AP Calculus BC - Divergence

Given the vector field

$\vec{F}=yz\vec{i}+xz\vec{j}+xy\vec{k}$

find the divergence of the vector field:

$div \vec{F}$ .

$div \vec{F}=0$

$div \vec{F}=x+y+z$

$div \vec{F}=xy+yz+xz$

$div \vec{F}=x+y$

$div \vec{F}=x+z$

Explanation

Given a vector field

$\vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}$

we find its divergence by taking the dot product with the gradient operator:

$div\vec{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$

We know that , so we have

$div\vec{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}=0+0+0=0$

Find the divergence of the vector $\left \langle x^2y^2z^2,3y^5,17z\right \rangle$

Explanation

To find the divergence a vector $a=\left \langle A,B,C\right \rangle$ , you use the following definition: $div(a)=\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}$ . Applying this to the vector from the problem statement, we get $\frac{\partial A}{\partial x}=2xy^2z^2, \frac{\partial B}{\partial y}=15y^4,\frac{\partial C}{\partial z}=17$ . Adding all of these up, according to the definition, will produce the correct answer.

Find the divergence of the following vector field:

$\vec{F}=\left \langle 7e^{3x}, 2yz, y^3\right \rangle$

$21e^{3x}+2z$

$21e^{x}+2z$

$21e^{3x}+2z+3y^2$

$21e^{3x}+2z+y^3$

Explanation

The divergence of the vector field is given by

$div \vec{F}=\bigtriangledown \cdot \vec{F}$

where $\bigtriangledown = \left \langle f_x, f_y, f_z\right \rangle$

Taking the dot product gives us the sum of the respective partial derivatives of the vector field. For higher order partial derivatives, we work from left to right for the given variables.

The partial derivatives are

$f_x=21e^{3x}$ , ,

Find the divergence of the following vector: $\left \langle yz^3,xy^5z,3xz\right \rangle$

$\left \langle 0,5xy^4z,3x\right \rangle$

$\left \langle 3yz^2,5xy^3z,3z\right \rangle$

$\left \langle 3yz^2,y^5z,3x\right \rangle$

$\left \langle 0,5xy^4z,3\right \rangle$

Explanation

To find the divergence a vector $a=\left \langle A,B,C\right \rangle$ , you use the following definition: $div(a)=\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}$ . Applying this to the vector from the problem statement, we get $\frac{\partial A}{\partial x}=0, \frac{\partial B}{\partial y}=5xy^4z,\frac{\partial C}{\partial z}=3x$ . Adding all of these up, according to the definition, will produce the correct answer.

Find $div \vec{F}$ , where F is given by

$\left \langle x^2z, z^2, ye^xz\right \rangle$

Explanation

The divergence of a vector is given by

$abla \cdot \vec{F}$ , where $abla = \left \langle f_x, f_y, f_z\right \rangle$

Taking the partial respective partial derivatives of the x, y, and z components of our curve, we get

The rules used to find the derivatives are as follows:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Find $div \vec{F}$ of the vector function below:

$\vec{F}=\cos(x)\hat{i}+(xyz)\hat{j}+z^2\sin(y)\hat{k}$

$-\sin(x)+xz+2z\sin(y)$

$\left \langle -\sin(x),xz,2z\sin(y) \right \rangle$

$\sin(x)+xz+2z\sin(y)$

$-\sin(x)+xz-2z\sin(y)$

Explanation

The divergence of a vector function is given by

$div \vec{F}=\bigtriangledown \cdot \vec{F}$

where $\bigtriangledown = \left \langle f_x, f_y, f_z\right \rangle$

So, in taking the dot product of the gradient and the function, we get the sum of the respective partial derivatives. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x=-\sin(x)$

$f_z=2z\sin(y)$

Given that F is a vector function and f is a scalar function, which of the following operations results in a vector?

$abla \times ( abla \times \mathbf{F})$

$abla \cdot ( abla \times \mathbf{F})$

$abla \cdot ( abla\cdot \mathbf{F})$

$abla \cdot( abla f)$

$abla( abla \cdot f)$

Explanation

For all the given answers:

$abla \cdot ( abla \times \mathbf{F})$ - The curl of a vector is also a vector, so the divergence of the term in parenthesis is scalar.

$abla \cdot ( abla \cdot \mathbf{F})$ - The divergence of a vector is a scalar. The divergence of a scalar is undefined, so this expression is undefined.

$abla \cdot( abla f)$ - The gradient of a scalar is a vector, so the divergence of the term in parenthesis is a scalar.

$abla( abla \cdot f)$ - The divergence of a scalar doesn't exist, so this expression is undefined.

The remaining answer is:

$abla \times ( abla \times \mathbf{F})$ - The curl of a vector is also a vector, so the curl of the term in parenthesis is a vector as well.

Find the divergence of the following vector field:

$\vec{F}=\left \langle xy,yz, \cos(z)\right \rangle$

$y+z-\sin(z)$

$y+z+\sin(z)$

$z-\sin(z)$

$y-\sin(z)$

Explanation

The divergence of a vector field is given by

$div\vec{F}=\bigtriangledown \cdot \vec{F}$

where $\bigtriangledown = \left \langle f_x, f_y, f_z\right \rangle$

In taking the dot product, we end up with the sum of the respective partial derivatives of the vector field. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

, . $f_z=-\sin(z)$

Find the divergence of the vector $\left \langle 12x^2y,4y^2z,xy^5\right \rangle$

Explanation

To find the divergence of a vector $v=\left \langle A,B,C\right \rangle$ , we apply the following definition: $div(v)=\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}$ . Applying the definition to the vector from the problem statement, we get $\frac{\partial }{\partial x}(12x^2y)+\frac{\partial }{\partial y}(4y^2z)+\frac{\partial }{\partial z}(xy^5)=24xy+8yz$

Find the divergence of the vector field:

$\vec{F}=\left \langle \tan(x^2z), y^2, xy\right \rangle$

$2xz\sec^2(x^2z)+2y$

$-2xz\sec^2(x^2z)+2y$

$x^2z\sec^2(x^2z)+2y$

$2xz\sec^2(x^2z)+y^2$

Explanation

The divergence of a vector field is given by

$div \vec{F}=\bigtriangledown \cdot \vec{F}$ , where $\bigtriangledown = \left \langle \frac{\partial }{\partial x}, \frac{\partial }{\partial y}, \frac{\partial }{\partial z} \right \rangle$

When we take the dot product of the gradient and the vector field, we get the sum of the respective partial derivatives. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x=2xs\sec^2(x^2z)$ , ,

Divergence

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