Your friends correctly calculate the gradient vector for f(x, y)=x²+y at the point (2,-3) as follows: 1. V{(2, – 3) = (2x, 2y ) =(4,-6) (3,-4) They say that (4,-6) is orthogonal to the surface at the point where x 2 and y=-3 (at the point (2,-3, 13)). Unfortunately, they are incorrect, and you will help them. (a) For f(x, y)=x²+y, what is (4,-6) orthogonal to when x=2 and y=-3? (Drawing a picture may help)
Your friends correctly calculate the gradient vector for f(x, y)=x²+y at the point (2,-3) as follows: 1. V{(2, – 3) = (2x, 2y ) =(4,-6) (3,-4) They say that (4,-6) is orthogonal to the surface at the point where x 2 and y=-3 (at the point (2,-3, 13)). Unfortunately, they are incorrect, and you will help them. (a) For f(x, y)=x²+y, what is (4,-6) orthogonal to when x=2 and y=-3? (Drawing a picture may help)
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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![Your friends correctly calculate the gradient vector for \( f(x, y) = x^2 + y^2 \) at the point \((2, -3)\) as follows:
\[
\nabla f(2, -3) = \langle 2x, 2y \rangle \bigg|_{(3, -4)} = \langle 4, -6 \rangle
\]
They say that \(\langle 4, -6 \rangle\) is orthogonal to the surface at the point where \( x = 2 \) and \( y = -3 \) (at the point \( (2, -3, 13) \)).
Unfortunately, they are incorrect, and you will help them.
(a) For \( f(x, y) = x^2 + y^2 \), what is \(\langle 4, -6 \rangle\) orthogonal to when \( x = 2 \) and \( y = -3 \)? (Drawing a picture may help)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6501bcec-69f0-41c7-aaf6-9d4563ecc7cd%2F16629576-366f-48fc-91ec-6b2f8a167daf%2Frebjov8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Your friends correctly calculate the gradient vector for \( f(x, y) = x^2 + y^2 \) at the point \((2, -3)\) as follows:
\[
\nabla f(2, -3) = \langle 2x, 2y \rangle \bigg|_{(3, -4)} = \langle 4, -6 \rangle
\]
They say that \(\langle 4, -6 \rangle\) is orthogonal to the surface at the point where \( x = 2 \) and \( y = -3 \) (at the point \( (2, -3, 13) \)).
Unfortunately, they are incorrect, and you will help them.
(a) For \( f(x, y) = x^2 + y^2 \), what is \(\langle 4, -6 \rangle\) orthogonal to when \( x = 2 \) and \( y = -3 \)? (Drawing a picture may help)

Transcribed Image Text:(b) Find a vector orthogonal to the surface at the point where \( x = 2 \) and \( y = -3 \). Show/explain your work please.
Hint: Define \( F(x, y, z) \).
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