Find the level curve through the point on the gradient

[Cocoleia]

Homework Statement

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]

 

[Mark44]

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 $\nabla f(1, 2) = <-2, 1> \ne \vec{j}$

 

[Cocoleia]

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

 

[Mark44]

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.

 

[Cocoleia]

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 ?

 

[Mark44]

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.

 

[LCKurtz]

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.