For what points p (x, y, z) does gradient of f = 25x 2 + 9y 2 + 16z 2 have the direction from p to origin
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For what points P: (x,y, z) does gradient of F where F = 25x^2 + 9y^2 + 16z^2 have the direction from P to the origin?
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Denis Schüle, smarter in Math than your average bodmas/Pemdas follower
Answered Sep 24 · Author has 151 answers and26.9k answer views
dF/dx=50x
dF/dy=18y
dF/dz=32z
So gradient is
grad(x,y,z)=(50x,18y,32z)
Since we merely care about the direction of it, let’s find
bgrad=1/(sqr((50x)^2+(18y)^2+(32z)^2))*grad(x,y,z)
the vector from P to the origin is
Q=(-x,-y,-z)=-P
for the mere direction of it, let’s find
bQ=1/(sqr(x^2+y^2+z^2))*Q
Now, so we got a vector from P to origin with length 1.
And we got a vector that has same direction as the gradient and has length 1.
So now if gradient and P-origin vector have same direcion, these 2 vectors should be identical, right? since their length is boith 1, we can say that
bgrad=bQ has to be fulfilled.
This will give us a really ugly system of 3 equations and 3 variables x,y,z.
This surely can be solved.
But I ain’t gonna do that.
I’ll leave it up to you :-)
Solve for x,y,z and you got your solution.
There also may have been much more beautiful ways to solve this.
But this is surely the safest way.
Still have a question? Ask your own!
What is your question?
2 ANSWERS

Denis Schüle, smarter in Math than your average bodmas/Pemdas follower
Answered Sep 24 · Author has 151 answers and26.9k answer views
dF/dx=50x
dF/dy=18y
dF/dz=32z
So gradient is
grad(x,y,z)=(50x,18y,32z)
Since we merely care about the direction of it, let’s find
bgrad=1/(sqr((50x)^2+(18y)^2+(32z)^2))*grad(x,y,z)
the vector from P to the origin is
Q=(-x,-y,-z)=-P
for the mere direction of it, let’s find
bQ=1/(sqr(x^2+y^2+z^2))*Q
Now, so we got a vector from P to origin with length 1.
And we got a vector that has same direction as the gradient and has length 1.
So now if gradient and P-origin vector have same direcion, these 2 vectors should be identical, right? since their length is boith 1, we can say that
bgrad=bQ has to be fulfilled.
This will give us a really ugly system of 3 equations and 3 variables x,y,z.
This surely can be solved.
But I ain’t gonna do that.
I’ll leave it up to you :-)
Solve for x,y,z and you got your solution.
There also may have been much more beautiful ways to solve this.
But this is surely the safest way.
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