If √ r=aeθ cot α where a and α are real numbers, then (d2r/dθ2) -4r Cot2α is

Question

If √ r=aeθ cot α where a and α are real numbers, then (d2r/dθ2) -4r Cot2α is (A) r (B) 1/r (C) 1 (D) 0

Tardigrade Admin · Accepted Answer

Given, √r=a eθ cot α...(i) Differentiating w.r.t. θ, (1/2 √r) (d r/d θ) =a cot α ⋅ eθ cot α (d r/d θ) =2 a √r cot α eθ cot α (d r/d θ) =2 a ⋅ a eθ cot α ⋅ cot α ⋅ eθ cot α [form Eq.(i)] (d r/d θ)=2 a2 cot α ⋅ α ⋅ e2 θ cot α Agian r differentiating w.r.t. θ (d2 r/d θ2)=2 a2 cot α ⋅ e2 θ cot α ⋅ 2 cot α (d2 r/d θ2)=4 a2 cot 2 α ⋅ e2 θ cot θ (d2 r/d θ2)=4 cot 2 α ⋅(a eθ cot α)2 (d2 r/d θ2)=4 cot 2 α ⋅(√r)2 [from Eq. (i)] (d2 r/d θ2)-4 r cot 2 α=0

Q. If $r = a e^{θ c o t α}$ where $a$ and $α$ are real numbers, then $\frac{d ^{2} r}{d θ ^{2}} - 4 r C o t^{2} α$ is

$r$

$1/ r$

$1$

$0$

Q. If r​=aeθcotα where a and α are real numbers, then dθ2d2r​−4rCot2α is

r

1/r

1

0

Q. If $r = a e^{θ c o t α}$ where $a$ and $α$ are real numbers, then $\frac{d ^{2} r}{d θ ^{2}} - 4 r C o t^{2} α$ is

$r$

$1/ r$

$1$

$0$