatives at \( t = -2 \) to find the slope of the tangent line.**
4. **Use the point-slope form of the line equation to get the equation of the tangent line.**
---
**Step 1: Compute \( x \) and \( y \) at \( t = -2 \).**
\[ x = -3(-2)^2 + 6(-2) + 1 \]
\[ x = -3(4) - 12 + 1 \]
\[ x = -12 - 12 + 1 \]
\[ x = -23 \]
\[ y = 5(-2)^3 - 4(-2)^2 \]
\[ y = 5(-8) - 4(4) \]
\[ y = -40 - 16 \]
\[ y = -56 \]
So, the coordinates are \( (-23, -56) \).
---
**Step 2: Find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).**
\[ \frac{dx}{dt} = \frac{d(-3t^2 + 6t + 1)}{dt} \]
\[ \frac{dx}{dt} = -6t + 6 \]
\[ \frac{dy}{dt} = \frac{d(5t^3 - 4t^2)}{dt} \]
\[ \frac{dy}{dt} = 15t^2 - 8t \]
---
**Step 3: Evaluate derivatives at \( t = -2 \).**
\[ \frac{dx}{dt} \Big|_{t = -
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