Consider the function f(x, y) = (eª — 4x) cos(y). Suppose S is the surface z = f(x, y). (a) Find a vector which is perpendicular to the level curve of f through the point (2, 3) in the direction in which f decreases most rapidly. vector = -(e^2 - 4) cos(3) i - ( e^2 - 12) sin (3) j (b) Suppose 7 = 27 +33 + ak is a vector in 3-space which is tangent to the surface S at the point P lying on the surface above (2, 3). What is a? a = 2(e^2 - 4) cos(3) -3(e^2 - 12)sin (3)
Consider the function f(x, y) = (eª — 4x) cos(y). Suppose S is the surface z = f(x, y). (a) Find a vector which is perpendicular to the level curve of f through the point (2, 3) in the direction in which f decreases most rapidly. vector = -(e^2 - 4) cos(3) i - ( e^2 - 12) sin (3) j (b) Suppose 7 = 27 +33 + ak is a vector in 3-space which is tangent to the surface S at the point P lying on the surface above (2, 3). What is a? a = 2(e^2 - 4) cos(3) -3(e^2 - 12)sin (3)
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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1.11
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![**Consider the function \( f(x, y) = (e^x - 4x) \cos(y) \). Suppose \( S \) is the surface \( z = f(x, y) \).**
### (a) Find a vector which is perpendicular to the level curve of \( f \) through the point \( (2,3) \) in the direction in which \( f \) decreases most rapidly.
**Solution:**
The vector perpendicular to the level curve of \( f \) through the point \( (2,3) \) is given by the gradient of \( f \) evaluated at \( (2,3) \). It is in the direction in which \( f \) decreases most rapidly.
\[
\text{vector} = -(e^2 - 4) \cos(3) \mathbf{i} - ( e^2 - 12) \sin (3) \mathbf{j}
\]
### (b) Suppose \( \vec{v} = 2\mathbf{i} + 3\mathbf{j} + a\mathbf{k} \) is a vector in 3-space which is tangent to the surface \( S \) at the point \( P \) lying on the surface above \( (2,3) \). What is \( a \)?
**Solution:**
We need to find the value of \( a \) such that the vector \( \vec{v} = 2\mathbf{i} + 3\mathbf{j} + a\mathbf{k} \) is tangent to the surface \( S \) at the point above \( (2,3) \).
\[
a = 2( e^2 - 4) \cos(3) - 3( e^2 - 12)\sin (3)
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffe70859f-d1ee-4d19-93f5-d3c21b44393e%2Fb9e18488-84ed-4b3d-aa93-4d4b65e12a69%2Fcjitlva_processed.png&w=3840&q=75)
Transcribed Image Text:**Consider the function \( f(x, y) = (e^x - 4x) \cos(y) \). Suppose \( S \) is the surface \( z = f(x, y) \).**
### (a) Find a vector which is perpendicular to the level curve of \( f \) through the point \( (2,3) \) in the direction in which \( f \) decreases most rapidly.
**Solution:**
The vector perpendicular to the level curve of \( f \) through the point \( (2,3) \) is given by the gradient of \( f \) evaluated at \( (2,3) \). It is in the direction in which \( f \) decreases most rapidly.
\[
\text{vector} = -(e^2 - 4) \cos(3) \mathbf{i} - ( e^2 - 12) \sin (3) \mathbf{j}
\]
### (b) Suppose \( \vec{v} = 2\mathbf{i} + 3\mathbf{j} + a\mathbf{k} \) is a vector in 3-space which is tangent to the surface \( S \) at the point \( P \) lying on the surface above \( (2,3) \). What is \( a \)?
**Solution:**
We need to find the value of \( a \) such that the vector \( \vec{v} = 2\mathbf{i} + 3\mathbf{j} + a\mathbf{k} \) is tangent to the surface \( S \) at the point above \( (2,3) \).
\[
a = 2( e^2 - 4) \cos(3) - 3( e^2 - 12)\sin (3)
\]
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