Vector analysis problem about a gradient

[patric44]

Homework Statement

if Φ=Φ(λx,λy,λz) = λ^n(x,y,z)
proof that r.grad(Φ) = nΦ

Relevant Equations

Φ=Φ(λx,λy,λz) = λ^n(x,y,z)
r.grad(Φ) = nΦ

hi guys i saw this problem in my collage textbook on vector calculus , i don't know if the statement is wrong because it don't make sense to me
so if anyone can help on getting a hint where to start i will appreciate it , basically it says :
$$ \phi =\phi(\lambda x,\lambda y,\lambda z)=\lambda^{n}(x,y,z) $$
prove that
$$\vec{r} . \vec{\nabla}\phi=n\phi$$

 

[wrobel]

The correct statement is $$f(\lambda x)=\lambda^s f(x),\quad \forall \lambda>0,\quad \forall x\in D\subset \mathbb{R}^m.$$
Here the domain $D$ is such that $x\in\ D\Longrightarrow \lambda x\in D$.Differentiate this equation in $\lambda$ and then put $\lambda=1$:
$$\Big(x,\frac{\partial f}{\partial x}\Big)=sf(x).$$

 

[patric44]

The correct statement is $$f(\lambda x)=\lambda^s f(x),\quad \forall \lambda>0,\quad \forall x\in D\subset \mathbb{R}^m.$$
Here the domain $D$ is such that $x\in\ D\Longrightarrow \lambda x\in D$.Differentiate this equation in $\lambda$ and then put $\lambda=1$:
$$\Big(x,\frac{\partial f}{\partial x}\Big)=sf(x).$$

why should i differentiate with respect to $\lambda$ the gradient of $\phi$ is with respect to the $(x,y,z)$ ?
and what about $\vec{r}.\vec{\nabla}\phi$

 

[PeroK]

why should i differentiate with respect to $\lambda$ the gradient of $\phi$ is with respect to the $(x,y,z)$ ?
and what about $\vec{r}.\vec{\nabla}\phi$

You can let $g(\lambda) = f(\lambda x, \lambda y, \lambda z) = \lambda^n f(x, y, z)$ and differentiate that with respect to $\lambda$.

 

[patric44]

You can let $g(\lambda) = f(\lambda x, \lambda y, \lambda z) = \lambda^n f(x, y, z)$ and differentiate that with respect to $\lambda$.

ok :
let $g(\lambda) = f(\lambda x ,\lambda y,\lambda z)=\lambda^{n}f(x,y,z)$ then by differentiating :
$$\frac{dg(\lambda)}{d\lambda}=n\lambda^{n-1}f(x,y,z)$$
and the gradient part :
$$\vec{r}.\vec{\nabla}f(\lambda x ,\lambda y,\lambda z) = \lambda^{n}(x\frac{df}{dx}+y\frac{df}{dy}+z\frac{df}{dz})$$
i don't get how this is related to my question!

 

[PeroK]

ok :
let $g(\lambda) = f(\lambda x ,\lambda y,\lambda z)=\lambda^{n}f(x,y,z)$ then by differentiating :
$$\frac{dg(\lambda)}{d\lambda}=n\lambda^{n-1}f(x,y,z)$$
and the gradient part :
$$\vec{r}.\vec{\nabla}f(\lambda x ,\lambda y,\lambda z) = \lambda^{n}(x\frac{df}{dx}+y\frac{df}{dy}+z\frac{df}{dz})$$
i don't get how this is related to my question!

There's another way you can differentiate $g(\lambda)$.

 

[patric44]

There's another way you can differentiate $g(\lambda)$.

a little more help , you mean using something like the chain rule ?

 

[PeroK]

a little more help , you mean using something like the chain rule ?

Yes, because the chain rule invokes the gradient.

 

[patric44]

Yes, because the chain rule invokes the gradient.

thanks , i guess i got it now :
if i let $u=\lambda x , v = \lambda y , w = \lambda z ,$then
$$\frac{dg}{d\lambda} = \frac{\partial\phi}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial\phi}{\partial v}\frac{\partial v}{\partial y}+\frac{\partial\phi}{\partial w}\frac{\partial w}{\partial z}$$
then
$$\frac{dg}{d\lambda} = \frac{\partial\phi}{\partial \lambda x} x +\frac{\partial\phi}{\partial \lambda y}y+\frac{\partial\phi}{\partial \lambda z}z=n\lambda^{n-1}\phi(x,y,z)$$
if i let $\lambda = 1$ :
$$ \frac{\partial\phi}{\partial x} x +\frac{\partial\phi}{\partial y}y+\frac{\partial\phi}{\partial z}z=n\phi(x,y,z)=\vec{r}.\vec{\nabla}\phi$$
thanks guys so much .

 

[PeroK]

thanks , i guess i got it now :
if i let $u=\lambda x , v = \lambda y , w = \lambda z ,$then
$$\frac{dg}{d\lambda} = \frac{\partial\phi}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial\phi}{\partial v}\frac{\partial v}{\partial y}+\frac{\partial\phi}{\partial w}\frac{\partial w}{\partial z}$$
then
$$\frac{dg}{d\lambda} = \frac{\partial\phi}{\partial \lambda x} x +\frac{\partial\phi}{\partial \lambda y}y+\frac{\partial\phi}{\partial \lambda z}z=n\lambda^{n-1}\phi(x,y,z)$$
if i let $\lambda = 1$ :
$$ \frac{\partial\phi}{\partial x} x +\frac{\partial\phi}{\partial y}y+\frac{\partial\phi}{\partial z}z=n\phi(x,y,z)=\vec{r}.\vec{\nabla}\phi$$
thanks guys so much .

This is not quite right. First, you should have:
$$\frac{dg}{d\lambda} = \frac{\partial\phi}{\partial u}\frac{\partial u}{\partial \lambda}+\frac{\partial\phi}{\partial v}\frac{\partial v}{\partial \lambda}+\frac{\partial\phi}{\partial w}\frac{\partial w}{\partial \lambda }$$
And then this becomes:
$$\frac{dg}{d\lambda} = \frac{\partial\phi}{\partial u}x +\frac{\partial\phi}{\partial v} y +\frac{\partial\phi}{\partial w} z = \vec{\nabla}\phi(u, v, w) \cdot \vec r = \vec{\nabla}\phi(\lambda x, \lambda y, \lambda z) \cdot \vec r$$

 

[patric44]

This is not quite right. First, you should have:
$$\frac{dg}{d\lambda} = \frac{\partial\phi}{\partial u}\frac{\partial u}{\partial \lambda}+\frac{\partial\phi}{\partial v}\frac{\partial v}{\partial \lambda}+\frac{\partial\phi}{\partial w}\frac{\partial w}{\partial \lambda }$$
And then this becomes:
$$\frac{dg}{d\lambda} = \frac{\partial\phi}{\partial u}x +\frac{\partial\phi}{\partial v} y +\frac{\partial\phi}{\partial w} z = \vec{\nabla}\phi(u, v, w) \cdot \vec r = \vec{\nabla}\phi(\lambda x, \lambda y, \lambda z) \cdot \vec r$$

oh my bad , but the book stated it explicitly as $\vec{r}.\vec{\nabla\phi}$ so i thought the book was right , does it matter since i can change the order of the product of the chain rule terms , and get the other reversed expression , or i cannot do that ?

 

[PeroK]

You still have to set $\lambda = 1$. That sorts everything out.

 

[patric44]

thanks so much

 

[wrobel]

the converse theorem also holds