prove-that-curl-r-n-c-r-n-2-r-n-c-nr-n-2-r-c-where-c-is-the-constant-vector-

Question Number 85865 by subhankar10 last updated on 25/Mar/20

prove that

curl (r^{n} \vec{c} \times \vec{r}) = (n + 2) r^{n} \vec{c} - {nr}^{n - 2} (\vec{r} . \vec{c}) .

where c is the constant vector .

Answered by TANMAY PANACEA. last updated on 25/Mar/20

\vec{r} = i x + j y + k z \to r^{2} = x^{2} + y^{2} + z^{2}

r^{n} = {(x^{2} + y^{2} + z^{2})}^{\frac{n}{2}}

\vec{c} = i a + i b + k d

c u r l \vec{A} = \vec{▽} \times \vec{A}

(i \frac{\partial}{\partial x} + j \frac{\partial}{\partial y} + k \frac{\partial}{\partial z}) \times \vec{A}

n o w

\vec{▽} \times (\vec{c} \times r^{n} \vec{r})

= (\vec{▽} . r^{n} \vec{r}) \vec{c} - (\vec{▽} . \vec{c}) r^{n} \vec{r}

= \vec{c} (i \frac{\partial}{\partial x} + j \frac{\partial}{\partial y} + k \frac{\partial}{\partial z}) . {(x^{2} + y^{2} + z^{2})}^{\frac{n}{2}} (i x + j y + k z)

= \vec{c} [\frac{\partial}{\partial x} {x . {(x^{2} + y^{2} + z^{2})}^{\frac{n}{2}}} + \frac{\partial}{\partial y} {y (x^{2} + y^{2} + z^{2})} + \frac{\partial}{\partial z} {z (x^{2} + y^{2} + z^{2})}]

c a l c u l a t i o n o f

\frac{\partial}{\partial x} {x {(x^{2} + y^{2} + z^{2})}^{\frac{n}{2}}}

= {(x^{2} + y^{2} + z^{2})}^{\frac{n}{2}} \times 1 + x \times \frac{n}{2} {(x^{2} + y^{2} + z^{2})}^{\frac{n}{2} - 1} \times 2 x

= r^{n} + x^{2} \times n {(r^{2})}^{\frac{n}{2} - 1}

= r^{n} + x^{2} \times n \times r^{n - 2}

a d d i n g t h r e e

3 r^{n} + {n r}^{n - 2} (x^{2} + y^{2} + z^{2})

= 3 r^{n} + {n r}^{n} = r^{n} (n + 3)

s o a n s w e r i s r^{n} (n + 3) \vec{c}

t i m o r r o w i s h a l l s o l v e i n p a p e r

. = . .

\vec{c} \times \vec{r}

∣ i j k ∣

∣ a b d ∣

∣ x y z ∣

= i (b z - y d) - j (a z - x d) + k (a y - b x)

n o w

∣ i j