I have no idea if it is just me, but for some reason I was completely unable to solve this problem even after consulting the book and lecture notes. I don't believe this concept was yet covered in class or know if it will be. Please help me escape the land of not knowing.
Thanks,
AND
#12 Dear Professor Taylor,
Similar to the last problem, for some reason I was completely unable to solve this problem as well and for the same reasons. I assume that this problem is related to the last as it asks for some sign in relation to some partial derivative. Again, I would like some help in understanding this concept.
Thanks,
(lets talk about #11 because #12 uses the same thinking)
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OK, so you are supposed to be familiar with these concepts from Calculus 1 :
1) derivatives are approximated by difference quotients,
2) difference quotients are a ratio [f(x_2)-f(x_1)]/[x_2-x_1]
3) you can read the values x_1, f(x_1), x_2, f(x_2) from the graph of the function.
If you are not familiar with or don't remember these concepts you need to pick them up, either reviewing on your own or in the tutoring center or in my office hours.
The new concepts you should have picked up in lecture, where we did indeed talk about them, are that
1) partial derivatives of two variable functions are approximated by directional difference quotients; the x-direction for f_x = ∂f/∂x and the y-direction for f_y = ∂f∂y.
2) that contours of a function in two dimensions, aka the level curves of the function, describe values taken by the function at different places
While you can eyeball the value of the function anywhere in the square 0≤x≤10, 0≤y≤10 from the above contours the easiest place to get the value of the function is right on the contour. So looking at the point (3,5), we see that the contour of 14 passes right through that point, so we know that f(3,5)=14. Now we need to find a point (x,y) that still has y=5 but x≠3, for which preferably it's easy to figure out the value of f(x,5). And sure enough, looking on the line y=5 we see that the contour for 16 passes almost right through the point (6,5)!
Therefore, I can tell you that ∂f/∂x is approximately equal to
Therefore, I can tell you that ∂f/∂x is approximately equal to
[f(6,5)-f(3,5)]/[6-3] = [16-14]/[6-3] = 2/3
By the same token if you had wanted to compute the ∂f/∂y at (3,5) you would have wanted to look for values of f(x,y) on the line x=3 to compute your difference quotient.
