Given a function $f$ defined in $R^2$. Let $F(r,\theta)=f(r\cos\theta,r\sin\theta).$ Verify a formula of the modulus of the gradient.

Question

Given a function $f$ defined in $R^2$. Let $$F(r,\theta)=f(r\operatorname{cos}\theta,r\operatorname{sin}\theta).$$

Verify the formula

$$|\nabla f(r\operatorname{cos}\theta, r\operatorname{sin}\theta)|^2=[D_1 F(r,\theta)]^2+\frac{1}{r^2}[D_2 F(r,\theta)]^2.$$

I have found that $$D_1 F(r,\theta)=\operatorname{cos}\theta D_1 f(x,y)+\operatorname{sin}\theta D_2 f(x,y)$$

and

$$D_2 F(r,\theta)=-r\operatorname{sin}\theta D_1 f(x,y)+r\operatorname{cos}\theta D_2 f(x,y)$$, where $x=rcos\theta, y=rsin\theta$.

Hence, we get $$[D_1 F(r,\theta)]^2+\frac{1}{r^2}[D_2 F(r,\theta)]^2=D_1 f(x,y)^2+D_2 f(x,y)^2$$.

However, $$\nabla f(r\operatorname{cos}\theta, r\operatorname{sin}\theta)=(D_1 f\operatorname{cos}\theta+D_2 f\operatorname{sin}\theta,-rD_1 f\operatorname{sin}\theta+rD_2 f\operatorname{cos}\theta)$$.

Hence I do not get the equality. Where am I getting this wrong? I've been checking my calculations for a long time but I just can't get the answer.

I would greatly appreciate any help.

Answer 1

I think the notation $\nabla f(r\cos \theta,r\sin\theta)$ here means $(D_1 f(x,y),D_2f(x,y))$. Hence

$$|\nabla f(r\cos \theta,r\sin\theta)|^2=(D_1f(x,y))^2+(D_2f(x,y))^2$$