# Find the level curve through the point on the gradient

1. Dec 13, 2016

### Cocoleia

1. The problem statement, all variables and given/known data

2. Relevant equations

3. 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?

2. Dec 13, 2016

### Staff: Mentor

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

3. Dec 13, 2016

### Cocoleia

I understand, but why is f(x,y) = y- 2/x

4. Dec 13, 2016

### Staff: Mentor

I'm defining that way. With this definition, f(x, y) = 0 is equivalent to y = 2/x.

5. Dec 13, 2016

### Cocoleia

Can we always define it as being this way ?

6. Dec 13, 2016

### Staff: Mentor

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. Dec 13, 2016

### 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.