Find the level curve through the point on the gradient

In summary: When you're talking about level curves, as this problem is, the equation y = 2/x represents the level curve f(x, y) = 0, with f(x, y) = y - 2/x.Here's another way to look at it. Each of your examples except (E) is of the form ##y=f(x)## for some ##f(x)##. It's not good form to use ##f(x,y)## in the same problem you have ##f(x)##, so call ##F(x,y) = y - f(x)##. Then ##\nabla F(x,y) = -f'(x)\
  • #1
Cocoleia
295
4

Homework Statement


upload_2016-12-13_18-25-48.png

Homework Equations

The Attempt at a Solution


The answer is F. I don't how to get this. I know that it is perpendicular and must have a horizontal tangent. How do I come to this answer?[/B]
 
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  • #2
Cocoleia said:

Homework Statement


View attachment 110363

Homework Equations

The Attempt at a Solution


The answer is F. I don't how to get this. I know that it is perpendicular and must have a horizontal tangent. How do I come to this answer?[/B]
Let's look at one choice and see why it isn't an answer.
D. y = 2/x
Let f(x, y) = y - 2/x
Then ##\nabla f = <-2x^{-2}, 1>##, so ##\nabla f(1, 2) = <-2, 1> \ne \vec{j}##
 
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  • #3
Mark44 said:
Let's look at one choice and see why it isn't an answer.
D. y = 2/x
Let f(x, y) = y - 2/x
Then ##\nabla f = <-2x^{-2}, 1>##, so ##\nabla f(1, 2) = <-2, 1> \ne \vec{j}##
I understand, but why is f(x,y) = y- 2/x
 
  • #4
Cocoleia said:
I understand, but why is f(x,y) = y- 2/x
I'm defining that way. With this definition, f(x, y) = 0 is equivalent to y = 2/x.
 
  • #5
Mark44 said:
I'm defining that way. With this definition, f(x, y) = 0 is equivalent to y = 2/x.
Can we always define it as being this way ?
 
  • #6
Cocoleia said:
Can we always define it as being this way ?
When you're talking about level curves, as this problem is, the equation y = 2/x represents the level curve f(x, y) = 0, with f(x, y) = y - 2/x.
 
  • #7
Here's another way to look at it. Each of your examples except (E) is of the form ##y=f(x)## for some ##f(x)##. It's not good form to use ##f(x,y)## in the same problem you have ##f(x)##, so call ##F(x,y) = y - f(x)##. Then ##\nabla F(x,y) = -f'(x)\hat i + 1\hat j## so ##\nabla F(1,2) = -f'(1)\hat i + 1\hat j =0 \hat i + \hat j## means ##f'(1) = 0##. That is quick and easy to check for each ##f(x)## and you will easily will see (F) works.
 

Related to Find the level curve through the point on the gradient

What is a level curve?

A level curve is a curve on a three-dimensional graph that represents the set of points where a function has a constant value. This value is known as the level or contour value.

What is the gradient of a function?

The gradient of a function is a vector that points in the direction of the steepest increase of the function at a given point. It is calculated by taking the partial derivatives of the function with respect to each independent variable.

How do you find the level curve through a given point?

To find the level curve through a given point, you must first calculate the gradient of the function at that point. Then, you can use the gradient vector to find the direction in which the level curve passes through the point. Finally, you can use this information to plot the level curve on the graph.

Why is it important to find the level curve through a point on the gradient?

Finding the level curve through a point on the gradient allows us to visualize the behavior of a function in a specific direction. This can be helpful in understanding the overall shape of a function and identifying any critical points or extrema.

Can the level curve through a point on the gradient intersect itself?

No, the level curve through a point on the gradient cannot intersect itself. This is because a level curve represents a set of points with the same value, and if it were to intersect itself, it would have two different values at the same point, which is not possible for a function.

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