Calculating Vector Length Using Contour Lines

[2RIP]

From this diagram, http://www34.homepage.villanova.edu/robert.jantzen/courses/mat2500/handouts/s14-6-38.pdf how can i find the length of the vector at point (4,6)

Answer is length = (-2-(-3))/0.5 =2. I don't understand why we use -3 and -2 from the contour lines. The gradient vector at (4,6) does not go to -3. So why do we input -3 to find length? Or do we use the closest level lines to find the approximate rate of change?

Thanks

 

[Prologue]

The gradient vector is found by changing the constant that the implicit function is equal to.

So for demonstrative purposes say that your implicit function looks like this (it isn't actually but just assume it is for now):

-2.5 = x - y - 0.5 = (4) - (6) - 0.5

So the constant at that point (4,6) is -2.5. Now, if you were to approximate what the value of the gradient would be, what would you do? Look at the definition of the magnitude gradient again. The magnitude of the gradient is found by looking at how much the magnitude of the constant changes compared to how much the magnitude of the inputs changed.

To estimate the magnitude of the gradient change at that spot, find a contour line close to it then jump over that spot (4,6) to the other contour line. In this way, you are looking at a change of 1 in the constant, and it is around the area of the point (estimations are ok at this point, these are not 'exact' gradients). You will always pick the contour line largest in value as your final measuring spot and the contour of lowest value is your starting point. This is because when finding the gradient we ALWAYS move in a positive direction, we change the constant to a MORE POSITIVE value. So initial -3, final -2. Now again, look at the contour lines that you just got the value of the constant from. Now you are going to measure the shortest physical distance between those lines and measured directly over the point (4,6) because this gives you the smallest change in the magnitude of the inputs to achieve the change in the constant, it gives you the change in sqrt(x+y). In this case it is about 0.5, that is your denominator.


So you use the definition of the magnitude of the gradient again to assemble this data:

(positive change in constant)/(distance it took to make that change in the constant)

or

(-2 - (-3))/(0.5)


Then draw the gradient vector starting at the point you evaluated the gradient at (4,6) and point it in the direction of a postive change in your constant. And now make it have a length/magnitude of 2 because this is what you found the magnitude of the gradient at the point (4,6) to be.


Good luck, this is a very difficult concept. Keep at trying to understand it in different ways. It always helped me to remember that it is the change in the CONSTANT that is the key. Everything about the gradient is rigged to make the change in the constant positive, and then the direction that achieves the largest change in the constant, given a certain movement in the input values, is the direction that the gradient is defined to point.