[latex]\frac{\partial z}{\partial x}(x_0, y_0)=\frac{\text{change in } z}{\text{change in }x}(\text{holding }y \text{ as a constant } y_0)[/latex]

[latex]\frac{\partial z}{\partial y}(x_0, y_0)=\frac{\text{change in } z}{\text{change in }y}(\text{holding }x \text{ as a constant } x_0)[/latex]

Besides the Leibniz notations above, you can also write the derivatives as [latex]\frac{\partial z}{\partial x}=f_x[/latex] and [latex]\frac{\partial z}{\partial y}=f_y[/latex]. Similar to the geometric meaning of [latex]\frac{dy}{dx}[/latex] in two-dimensional (2D), [latex]\frac{\partial z}{\partial x}[/latex] and [latex]\frac{\partial z}{\partial y}[/latex] in three-dimensional (3D) represent the slopes of tangent lines as well.- Input a function of two variables, then set [latex]y[/latex] as a constant, e.g., [latex]-1[/latex]. A cross-section plane [latex]y=-1[/latex] is plotted. Recall that the function [latex]y = y_0[/latex] (or [latex]x=x_0[/latex]) in 3D represents the planes that are perpendicular to the
*[latex]xy[/latex]*-plane. - A tangent line passing through the point ([latex]x_0[/latex], [latex]y_0[/latex]) and also on the cross-section plane [latex]y=y_0[/latex] is plotted. Change the [latex]y[/latex]-values using the slider, and you will see the cross-section and the tangent line changes. You can also rotate the graph to get a better view. Since the particle derivative is the slope of the tangent line, the partial derivative [latex]\frac{\partial z}{\partial x}[/latex] changes as well.
- Repeat the same steps in (1) and (2) for [latex]\frac{\partial z}{\partial y}[/latex].

Check your understanding by solving the following questions[footnote]Gilbert Strang, Edwin “Jed” Herman, OpenStax, Calculus Volume 3, Calculus Volumes 1, 2, and 3 are licensed under an Attribution-NonCommercial-Sharealike 4.0 International License (CC BY-NC-SA).[/footnote]:

- Let [latex]f(x,y)=\frac{xy}{x-y}[/latex]. Find [latex]f_x(2,-2)[/latex] and [latex]f_y(2,-2)[/latex].
- The apparent temperature index, [latex]A[/latex], is a measure of how the temperature feels,

[latex]A=0.885x -22.4 y +1.2 xy -0.544[/latex]

where [latex]x[/latex] is relative humidity and [latex]y[/latex] is the air temperature. Find [latex]\frac{\partial A}{\partial x}[/latex] and [latex]\frac{\partial A}{\partial y}[/latex] when [latex]x=20°F[/latex] and [latex]y=1[/latex].

[h5p id="7"][latex]\int_a^bf(x)dx \approx \sum_{i=1}^n f(x_i) \Delta x[/latex], where [latex]\Delta x = \frac{b-a}{n}[/latex] and [latex]x_i=a + i\Delta x[/latex]. The larger [latex]n[/latex] is, the better the estimation is. Thus, the limit of the Riemann sum defines the definite integral,

[latex]\int_a^bf(x)dx =\lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x = \sum_{i=1}^{\infty} f(x_i)\,\Delta x[/latex].

Similar to the single integral, the[latex]\iint_R f(x,y) dA=\lim_{m,n \to \infty}\sum_{i=1}^{m} \sum_{j=1}^{n} f(x_{i},y_{j}) \Delta A[/latex]

where [latex]\Delta A=\Delta x \Delta y, \Delta x=\frac{b-a}{n}, \Delta y=\frac{d-c}{m}, x_i=a + i\Delta x[/latex] and [latex]y_i=c + i\Delta y[/latex].- Input a function of two variables into the [latex]f(x,y)[/latex] input function section.
- Move the [latex]n[/latex]-slide around to decide the subregions of the rectangular region, [latex]R[/latex], and we consider the subregions are squares.
- Pick xmin, xmax, ymin, and ymax points for your domain/bounds of the rectangular region, [latex]R[/latex].
- Use the [latex]k[/latex]-slider to choose which square-shaped subregion you’d like to highlight.
- Use the checkboxes to show either all of the rectangular prisms compared to just the one you are highlighting, as well as whether to see the graph or not.
- By changing your view and hovering over the plot, you can see a 2D representation of the rectangular area. Additionally, the double integral is dynamically calculated at the bottom.

