Vector Calculus Gradient - are my answers correct?

Question

Could someone please clarify if I have the correct answers for these following questions?

[Assume $r=x$i +yj+zk and $a=a_1$i +$a_2$j+$a_3$k for some constants $a_1,a_2,a_3$]

1. $\nabla f forf=cos(x)+3y^2sin^3z $

Answer: $\nabla f = -sin(x)i+6ysin^3(z)j+9y^2sin^2(z)cos(z)k$ [edited]

2. $\nabla f forf=r\cdot r$

Answer: $\nabla f =\nabla(x^2i+y^2j+z^2k)+(2xy+2xz+2yz) $ [edited]

3. $\nabla \cdot (a\times r - r)$

Answer: $\nabla \cdot ((a_2z-a_3y)i-(a_1z-a_3x)j+(a_1y-a_2x)k)-(xi+yj+zk)=0$

Answer 1

The gradient of f(x,y,z) is $\nabla f= \frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}{\partial z}\vec{j}$.

In (i) $f(x,y,z)= \cos(x)+ 3y^2 \sin^3(z)$ so $\frac{\partial f}{\partial x}= -\sin(x)$, $\frac{\partial f}{\partial y}= 6y \sin^3(z)$, and $\frac{\partial f}{\partial z}= 9y^2\sin^2(z)\cos(z)$

So $\nabla f= -\sin(x)\vec{i}+ 6y \sin^3(z)\vec{j}+ 9y^2\sin^2(z)\cos(z)\vec{k}$

I don't know how you got "$\cos(i)$"! I'm hoping it was a typo but I am afraid you decided that the derivative of cos(x) with respect to x was "cos(1)" and then just replaced the "1" with "i"!

Answer 2

No, these are not correct. Check the definition of the gradient, as well as dot product. For example, for 2):
$r\cdot r=x^2+y^2+z^2$. The gradient of this function is a vector of the partial derivative of the function with respect to each variable, in this case $[2x \space 2y\space 2z]$.

Answer 3

1.: You are not differentiating the trigonometric functions correctly.

2.: The result is not correct, because $r \cdot r=x^2+y^2+z^2$.

3.: Your cross product is good, but the calculation of divergence is not correct.