### Topic: Calculus - Tangent Planes **Find an equation of the tangent plane to the surface** Given the surface equation: \[ x^2 + 8y^2 + z^2 = 468 \] at the point \( (6, -6, 12) \). **Exercise Progression:** Here are five potential forms of the equation for the tangent plane: A. \( 12(x - 6) - 96(y + 6) + 12(z - 12) = 0 \) B. \( 12(x - 6) - 96(y + 6) + 24(z - 12) = 0 \) C. \( 12(x - 6) - 96(y + 6) - 24(z - 12) = 0 \) D. \( 12(x - 6) - 48(y + 6) + 24(z - 12) = 0 \) E. \( 12(x - 6) + 96(y + 6) + 24(z - 12) = 0 \) **Approach to the Problem:** To solve for the correct tangent plane equation, apply the concept of gradients to find the normal vector. Calculate partial derivatives of the surface equation with respect to \(x\), \(y\), and \(z\) to form these gradients at the given point. Select the equation where the computed normal vector aligns with the coefficients in the standard form of the tangent plane: \[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \] where \((x_0, y_0, z_0)\) is the point of tangency, and \(a\), \(b\), and \(c\) are components of the normal vector.
### Topic: Calculus - Tangent Planes **Find an equation of the tangent plane to the surface** Given the surface equation: \[ x^2 + 8y^2 + z^2 = 468 \] at the point \( (6, -6, 12) \). **Exercise Progression:** Here are five potential forms of the equation for the tangent plane: A. \( 12(x - 6) - 96(y + 6) + 12(z - 12) = 0 \) B. \( 12(x - 6) - 96(y + 6) + 24(z - 12) = 0 \) C. \( 12(x - 6) - 96(y + 6) - 24(z - 12) = 0 \) D. \( 12(x - 6) - 48(y + 6) + 24(z - 12) = 0 \) E. \( 12(x - 6) + 96(y + 6) + 24(z - 12) = 0 \) **Approach to the Problem:** To solve for the correct tangent plane equation, apply the concept of gradients to find the normal vector. Calculate partial derivatives of the surface equation with respect to \(x\), \(y\), and \(z\) to form these gradients at the given point. Select the equation where the computed normal vector aligns with the coefficients in the standard form of the tangent plane: \[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \] where \((x_0, y_0, z_0)\) is the point of tangency, and \(a\), \(b\), and \(c\) are components of the normal vector.
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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![### Topic: Calculus - Tangent Planes
**Find an equation of the tangent plane to the surface**
Given the surface equation:
\[ x^2 + 8y^2 + z^2 = 468 \]
at the point \( (6, -6, 12) \).
**Exercise Progression:**
Here are five potential forms of the equation for the tangent plane:
A. \( 12(x - 6) - 96(y + 6) + 12(z - 12) = 0 \)
B. \( 12(x - 6) - 96(y + 6) + 24(z - 12) = 0 \)
C. \( 12(x - 6) - 96(y + 6) - 24(z - 12) = 0 \)
D. \( 12(x - 6) - 48(y + 6) + 24(z - 12) = 0 \)
E. \( 12(x - 6) + 96(y + 6) + 24(z - 12) = 0 \)
**Approach to the Problem:**
To solve for the correct tangent plane equation, apply the concept of gradients to find the normal vector. Calculate partial derivatives of the surface equation with respect to \(x\), \(y\), and \(z\) to form these gradients at the given point. Select the equation where the computed normal vector aligns with the coefficients in the standard form of the tangent plane:
\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]
where \((x_0, y_0, z_0)\) is the point of tangency, and \(a\), \(b\), and \(c\) are components of the normal vector.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7dbb4ae4-0d65-4baa-9481-63f79be91eca%2F053eaa28-18cc-4bfb-a4fb-8845b6643465%2F26qr6vh.jpeg&w=3840&q=75)
Transcribed Image Text:### Topic: Calculus - Tangent Planes
**Find an equation of the tangent plane to the surface**
Given the surface equation:
\[ x^2 + 8y^2 + z^2 = 468 \]
at the point \( (6, -6, 12) \).
**Exercise Progression:**
Here are five potential forms of the equation for the tangent plane:
A. \( 12(x - 6) - 96(y + 6) + 12(z - 12) = 0 \)
B. \( 12(x - 6) - 96(y + 6) + 24(z - 12) = 0 \)
C. \( 12(x - 6) - 96(y + 6) - 24(z - 12) = 0 \)
D. \( 12(x - 6) - 48(y + 6) + 24(z - 12) = 0 \)
E. \( 12(x - 6) + 96(y + 6) + 24(z - 12) = 0 \)
**Approach to the Problem:**
To solve for the correct tangent plane equation, apply the concept of gradients to find the normal vector. Calculate partial derivatives of the surface equation with respect to \(x\), \(y\), and \(z\) to form these gradients at the given point. Select the equation where the computed normal vector aligns with the coefficients in the standard form of the tangent plane:
\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]
where \((x_0, y_0, z_0)\) is the point of tangency, and \(a\), \(b\), and \(c\) are components of the normal vector.
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