Check your understanding by solving the following questions[footnote]Gilbert Strang, Edwin “Jed” Herman, OpenStax, Calculus Volume 3, Calculus Volumes 1, 2, and 3 are licensed under an Attribution-NonCommercial-Sharealike 4.0 International License (CC BY-NC-SA).[/footnote]:
Calculate the integrals by interchanging the order of integration:

- [latex]\int_{-1}^{1} \int_{-1}^{2} 2x + 3y + 5 \, dx dy [/latex]
- [latex]\int_{0}^{\pi} \int_{0}^{\pi/2}sin(2x) cos(3y)\, dx dy [/latex]

[latex]J=\begin{pmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} \end{pmatrix}=\begin{pmatrix} f_{xx} & f_{xy} \\ f_{yx}& f_{yy} \end{pmatrix}[/latex].

Note that [latex]f_{xy}=f_{yx}[/latex]. We plug in the critical points from the first derivative into the Jacobian and calculate[latex]D=\begin{vmatrix} f_{xx}(x_0, y_0) & f_{xy}(x_0, y_0) \\ f_{yx}(x_0, y_0)& f_{yy}(x_0, y_0) \end{vmatrix}=f_{xx}(x_0, y_0) f_{yy}(x_0, y_0) - ( f_{xy}(x_0, y_0) )^2[/latex]

Then we use the following rules to conduct the second derivative test:- If [latex]D>0[/latex] and [latex]f_{xx}(x_0, y_0)>0[/latex], then [latex]f[/latex] has a
**local minimum**at [latex](x_0, y_0)[/latex]. - If [latex]D>0[/latex] and [latex]f_{xx}(x_0, y_0)<0[/latex], then [latex]f[/latex] has a
**local maximum**at [latex](x_0, y_0)[/latex]. - If [latex]D<0[/latex], then [latex]f[/latex] has a
**saddle point**at [latex](x_0, y_0)[/latex]. - If [latex]D=0[/latex], then the test is inconclusive.

- Input a function of two variables into the [latex]f(x,y)[/latex] input function section.
- Move the point on the plane around and the Jacobian determinant will automatically be calculated for you. The equation for each is provided, where the determinant of the jacobian represents the D value from the formula above.
- Once you hover over a local maximum, local minimum or a saddle point, a text will appear notifying you of the answer.

Check your understanding by solving the following questions[footnote]Gilbert Strang, Edwin “Jed” Herman, OpenStax, Calculus Volume 3, Calculus Volumes 1, 2, and 3 are licensed under an Attribution-NonCommercial-Sharealike 4.0 International License (CC BY-NC-SA).[/footnote]:
Use the first and second derivative tests to identify any critical points and determine whether each critical point is a maximum, minimum, saddle point or none of these.

- [latex]f(x,y) = -x^3+4xy-2y^2+1[/latex]
- [latex]f(x,y) = 2xye^{-x2-y2}[/latex]

[latex]y=y_0+f'(x_0)(x-x_0)[/latex].

The[latex] z = z_0 + f_x(x_0,y_0)(x - x_0) + f_y(x_0,y_0)(y-y_0)[/latex].

- Input a 3D surface function in the function box in the plot. The function can be a single variable function or a double variable function.
- Adjust point [latex]P[/latex] using the sliders or by dragging the point on the graph below.
- The tangent plane equation will be depicted on the plot.

Check your understanding by solving the following questions[footnote]Gilbert Strang, Edwin “Jed” Herman, OpenStax, Calculus Volume 3, Calculus Volumes 1, 2, and 3 are licensed under an Attribution-NonCommercial-Sharealike 4.0 International License (CC BY-NC-SA).[/footnote]:

- Find the equation of the tangent plane to the surface defined by the function [latex]x^2+10xyz+y^2+8z^2=0,P(−1,−1,−1)[/latex]
- Find the equation of the tangent plane to the surface defined by the function [latex]h(x,y) = ln(x^2) + y^2[/latex] at Point [latex](x_0,y_0) = (3,4)[/latex].

[h5p id="9"]

**Directional derivatives** look to extend the concept of **partial derivative**s by finding the tangent line parallel to neither the [latex]x[/latex]-axis or [latex]y[/latex]-axis.

