11.6 problem 20 (updated)

Dear Professor Taylor,

I'm not really sure how to do this. I'm fairly
certain the grad f vector point towards the upper right of the graph at
every point. But I think I just happened to guess correctly to get the
first two answers. Can you explain what how to do this?

Thank you,


At the point P=(1,4) in the direction v=(i+j)/√2, the directional derivative is approximately ______

The first step is to know where you are; in this case at the intersection of the vertical line x=1 and the horizontal line y=4. If you go up from the intersection of the two red lines, the contour changes from about 4.0 to half way between 4.0 and 6.0--i.e. 5.0. while the y-value changes from 4 to 5--hence the rise over the run is 1/1=1. Likewise the x-value changes from 1 to 2.5 while the contour changes from 4.0 and 10.0--the intersection of the red horizontal line with the yellow vertical line--which gives you another rise and a run to compute a difference quotient. For the direction vector v you add the two difference quotients and divide by √2.

Btw, grad(f) points to the upper right at a few points on the graph, but not at most points: toward the top of the graph near the y-axis it points mostly to the right, on the x-axis except near x=0 it points mostly up, on the diagonal line y=x it points to the upper right corner.

Hello Professor Taylor,

I did what you said to do for this problem.
I get
that
df/dx= (10-4)/(3-2)=4i
and
df/dy=(5-4)/(5-4)=1j

so the answer along
the vector v I get
(4i+1j)/sqrt(2) I kept getting 5/sqrt(2) but I changed
it to six/sqrt(2) and that happened to be the right. What am I doing wrong?

you've got a little problem using the dot product correctly, since you kept the i and j after you did the dot product, and they're supposed go away. basically you did this correctly though, the answer 5/√2 is correct and 6/√2 is incorrect.  The problem was in the software and I fixed it, it should work now