a) Find, in terms of a, the equations of the lines tangent to these curves ) Find, in terms of a, the y-intercepts of the tangent lines at r = -1. c) Find the .r-intercepts of the tangent lines at a = -1.
a) Find, in terms of a, the equations of the lines tangent to these curves ) Find, in terms of a, the y-intercepts of the tangent lines at r = -1. c) Find the .r-intercepts of the tangent lines at a = -1.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Part d: Find, in terms of a, the area enclosed by the graph of f(x), the tangent line at x=-1 and the y-axis.
![**Calculus Problem on Tangent Lines and Areas**
**Problem Statement:**
Let \( f(x) = a(7 - x^2) \) for all \( a \neq 0 \).
1. **Find, in terms of \( a \), the equations of the lines tangent to these curves at \( x = -1 \).**
2. **Find, in terms of \( a \), the \( y \)-intercepts of the tangent lines at \( x = -1 \).**
3. **Find the \( x \)-intercepts of the tangent lines at \( x = -1 \).**
4. **Find, in terms of \( a \), the area enclosed by the graph of \( f(x) \), the tangent line at \( x = -1 \), and the \( y \)-axis.**
**Solutions:**
_(a) Find, in terms of \( a \), the equations of the lines tangent to these curves at \( x = -1 \)._
1. **Deriving the tangent lines:**
Since \( a > 0 \), that means \( a \neq 0 \) and these will become our two equations.
2. **Find the slope by finding the derivative:**
\[
f'(x) = (7a - ax^2)' = -2ax
\]
\(*Note: The term \( 7a \) becomes zero since \( a \) is considered to be a constant\)
3. **Plug in \( x = -1 \) to find the slope:**
\[
f'(-1) = -2a(-1) \Rightarrow m = 2a
\]
4. **Find the y-coordinate at \( x = -1 \):**
\[
f(-1) = a(7 - (-1)^2) = 6a
\]
5. **With the coordinates \((-1, 6a)\), use the point-slope form of the equation of a line:**
\[
y - 6a = 2a(x + 1)
\]
Simplifying, we get:
\[
y = 2ax + 8a
\]
_(b) Find, in terms of \( a \), the \( y \)-intercepts of the tangent lines at \( x = -](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa4a8f500-8098-499d-b584-67cf794018c0%2Ffe801b6a-0dfd-4476-9b56-5c15eb5f8207%2Fgl1h2qs9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Calculus Problem on Tangent Lines and Areas**
**Problem Statement:**
Let \( f(x) = a(7 - x^2) \) for all \( a \neq 0 \).
1. **Find, in terms of \( a \), the equations of the lines tangent to these curves at \( x = -1 \).**
2. **Find, in terms of \( a \), the \( y \)-intercepts of the tangent lines at \( x = -1 \).**
3. **Find the \( x \)-intercepts of the tangent lines at \( x = -1 \).**
4. **Find, in terms of \( a \), the area enclosed by the graph of \( f(x) \), the tangent line at \( x = -1 \), and the \( y \)-axis.**
**Solutions:**
_(a) Find, in terms of \( a \), the equations of the lines tangent to these curves at \( x = -1 \)._
1. **Deriving the tangent lines:**
Since \( a > 0 \), that means \( a \neq 0 \) and these will become our two equations.
2. **Find the slope by finding the derivative:**
\[
f'(x) = (7a - ax^2)' = -2ax
\]
\(*Note: The term \( 7a \) becomes zero since \( a \) is considered to be a constant\)
3. **Plug in \( x = -1 \) to find the slope:**
\[
f'(-1) = -2a(-1) \Rightarrow m = 2a
\]
4. **Find the y-coordinate at \( x = -1 \):**
\[
f(-1) = a(7 - (-1)^2) = 6a
\]
5. **With the coordinates \((-1, 6a)\), use the point-slope form of the equation of a line:**
\[
y - 6a = 2a(x + 1)
\]
Simplifying, we get:
\[
y = 2ax + 8a
\]
_(b) Find, in terms of \( a \), the \( y \)-intercepts of the tangent lines at \( x = -
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