[latex] D_u f(x_0,y_0) = ∇f \cdot \vec{u}=f_x(x_0, y_0) \, a+ f_y(x_0, y_0) \,b [/latex],

where [latex]a=\cos\theta[/latex] and [latex]b=\sin \theta[/latex]. Consider two special cases of directional derivatives:- When [latex] \theta = 0 [/latex], we travel in the direction that is parallel to positive [latex]x[/latex]-axis, so the direction [latex]\vec{u} = \langle \cos0, \, \sin0\rangle = \langle 1,\,0\rangle [/latex] and the corresponding directional derivative is [latex]D_u f(x,y) = f_x(x_0,y_0)\,1+ f_y(x_0,y_0)\,0= f_x(x_0,y_0)[/latex].
- When [latex] \theta = \frac{\pi}{2}[/latex], we travel in the direction that is parallel to positive [latex]y[/latex]-axis, so the direction [latex] \vec{u} = \langle \cos \frac{\pi}{2} , \, \sin \frac{\pi}{2} \rangle = \langle 0,\, 1 \rangle[/latex] and the directional derivative is [latex] D_u f(x_0,y_0) = f_x(x_0,y_0)\,0+ f_y(x_0,y_0)\,1= f_y(x_0,y_0)[/latex].

[latex] D_u f(x_0,y_0, z_0) = ∇f \cdot \vec{u}=f_x(x_0, y_0, z_0)\, a + f_y(x_0, y_0, z_0) \,b + f_y(x_0, y_0, z_0) \,c[/latex].

- There are two separate plots where the direction vector (i.e., the direction of the derivative, denoted by [latex]\vec{u}[/latex]) is defined by either an angle in radians or a vector.
- Point (P) is adjusted with the [latex]x[/latex] and [latex]y[/latex] sliders.
- [latex]\vec{u}[/latex] is selected by either the angle or vector and is indicated by the red arrow on the graph.

Check your understanding by solving the following questions[footnote]Gilbert Strang, Edwin “Jed” Herman, OpenStax, Calculus Volume 3, Calculus Volumes 1, 2, and 3 are licensed under an Attribution-NonCommercial-Sharealike 4.0 International License (CC BY-NC-SA).[/footnote]:

- Find the gradient, [latex]∇f(x,y)[/latex], of the function: [latex] f(x,y) = x^2 - xy + 3y^2[/latex]
- Find the directional derivative, [latex] D_u f[/latex], of the function: [latex] f(x,y,z) = e^{-2z} sin(2x)cos(2y)[/latex] at point (0,1).

**2D Notation:**[latex] \vec{F}(x, y) = P(x, y)\vec{i} + Q(x, y)\vec{j} = \langle P(x, y) , Q(x, y)\rangle[/latex]**3D Notation:**[latex] \vec{F}(x,y,z) = P(x, y,z)\vec{i}+ Q(x,y,z)\vec{j} + R(x,y,z)\vec{k} = \langle P(x, y,z) , Q(x, y,z) , R(x,y,z)\rangle[/latex]

- The vector definition is done using [latex]P[/latex] and [latex]Q[/latex].
- Y Grid and X Grid control the number of arrows that will appear in the 2D plot.
- Xmin and Ymin set the minimum boundaries for the plot.
- YMax and Xmax set the maximum boundaries for the plot.

Check your understanding by solving the following questions[footnote]Gilbert Strang, Edwin “Jed” Herman, OpenStax, Calculus Volume 3, Calculus Volumes 1, 2, and 3 are licensed under an Attribution-NonCommercial-Sharealike 4.0 International License (CC BY-NC-SA).[/footnote]:

- Draw the following vector field [latex]\vec{F}(x,y)=x\vec{i} + y\vec{j}[/latex]
- Draw the following vector field [latex] \vec{F}(x, y, z) = 2x\vec{i} − 2y\vec{j} − 2z\vec{k} [/latex]

**Type I double integral**: [latex]\int_{a}^{b} \int_{h(x)}^{g(x)} f(x,y)\,dy dx[/latex], where [latex]x=a[/latex] and [latex]x=b[/latex] are the lower and upper bounds of [latex]x[/latex]; [latex]y=h(x)[/latex] and [latex]y=g(x)[/latex] are the lower and upper bounds of [latex]y[/latex].**Type II double integral:**[latex]\int_{a}^{b} \int_{h(x)}^{g(x)} f(x,y)\,dx dy[/latex], where [latex]y=a[/latex] and [latex]y=b[/latex] are the lower and upper bounds of [latex]y[/latex]; [latex]x=h(y)[/latex] and [latex]x=g(y)[/latex]are the lower and upper bounds of [latex]y[/latex].

