Vector calculus evaluation question

[641354337]

Homework Statement

Q1. Evaluate grad(f) for the function f($$\underline{r}$$)=($$\underline{a}$$ dot $$\underline{r}$$) ($$\underline{b}$$ dot $$\underline{r}$$)

Q2. If $$\underline{c}$$ is a constant vector, show that grad |$$\underline{c}$$ cross $$\underline{r}$$| ^n = n |$$\underline{c}$$ cross $$\underline{r}$$| ^(n-2) *$$\underline{c}$$ cross ($$\underline{r}$$ cross $$\underline{c}$$ )

Homework Equations

dot product and cross product identities
formula for grad f

The Attempt at a Solution

Q1, I tried to expand it use product rule but simplified it back to where I started, I don't see how I can simplify the result.
Q2, used chain rule got n |$$\underline{c}$$ cross $$\underline{r}$$| ^(n-2) *($$\underline{c}$$ cross $$\underline{r}$$ )
btw I have proven that grad r^n = n * r^ (n-2) * $$\underline{r}$$, not sure if this is correct.

This is my first post, hopefully it is understandable..

 

[eshaw]

$$\nabla(\vec{a}\cdot\vec{r})(\vec{b}\cdot\vec{r})=<br /> \frac{d}{dx_{i}}a_{j}x_{j}b_{k}x_{k}=<br /> a_{j}b_{k}(\frac{dx_{j}}{dx_{i}}x_{k}+\frac{dx_{k}}{dx_{i}}x_{j})=<br /> a_{j}b_{k}(\delta_{ij}x_{k}+\delta_{ik}x_{j})=<br /> \vec{a}(\vec{b}\cdot\vec{r})+\vec{b}(\vec{a}\cdot\vec{r})$$

 

[641354337]

$$\nabla(\vec{a}\cdot\vec{r})(\vec{b}\cdot\vec{r})=<br /> \frac{d}{dx_{i}}a_{j}x_{j}b_{k}x_{k}=<br /> a_{j}b_{k}(\frac{dx_{j}}{dx_{i}}x_{k}+\frac{dx_{k}}{dx_{i}}x_{j})=<br /> a_{j}b_{k}(\delta_{ij}x_{k}+\delta_{ik}x_{j})=<br /> \vec{a}(\vec{b}\cdot\vec{r})+\vec{b}(\vec{a}\cdot\vec{r})$$

but i got $$[\vec{a}(\vec{b}\cdot\vec{r})+\vec{b}(\vec{a}\cdot\vec{r})]\nabla\vec{r}$$

btw i havnt done tensors yet, so I've used the basic property of $$\nabla$$

 

[eshaw]

The only honest way you can do it without tensor analysis, is by breaking it up into all the pieces. That is, take the dot product of the vectors a and r which is a1x+a2y+a3z and multiply it by the dot product of b and r. Then the gradient of that is (df/dx,df/dy,df/dz) and then if you rearrange the terms and you will get the answer that I got. I'm kind of curious how you got the nabla r term. I'll try to give you some useful relations that I used to solve the second problem. I'm not very adept at using the Latex, so it would take me to long to write out the whole solution.

 

[eshaw]

$$\vec{c} \times \vec{r} = \left| \vec{c} \right| \left| \vec{r} \right| \sin \theta \vec{n}$$
$$\left| \vec{c} \times \vec{r} \right| = \left| \vec{c} \right| \left| \vec{r}\right| \sin \theta$$
$$\left| \vec{r} \right| = \sqrt{\vec{r} \cdot \vec{r}} = r$$
$$\vec{r} = r\frac{\vec{r}}{r}=r\vec{n}=\left| \vec{r} \right| \vec{n}$$

 

[641354337]

The only honest way you can do it without tensor analysis, is by breaking it up into all the pieces. That is, take the dot product of the vectors a and r which is a1x+a2y+a3z and multiply it by the dot product of b and r. Then the gradient of that is (df/dx,df/dy,df/dz) and then if you rearrange the terms and you will get the answer that I got. I'm kind of curious how you got the nabla r term. I'll try to give you some useful relations that I used to solve the second problem. I'm not very adept at using the Latex, so it would take me to long to write out the whole solution.

I have used chain rule so my first step is
$$\nabla(\vec{a}\cdot\vec{r})(\vec{b}\cdot\vec{r})=(\vec{b}\cdot\vec{r})\nabla(\vec{a}\cdot\vec{r})+(\vec{a}\cdot\vec{r})\nabla(\vec{b}\cdot\vec{r})$$

I think this is correct, so if ur result is true it means that $$\nabla(\vec{a}\cdot\vec{r}=\vec{a}$$ ?

 

[641354337]

$$\vec{c} \times \vec{r} = \left| \vec{c} \right| \left| \vec{r} \right| \sin \theta \vec{n}$$
$$\left| \vec{c} \times \vec{r} \right| = \left| \vec{c} \right| \left| \vec{r}\right| \sin \theta$$
$$\left| \vec{r} \right| = \sqrt{\vec{r} \cdot \vec{r}} = r$$
$$\vec{r} = r\frac{\vec{r}}{r}=r\vec{n}=\left| \vec{r} \right| \vec{n}$$

Thanks! i think I've got it now

 

[eshaw]

$$\nabla(\vec{a}\cdot\vec{r})=\nabla(a_{1}x+a_{2}y+a_{3}z)=a_{1}\vec{i}+a_{2}\vec{j}+a_{3}\vec{k}=\vec{a}$$