Physics Exam > Physics Questions > Find the directional derivative of v^(2) at t...
Start Learning for Free
Find the directional derivative of v^(2) at the point (2 0 3) in the direction of the outer normal of the sphere x^(2) y^(2) z^(2)=14 at the point (3 2 1) for v=xy^(2)y zy^(2j xz^(2)k.?
Most Upvoted Answer
Find the directional derivative of v^(2) at the point (2 0 3) in the d...
Step 1: Understanding the Directional Derivative
The directional derivative of a function at a point in a specified direction indicates how the function changes as you move in that direction. It is calculated using the gradient of the function and the unit vector of the direction.
Step 2: Given Function and Point
- The function v = xy^2 + zy^2 + xz^2.
- Evaluate at the point (2, 0, 3).
Step 3: Calculating the Gradient
- The gradient (∇v) is derived by taking partial derivatives:
- ∂v/∂x = y^2 + z^2
- ∂v/∂y = 2xy + 2zy
- ∂v/∂z = y^2 + 2xz
- At the point (2, 0, 3):
- ∇v(2, 0, 3) = (0 + 3^2, 2(2)(0) + 2(3)(0), 0 + 2(2)(3)) = (9, 0, 12).
Step 4: Finding the Outer Normal of the Sphere
- The sphere equation is x^2 + y^2 + z^2 = 14.
- The gradient of the sphere function gives the normal vector:
- ∇g = (2x, 2y, 2z).
- At the point (3, 2, 1):
- ∇g(3, 2, 1) = (6, 4, 2).
Step 5: Normalizing the Normal Vector
- Calculate the magnitude of ∇g:
- ||∇g|| = √(6^2 + 4^2 + 2^2) = √(56) = 2√14.
- The unit normal vector (n):
- n = (6/2√14, 4/2√14, 2/2√14) = (3/√14, 2/√14, 1/√14).
Step 6: Calculating the Directional Derivative
- The directional derivative D_n v at (2, 0, 3) in the direction of n:
- D_n v = ∇v · n.
- Calculate:
- D_n v = (9, 0, 12) · (3/√14, 2/√14, 1/√14) = (27/√14 + 0 + 12/√14) = 39/√14.
Final Result
The directional derivative of v at (2, 0, 3) in the direction of the outer normal of the sphere at (3, 2, 1) is 39/√14.
The directional derivative of a function at a point in a specified direction indicates how the function changes as you move in that direction. It is calculated using the gradient of the function and the unit vector of the direction.
Step 2: Given Function and Point
- The function v = xy^2 + zy^2 + xz^2.
- Evaluate at the point (2, 0, 3).
Step 3: Calculating the Gradient
- The gradient (∇v) is derived by taking partial derivatives:
- ∂v/∂x = y^2 + z^2
- ∂v/∂y = 2xy + 2zy
- ∂v/∂z = y^2 + 2xz
- At the point (2, 0, 3):
- ∇v(2, 0, 3) = (0 + 3^2, 2(2)(0) + 2(3)(0), 0 + 2(2)(3)) = (9, 0, 12).
Step 4: Finding the Outer Normal of the Sphere
- The sphere equation is x^2 + y^2 + z^2 = 14.
- The gradient of the sphere function gives the normal vector:
- ∇g = (2x, 2y, 2z).
- At the point (3, 2, 1):
- ∇g(3, 2, 1) = (6, 4, 2).
Step 5: Normalizing the Normal Vector
- Calculate the magnitude of ∇g:
- ||∇g|| = √(6^2 + 4^2 + 2^2) = √(56) = 2√14.
- The unit normal vector (n):
- n = (6/2√14, 4/2√14, 2/2√14) = (3/√14, 2/√14, 1/√14).
Step 6: Calculating the Directional Derivative
- The directional derivative D_n v at (2, 0, 3) in the direction of n:
- D_n v = ∇v · n.
- Calculate:
- D_n v = (9, 0, 12) · (3/√14, 2/√14, 1/√14) = (27/√14 + 0 + 12/√14) = 39/√14.
Final Result
The directional derivative of v at (2, 0, 3) in the direction of the outer normal of the sphere at (3, 2, 1) is 39/√14.
|
Explore Courses for Physics exam
|
|
Similar Physics Doubts
Find the directional derivative of v^(2) at the point (2 0 3) in the direction of the outer normal of the sphere x^(2) y^(2) z^(2)=14 at the point (3 2 1) for v=xy^(2)y zy^(2j xz^(2)k.?
Question Description
Find the directional derivative of v^(2) at the point (2 0 3) in the direction of the outer normal of the sphere x^(2) y^(2) z^(2)=14 at the point (3 2 1) for v=xy^(2)y zy^(2j xz^(2)k.? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about Find the directional derivative of v^(2) at the point (2 0 3) in the direction of the outer normal of the sphere x^(2) y^(2) z^(2)=14 at the point (3 2 1) for v=xy^(2)y zy^(2j xz^(2)k.? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Find the directional derivative of v^(2) at the point (2 0 3) in the direction of the outer normal of the sphere x^(2) y^(2) z^(2)=14 at the point (3 2 1) for v=xy^(2)y zy^(2j xz^(2)k.?.
Find the directional derivative of v^(2) at the point (2 0 3) in the direction of the outer normal of the sphere x^(2) y^(2) z^(2)=14 at the point (3 2 1) for v=xy^(2)y zy^(2j xz^(2)k.? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about Find the directional derivative of v^(2) at the point (2 0 3) in the direction of the outer normal of the sphere x^(2) y^(2) z^(2)=14 at the point (3 2 1) for v=xy^(2)y zy^(2j xz^(2)k.? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Find the directional derivative of v^(2) at the point (2 0 3) in the direction of the outer normal of the sphere x^(2) y^(2) z^(2)=14 at the point (3 2 1) for v=xy^(2)y zy^(2j xz^(2)k.?.
Solutions for Find the directional derivative of v^(2) at the point (2 0 3) in the direction of the outer normal of the sphere x^(2) y^(2) z^(2)=14 at the point (3 2 1) for v=xy^(2)y zy^(2j xz^(2)k.? in English & in Hindi are available as part of our courses for Physics.
Download more important topics, notes, lectures and mock test series for Physics Exam by signing up for free.
Here you can find the meaning of Find the directional derivative of v^(2) at the point (2 0 3) in the direction of the outer normal of the sphere x^(2) y^(2) z^(2)=14 at the point (3 2 1) for v=xy^(2)y zy^(2j xz^(2)k.? defined & explained in the simplest way possible. Besides giving the explanation of
Find the directional derivative of v^(2) at the point (2 0 3) in the direction of the outer normal of the sphere x^(2) y^(2) z^(2)=14 at the point (3 2 1) for v=xy^(2)y zy^(2j xz^(2)k.?, a detailed solution for Find the directional derivative of v^(2) at the point (2 0 3) in the direction of the outer normal of the sphere x^(2) y^(2) z^(2)=14 at the point (3 2 1) for v=xy^(2)y zy^(2j xz^(2)k.? has been provided alongside types of Find the directional derivative of v^(2) at the point (2 0 3) in the direction of the outer normal of the sphere x^(2) y^(2) z^(2)=14 at the point (3 2 1) for v=xy^(2)y zy^(2j xz^(2)k.? theory, EduRev gives you an
ample number of questions to practice Find the directional derivative of v^(2) at the point (2 0 3) in the direction of the outer normal of the sphere x^(2) y^(2) z^(2)=14 at the point (3 2 1) for v=xy^(2)y zy^(2j xz^(2)k.? tests, examples and also practice Physics tests.
|
|
Explore Courses for Physics exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup