Divergence can be viewed as a measure of the magnitude of a vector field's source or sink at a given point.

To visualize this, picture an open drain in a tub full of water; this drain may represent a 'sink,' and all of the velocities at each specific point in the tub represent the vector field. Close to the drain, the velocity will be greater than a spot farther away from the drain.

What divergence can calculate is what this velocity is at a given point. Again, the magnitude of the vector field.

We're given the function

$\displaystyle F(x,y,z)=4xz^2\widehat{i}+3xyz\widehat{j}+2yz^3\widehat{k}$ 

What we will do is take the derivative of each vector element with respect to its variable $\displaystyle (\widehat{i}:x,\widehat{j}:y,\widehat{k}:z)$

Then sum the results together:

$\displaystyle divF(x,y,z)=4z^2+3xz+6yz^2$

At the point $\displaystyle (4,1,1)$

$\displaystyle divF(4,1,1)=4(1)^2+3(4)(1)+6(1)(1)^2=22$

Divergence can be viewed as a measure of the magnitude of a vector field's source or sink at a given point.

To visualize this, picture an open drain in a tub full of water; this drain may represent a 'sink,' and all of the velocities at each specific point in the tub represent the vector field. Close to the drain, the velocity will be greater than a spot farther away from the drain.

What divergence can calculate is what this velocity is at a given point. Again, the magnitude of the vector field.

We're given the function

$\displaystyle F(x,y,z)=y^2z^3\widehat{i}+3x^3y^2\widehat{j}+x^2z\widehat{k}$ 

What we will do is take the derivative of each vector element with respect to its variable $\displaystyle (\widehat{i}:x,\widehat{j}:y,\widehat{k}:z)$

Then sum the results together:

$\displaystyle divF(x,y,z)=0+6x^3y+x^2$

At the point $\displaystyle (3,2,4)$

$\displaystyle divF(3,2,4)=6(3)^3(2)+3^2=333$

Divergence can be viewed as a measure of the magnitude of a vector field's source or sink at a given point.

To visualize this, picture an open drain in a tub full of water; this drain may represent a 'sink,' and all of the velocities at each specific point in the tub represent the vector field. Close to the drain, the velocity will be greater than a spot farther away from the drain.

What divergence can calculate is what this velocity is at a given point. Again, the magnitude of the vector field.

We're given the function

$\displaystyle F(x,y)=ln(x^2y)\widehat{i}+3^{xy^2}\widehat{j}$ 

What we will do is take the derivative of each vector element with respect to its variable $\displaystyle (\widehat{i}:x,\widehat{j}:y,\widehat{k}:z)$

Then sum the results together:

Derivative of an exponential: 

$\displaystyle d[a^u]=a^uduln(a)$

Derivative of a natural log: 

$\displaystyle d[ln(u)]=\frac{du}{u}$

Note that u may represent large functions, and not just individual variables!

$\displaystyle divF(x,y)=\frac{2xy}{x^2y}+2xy(3^{xy^2})ln(3)$

$\displaystyle divF(x,y)=\frac{2}{x}+2xy(3^{xy^2})ln(3)$

At the point $\displaystyle (1.3,1.7)$

$\displaystyle divF(1.3,1.7)=\frac{2}{1.3}+2(1.3)(1.7)(3^{(1.3)(1.7)^2})ln(3)$

$\displaystyle divF(1.3,1.7)=302.71$

Divergence can be viewed as a measure of the magnitude of a vector field's source or sink at a given point.

To visualize this, picture an open drain in a tub full of water; this drain may represent a 'sink,' and all of the velocities at each specific point in the tub represent the vector field. Close to the drain, the velocity will be greater than a spot farther away from the drain.

What divergence can calculate is what this velocity is at a given point. Again, the magnitude of the vector field.

We're given the function

$\displaystyle F(x,y)=ln(2^{xy})\widehat{i}+4x^3y^3\widehat{j}$ 

What we will do is take the derivative of each vector element with respect to its variable $\displaystyle (\widehat{i}:x,\widehat{j}:y,\widehat{k}:z)$

Then sum the results together:

Derivative of an exponential: 

$\displaystyle d[a^u]=a^uduln(a)$

Derivative of a natural log: 

$\displaystyle d[ln(u)]=\frac{du}{u}$

Note that u may represent large functions, and not just individual variables!

$\displaystyle divF(x,y)=\frac{y2^{xy}ln(2)}{2^{xy}}+12x^3y^2$

$\displaystyle divF(x,y)=yln(2)+12x^3y^2$

At the point $\displaystyle (1.2,1.5)$

$\displaystyle divF(1.2,1.5)=1.5ln(2)+12(1.2)^3(1.5)^2$

$\displaystyle divF(1.2,1.5)=47.7$

Divergence can be viewed as a measure of the magnitude of a vector field's source or sink at a given point.

To visualize this, picture an open drain in a tub full of water; this drain may represent a 'sink,' and all of the velocities at each specific point in the tub represent the vector field. Close to the drain, the velocity will be greater than a spot farther away from the drain.

