Nikki C asked in Science & MathematicsMathematics · 10 years ago

# how to calculate curl((r_)/r^3)?

the r_ inside the braket means vector r,I dont know how to write it proper.

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• kb
Lv 7
10 years ago

r = <x, y, z>, ||r|| = (x^2 + y^2 + z^2)^(-1/2)

So, we need to compute

curl <x(x^2 + y^2 + z^2)^(-1/2), y(x^2 + y^2 + z^2)^(-1/2), z(x^2 + y^2 + z^2)^(-1/2)>

= < -yz(x^2 + y^2 + z^2)^(-3/2) + -yz(x^2 + y^2 + z^2)^(-3/2),

-zx(x^2 + y^2 + z^2)^(-3/2) + zx(x^2 + y^2 + z^2)^(-3/2),

-xy(x^2 + y^2 + z^2)^(-3/2) + xy(x^2 + y^2 + z^2)^(-3/2)>

= <0, 0, 0>.

I hope this helps!

• 5 years ago

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• Anonymous
10 years ago

I got the same except for the II r II is II r II^3 so wouldnt it be (x^2 + y^2 + z^2)^-3/2 to start with and then (x^2 + y^2 + z^2)^-5/2 when differentiated?? The answer comes out to the same <0,0,0> but working is slightly different? Is this right? :)

Source(s): my brain
• 10 years ago

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