Gradients and level curves, getting started with my homework

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Homework Statement


f(x,y)=y-x
i) compute the gradient
ii) at the given point (2,1) compute the gradient vector gradf(2,1)
iii) compute the level curve that passes through the point (2,1)
iiii) on the xy-plane, sketch the level curve from iii and the gradient vector gradf(2,1) you found in ii. Verify that they meet orthogonally


Homework Equations


gradients, derivatives, etc.


The Attempt at a Solution



I was out sick all last week from class so I need a little leg up on what we are doing...

for i) wouldn't this just be <-1,1> ?
and if so, I'm a little confused on ii) which I assumed to be <-1,1> still regardless of what point it was 'evaluated' at.

And then from here I'm not sure how to compute the level curve for this. Am I looking for a vector, equation, scalar? Before this when we talked about level curves in class, we were always given a set of constants to set them equal to.

And as far as iiii) goes, I'm a bit confused what this is supposed to look like since ii) would always be the vector <-1,1>
 
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sheepcountme said:

Homework Statement


f(x,y)=y-x
i) compute the gradient
ii) at the given point (2,1) compute the gradient vector gradf(2,1)
iii) compute the level curve that passes through the point (2,1)
iiii) on the xy-plane, sketch the level curve from iii and the gradient vector gradf(2,1) you found in ii. Verify that they meet orthogonally


Homework Equations


gradients, derivatives, etc.


The Attempt at a Solution



I was out sick all last week from class so I need a little leg up on what we are doing...

for i) wouldn't this just be <-1,1> ?
and if so, I'm a little confused on ii) which I assumed to be <-1,1> still regardless of what point it was 'evaluated' at.

You are correct. In this example the gradient field is constant

And then from here I'm not sure how to compute the level curve for this. Am I looking for a vector, equation, scalar? Before this when we talked about level curves in class, we were always given a set of constants to set them equal to.

The level curves are f(x,y) = C for various constants C. But you are being asked for just one of them. Figure out C for your point and what equation do you get for your single level curve?

And as far as iiii) goes, I'm a bit confused what this is supposed to look like since ii) would always be the vector <-1,1>

So show the vector is perpendicular to the level curve in ii.
 
Okay, so for iii, setting y-x=c c is -1 when we plug in the given point which gives us the equation y=x-1.

but for part 4, y=x-1 is pretty much the line y=x but shifted 1 down and to the right, and the line I get for part 2 is a straight horizontal line across the plane, which isn't perpedicular to the level curve.
 
You don't get a line for part 2, you get a vector. Prove that the vector -1i+ 1j is perpendicular to the line y= x- 1. (One point on the line y= x- 1 is (0, -1). Another is (2, 1). What is the vector from (0, -1), to (2, 1)? It points in the same direction as the line.)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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