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Quick Calculus question

andrey

Diamond Member
Here is the problem:

If contours of f(x,y) with values of f on the contours omitted, and we have 2 points (P and Q)-- which are on these contours and we also know that fx(P) (partial derivative) >0. The goal of this problem is to find the sign of fy(P), f(y)Q, and fx(Q). What would be the best approach here?

Thank you in advance!
 
Originally posted by: silverpig
Is that exactly how the question is written? I don't understand the first part of it.

f (x,y) describes a contour

P = f(x1, y1) some arbitrary x1, y1 values

Q = f(x2, y2) another set of arbitrary x and y values

the partial of f with respect to x at point P is > 0

finding signs fo the partial of f with respect to Y at point P and the partials with respect to x and y and point Q

i dont think there is enough info but there is probably some theoremn that i just forgot

i havent taken calc 3 in 2 years and i already forget

this is calc 3 correct?

 
There isn't enough there. You could choose a certain f(x,y) and a certain P and Q so that the partials you have to find are all zero, or all negative, or all positive.
 
Edit: sorry I misread the problem.

1) Make yourself a graph (y-axis vertical and x-axis horizontal, with the positive values on the top right - this is a standard graph but I'm being specific just in case some bozo comes here complaining about my post).
2) Put the two points on your graph. P will be in the lower right quadrant and Q will be in the upper left quadrant.
3) We know P and Q are on the same contour. Contours are continuous (they are smooth). Thus draw a smooth curve through both points (it could be a straight line, or a squiggly one, just make sure it doesn't intersect itself and it is smooth).
4) You will see your graph is divided into two parts (upper right and lower left).
5) Look at point P. We know that if you move a bit to the right, f will increase. Thus every point in that upper right section has greater values of f. Place a few + signs in that region so you know it has larger values of f.
6) Obviously if you move a bit to the left of P every point is has a lower value for f. Put - signs in that region so you know that it has a smaller value of f.

Now to answer your question about point P.

7a) We know that fy(P)<0 if your contour is sloped like this line: /
7b) We know that fy(P)>0 if your contour is sloped like this line: \
7c) We know that fy(P)=0 if your contour is sloped like this line: -
7d) We know that fy(P)=0 if your contour is sloped like this line: |

Since I can draw contours that have any of the slopes given in part (7), then any answer can occur - we cannot answer the question. However most likely part (7b) will occur.

Do the same for point Q. Most likely (7b) is the answer, but I can draw contour lines that obey any of those - we cannot answer the question.

Do you have a graph of the contours?
 
Originally posted by: dullard
Along a contour, f remains constant.

At point P, we know fx(P)>0. The only way to remain on the contour if f increases in the x direction is if fy(P)<0. These two effects cancel out: f increasing in the x direction and f decreasing in the y direction. Note: this is only true along the contour.

At point Q, we have no information. We cannot answer your question.

We also know that P and Q are on the same contour. Coordinates of P are (2.5,-1) and coordinate of Q are (-0.5, 0.5). Is that sufficient info or is there anyhing else missing?

Thanks again!
 
We also know that P and Q are on the same contour. Coordinates of P are (2.5,-1) and coordinate of Q are (-0.5, 0.5). Is that sufficient info or is there anyhing else missing?

Thanks again!

Edited above, sorry for the mistake.

 
Originally posted by: dullard

Do you have a graph of the contours?

At the place of of P (2.5,-1) and Q (-0.5, 0.5) contours look like parabola in the direction of the positive 'y', which makes me believe that 7a is the answer.

 
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