The general form of Euler's method, when a derivative function, initial value, and step size are known, is:

\(\displaystyle y_n=y_{n+1} +\Delta x f'(x_n,y_n)\)

To calculate the step size find the distance between the final and initial \(\displaystyle x\) value and divide by the number of steps to be used:

\(\displaystyle \Delta x = \frac{x_f-x_i}{Steps}\)

For this problem, we are told \(\displaystyle f'(x)=1.4^{cos(\pi e^{x})}\) and \(\displaystyle y(0)=0\)

Knowing this, we may take the steps to estimate our function value at our final \(\displaystyle x\) value:

\(\displaystyle \Delta x = \frac{4-0}{5}=0.8\)

\(\displaystyle y_0=0;x_0=0\)

\(\displaystyle y_1=0+(0.8)1.4^{cos(\pi e^{0})}=0.571\)

\(\displaystyle y_2=0.571+(0.8)1.4^{cos(\pi e^{0.8})}=1.604\)

\(\displaystyle y_3=1.604+(0.8)1.4^{cos(\pi e^{1.6})}=2.178\)

\(\displaystyle y_4=2.178+(0.8)1.4^{cos(\pi e^{2.4})}=2.750\)

\(\displaystyle y_5=2.750+(0.8)1.4^{cos(\pi e^{3.2})}=3.523\)

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

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

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: 

\(\displaystyle d[e^u]=e^udu\)

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

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

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

At the point \(\displaystyle (2,1.2)\)

 \(\displaystyle divF(2,1.2)=1.2^2e^{2(1.2^2)}+2(2)(1.2)(2^{(2)(1.2^2)})ln(2)\)

\(\displaystyle divF(2,1.2)=50.15\)

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

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

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:

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

At the point \(\displaystyle (-1,2)\)

\(\displaystyle divF(-1,2)=3(-1)^2(2)+2(-1)^2(2)\)

\(\displaystyle divF(-1,2)=10\)

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

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

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:

\(\displaystyle divF(x,y)=0+0\)

\(\displaystyle divF(x,y)=0\)

Since no variables were paired with their respective variable, the divergence goes to zero.

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

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

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: 

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

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

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

At the point \(\displaystyle (-2,1)\)

\(\displaystyle divF(-2,1)=3^{-2}ln(3)+2^{(-2)(1)}(-2)ln(2)\)

\(\displaystyle divF(-2,1)=-0.225\)

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

\(\displaystyle F(x,y)=ytan(x)\widehat{i}+ln(xy)\widehat{j}\)

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 a natural log: 

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

Trigonometric derivative: 

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

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

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

At the point \(\displaystyle (3,2)\)

\(\displaystyle divF(3,2)=\frac{2}{cos^2(3)}+\frac{1}{2}\)

\(\displaystyle divF(3,2)=2.541\)

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

\(\displaystyle F(x,y)=xsin(xy)\widehat{i}+xcos(xy)\widehat{j}\)

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:

Trigonometric derivative: 

\(\displaystyle d[sin(u)]=cos(u)du\)

\(\displaystyle d[cos(u)]=-sin(u)du\)

Product rule: \(\displaystyle d[uv]=udv+vdu\)

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

\(\displaystyle divF(x,y)=sin(xy)+xycos(xy)-x^2sin(xy)\)

At the point \(\displaystyle (\pi,2)\)

\(\displaystyle divF(\pi,2)=sin(2\pi)+2\pi cos(2\pi)-\pi^2sin(2\pi)\)

\(\displaystyle divF(\pi,2)=2\pi\) 

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

\(\displaystyle F(x,y)=\frac{x}{y}\widehat{i}+\frac{y}{x}\widehat{j}\)

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:

\(\displaystyle divF(x,y)=\frac{1}{y}+\frac{1}{x}\)

At the point \(\displaystyle (\frac{1}{2},\frac{1}{3})\)

\(\displaystyle divF(\frac{1}{2},\frac{1}{3})=\frac{1}{\frac{1}{3}}+\frac{1}{\frac{1}{2}}\)

\(\displaystyle divF(\frac{1}{2},\frac{1}{3})=5\)

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

\(\displaystyle F(x,y)=\frac{y}{x}\widehat{i}+\frac{x}{y}\widehat{j}\)

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:

Quotient rule: \(\displaystyle d[\frac{u}{v}]=\frac{vdu-udv}{v^2}\)

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

\(\displaystyle divF(x,y)=-\frac{y}{x^2}-\frac{x}{y^2}\)

At the point \(\displaystyle (\frac{1}{2},\frac{1}{3})\)

\(\displaystyle divF(\frac{1}{2},\frac{1}{3})=-\frac{\frac{1}{3}}{(\frac{1}{2})^2}-\frac{\frac{1}{2}}{(\frac{1}{3})^2}\)

\(\displaystyle divF(\frac{1}{2},\frac{1}{3})=\frac{-35}{6}\)

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

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

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:

\(\displaystyle divF(x,y)=ln(y)x^{ln(y)-1}+ln(x)y^{ln(x)-1}\)

At the point \(\displaystyle (3,1)\)

\(\displaystyle divF(3,1)=ln(1)(3^{ln(1)-1})+ln(3)(1^{ln(3)-1})\) 

\(\displaystyle divF(3,1)=ln(3)\)