What divergence can calculate is what this velocity is at a given point. Again, the magnitude of the vector field.

We're given the function

$\displaystyle F(x,y)=tan(xy^2)\widehat{i}+e^{x^2y^2}\widehat{j}$

What we will do is take the derivative of each vector element with respect to its variable $\displaystyle (\widehat{i}:x,\widehat{j}:y,\widehat{k}:z)$

Then sum the results together:

Derivative of an exponential: 

$\displaystyle d[e^u]=e^udu$

Trigonometric derivative: 

$\displaystyle d[tan(u)]=\frac{du}{cos^2(u)}$

Note that u v may represent large functions, and not just individual variables!

$\displaystyle divF(x,y)=\frac{y^2}{cos^2(xy^2)}+2x^2ye^{x^2y^2}$

At the point $\displaystyle (-1,1)$

$\displaystyle divF(x,y)=\frac{1^2}{cos^2((-1)1^2)}+2(-1)^2(1)e^{(-1)^2(1)^2}=8.86$ 

Divergence can be viewed as a measure of the magnitude of a vector field's source or sink at a given point.

To visualize this, picture an open drain in a tub full of water; this drain may represent a 'sink,' and all of the velocities at each specific point in the tub represent the vector field. Close to the drain, the velocity will be greater than a spot farther away from the drain.

What divergence can calculate is what this velocity is at a given point. Again, the magnitude of the vector field.

We're given the function

$\displaystyle F(x,y)=x^{ycos(y)}\widehat{i}+y^{xtan(x)}\widehat{j}$ 

What we will do is take the derivative of each vector element with respect to its variable $\displaystyle (\widehat{i}:x,\widehat{j}:y,\widehat{k}:z)$

Then sum the results together:

$\displaystyle divF(x,y)=ycos(y)x^{ycos(y)-1}+xtan(x)y^{xtan(x)-1}$

Note that for the complexity of the function, the derivatives are rather simple, given that when deriving with respect to one variable, we treat the other as constant!

At the point $\displaystyle (2,4)$

$\displaystyle divF(2,4)=4cos(4)2^{4cos(4)-1}+2tan(2)4^{2tan(2)-1}$

$\displaystyle divF(2,4)=-0.216$

Using chain rule, we get $\displaystyle g'(x)=\frac{1}{x^2+e^x} \cdot (2x+e^x) = \frac{2x+e^x}{x^2+e^x}.$

To consider finding the slope, let's discuss the topic of the gradient.

For a function $\displaystyle f(x_1,x_2,...,x_n)$, the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\displaystyle \frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point

Knowledge of the following derivative rules will be necessary:

$\displaystyle d[tan(u)]=\frac{du}{cos^2(u)}$

Note that u may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

$\displaystyle f(x,y)=x^3y+ tan(xy)$ at $\displaystyle (1.5,3.5)$

x:

$\displaystyle \frac{\delta f}{\delta x}=3x^2y+\frac{y}{cos^2(xy)};3(1.5)^2(3.5)+\frac{3.5}{cos^2((1.5)(3.5))}=36.97$

y:

$\displaystyle \frac{\delta f}{\delta y}=x^3+\frac{x}{cos^2(xy)};1.5^3+\frac{1.5}{cos^2((1.5)(3.5))}=9.10$

The slope is $\displaystyle (36.97,9.10)$

To consider finding the slope, let's discuss the topic of the gradient.

For a function $\displaystyle f(x_1,x_2,...,x_n)$, the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\displaystyle \frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point

Knowledge of the following derivative rules will be necessary:

Derivative of an exponential: 

$\displaystyle d[a^u]=a^uduln(a)$

Note that u may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

$\displaystyle f(x,y)=x^{2\pi y}$ at $\displaystyle (3.3,2.1)$

x:

$\displaystyle \frac{\delta f}{\delta x}=(2\pi y)x^{2\pi y -1};(2\pi (2.1))(3.3)^{2\pi (2.1) -1}=2.78 \cdot 10^{7}$

y:

$\displaystyle \frac{\delta f}{\delta y}=(2\pi)(x^{2\pi y})ln(x);(2\pi)(3.3^{2\pi (2.1)})ln(3.3)=5.21 \cdot 10^{7}$

The slope is $\displaystyle (2.78 \cdot 10^{7},5.21 \cdot 10^{7})$

$\displaystyle y=(4-x+5x^{3})^{14}$

Let $\displaystyle \left(4-x+5x^3\right)=u$

By applying the chain rule,$\displaystyle \quad \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}$ we get, 

$\displaystyle \frac{d}{du}(u^{14})\frac{d}{dx}\left(4-x+5x^3\right)$ implies, $\displaystyle 14u^{13}\left(15x^2-1\right)$

By substituting back $\displaystyle u=\left(4-x+5x^3\right)$ we get,

$\displaystyle 14(4-x+5x^3)^{13}(15x^2-1))$