- Assume you have a Type I integral [latex]\int_{0}^{1} \int_{-x}^{x^2} y^2 x\,dy dx[/latex]. Input [latex]y^2 x[/latex] into the [latex]f(x,y)[/latex] input function section.
- Input [latex]\textrm{If} (0 \leq x \leq 1, x^2)[/latex] into the Upper [latex]y[/latex] function, i.e., [latex]g(x)[/latex] section.
- Input [latex]\textrm{If} (0 \leq x \leq 1, -x)[/latex] into the lower [latex]y[/latex] function, i.e., [latex]h(x)[/latex] section.
- Use the slider for the value of x to see the change of the area of the cross-section, [latex]A(x)[/latex].
- The result of this double integral is dynamically calculated at the bottom.
- You can also use this plot for the Type II integral by switching [latex]x[/latex] and [latex]y.[/latex]

Check your understanding by solving the following questions[footnote]Gilbert Strang, Edwin “Jed” Herman, OpenStax, Calculus Volume 3, Calculus Volumes 1, 2, and 3 are licensed under an Attribution-NonCommercial-Sharealike 4.0 International License (CC BY-NC-SA).[/footnote]:

- [latex]\int_{0}^{1} \int_{2\sqrt{x}}^{2\sqrt{x}+1} xy+1\,dy dx[/latex].
- [latex] \int_{0}^{1} \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} 2x+4x^3\,dx dy[/latex]

[latex]r =\sqrt{x^2+y^2}[/latex] and [latex]\tan(\theta) = \frac{y}{x}[/latex],

or convert a point from polar to rectangular coordinates using the following equations:[latex]x =r \cos\theta[/latex] and [latex]y = r \sin\theta[/latex].

The double integral [latex]\iint_D f(x,y)\,dA[/latex] in rectangular coordinates can be converted to a double integral in polar coordinates as [latex]\iint_D f(r \cos\theta, r \sin\theta)\,r\,dr d\theta[/latex].- Change the bounds on the double integral in polar coordinates for both the [latex]r[/latex] and [latex]\theta[/latex] bounds. The bounded region will be shown in the plot and [latex]t[/latex] in the plot represents [latex]\theta[/latex].
- The result of the double integral in polar coordinates will be shown too.

Check your understanding by solving the following questions[footnote]Gilbert Strang, Edwin “Jed” Herman, OpenStax, Calculus Volume 3, Calculus Volumes 1, 2, and 3 are licensed under an Attribution-NonCommercial-Sharealike 4.0 International License (CC BY-NC-SA).[/footnote]:

- [latex]\iint_{D} 3x \, dA [/latex] where [latex]R={(r,\theta)| 0 \leq r \leq 1, 0 \leq \theta \leq 2}[/latex].
- [latex]\iint_{D} 1-x^2-y^2 \, dA [/latex] where [latex]R={(r,\theta)| 0 \leq r \leq 1, 0 \leq \theta \leq 2\pi}[/latex].

- You are able to change the bounds on the triple integral in rectangular coordinates.
- You may input your function, [latex]f(x,y,z)[/latex], to be integrated at the bottom as well, in which the triple integral of said function will be presented at the top of the screen in the beige area.
- You may also change the grid size of the 3D solid depicted on the screen for the function [latex]f(x,y,z) = 1[/latex].

Check your understanding by solving the following questions[footnote]Gilbert Strang, Edwin “Jed” Herman, OpenStax, Calculus Volume 3, Calculus Volumes 1, 2, and 3 are licensed under an Attribution-NonCommercial-Sharealike 4.0 International License (CC BY-NC-SA).[/footnote].
Evaluate the triple integrals over the rectangular solid box B.

