Cal 3 Problem:
\[f(x,y)=\sin(xy ^{2}), P=(\frac{ \pi }{ 4 }, 2) southeast\] find the directional derivative of f at P in the given direction
would I have to square pi/4 and 2 then take its um under a square root...put that under a 1 and multiply it by the given P to find what u equals?
*not um, I meant sum
I wonder whether your problem is lack of the direction of vector u? , without it, how can you take directional derivative? or south east is that direction?
yes SE is the direction, I'm having trouble cus idk if i use P to find the unit vector. if i had u it would be wayyy easier man. stupid prof assigning crazy problems
i think if south east that means u =<1,1,0>
no, no, since it has 2 variables , so SE is <1,-1> only
yeah, put the Cartesian coordinate system, south east is at quadrant 4, 45degree from x axis.
its 2 dimensional. I learned in Statics the unit vector is\[u=\frac{ x _{r}+y _{r} }{ \left| \sqrt{x _{r }^{2}+y _{r}^{2}} \right| }\]. but idk if this applies
i think you're right now that i think about it.
my professor told me that, the directional derivative get max when u // gradient and the value of it is | gradient|. do you have answer from your book? if so, try 2 ways, take gradient , replace the point and then take absolute value of it to get the answer. Or, take as if u is <1,-1> . good luck
i'm gonna evaluate it using <1,-1>. sounds correct and looking at some examples it seems correct. SE in 2-D is what you said, think i tried to figure it so bad i made it more difficult than it really is
did you get the answer to your problem you posted?
what is mine? I posted long time ago, no one reply. so I took it off
i replied. at least i think that was you. did you post trying to find the partial of G(s,t)
no. i'm not. another one. I saw you replied, and let you help him. we are in the same boat, help each other is studying again. but i don't think that is the problem to him or you
just because of sometimes, we are too tired when struggling with the problem and temporarily being blank
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