- [latex]\iiint_D (2x + 3y^2 + 4z^3)\, dV[/latex], where [latex]B={(x,y,z)| 0 \leq x \leq 1, 0 \leq y \leq 2, 0 \leq z \leq 3}[/latex]
- [latex]\iiint_D z \sin(x) + y2) \, dV[/latex], where [latex]B={(x,y,z)| 0 \leq x \leq \pi, 0 \leq y \leq 12, -1 \leq z \leq 2}[/latex]

- Change the bounds on the triple integral in cylindrical coordinates where and represent the outermost bounds. [latex]\alpha[/latex] and [latex]\beta[/latex] are constants and they are the lower and upper bounds of angle [latex]\theta[/latex]. [latex]r_1[/latex] and [latex]r_2[/latex] are functions of [latex]\theta[/latex]), and they are the lower and upper bounds of [latex]r[/latex]. [latex]u_1[/latex] and [latex]u_2[/latex] are functions of [latex]r[/latex] and [latex]\theta[/latex], and they are the lower and upper bounds of [latex]z[/latex].
- You may also change the grid size of the 3D solid depicted on the screen for the function [latex]f(x,y,z)[/latex].

Check your understanding by solving the following questions[footnote]Gilbert Strang, Edwin “Jed” Herman, OpenStax, Calculus Volume 3, Calculus Volumes 1, 2, and 3 are licensed under an Attribution-NonCommercial-Sharealike 4.0 International License (CC BY-NC-SA).[/footnote]:
Evaluate the triple integrals [latex]f(x,y,z)[/latex] over the solid [latex]E[/latex].

- [latex]E = {(x,y,z)| x^2+y^2 \leq 9, x \geq 0, y\geq 0, 0 \leq z \leq 1}, f(x,y,z) = z[/latex]
- [latex]E = {(x,y,z)| 1 \leq x^2+y^2 \leq 9, y \geq 0, 0 \leq z \leq 1}, f(x,y,z) = x^2+y^2[/latex]

- Fill in function 1 (i.e., [latex]f(x,y,z)[/latex]) and function 2 (i.e., [latex]g(x,y,z)[/latex]) with your desired quadric surfaces.
- The graph depicted on the right shows their intersection.

Check your understanding by solving the following questions[footnote]Gilbert Strang, Edwin “Jed” Herman, OpenStax, Calculus Volume 3, Calculus Volumes 1, 2, and 3 are licensed under an Attribution-NonCommercial-Sharealike 4.0 International License (CC BY-NC-SA).[/footnote]:
Plot the given quadric surface and specify the name of said quadric surface:

- [latex]x^2/4 + y^2/9 - z^2/12 = 1[/latex]
- [latex]z^2 = 4x^2 + 3y^2[/latex]

[latex]\int_C f(x,y)\,ds = \int_a^b f(x(t),y(t))\, \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \,\,dt[/latex].

The value of [latex]\int_C f(x,y)\,ds[/latex] is the area of the “wall”, "fence" or “curtain” whose base is the 2D curve [latex]C[/latex] on the [latex]xy[/latex]-plane and and whose height is given by the function [latex]z=f(x,y)[/latex]. The concept of line integral can be extended to high dimensions. For example, [latex]\int_C f(x,y,z)\,ds [/latex] integrates the function with three variables, [latex]w=f(x,y,z)[/latex], along a 3D curve C that is parameterized by [latex]r(t) = \langle x(t),y(t),z(t) \rangle [/latex]. It can be evaluated by a single integral as well, that is,[latex]\int_C f(x,y,z)\,ds = \int_a^b f(x(t),y(t),z(t))\, \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2+(\frac{dz}{dt})^2} \,\,dt[/latex].

- Input the function [latex]f(x, y)[/latex].
- Adjust the 2D curve [latex]C[/latex] on the [latex]xy[/latex]-plane.
- Adjust the number of rectangular subareas, [latex]n[/latex].
- The estimation of the line integral is shown. The larger [latex]n[/latex] is, the better the estimation is.

Check your understanding by solving the following questions[footnote]Gilbert Strang, Edwin “Jed” Herman, OpenStax, Calculus Volume 3, Calculus Volumes 1, 2, and 3 are licensed under an Attribution-NonCommercial-Sharealike 4.0 International License (CC BY-NC-SA).[/footnote]:

- Find the value of integral [latex]\int_C(x^2+y^2)\, ds[/latex], where [latex]C[/latex] is part of the helix parameterized by [latex]r(t)=\langle cos t, sin t \rangle, 0 \leq t \leq 2[/latex].
- Evaluate [latex]\int_C \frac{1}{x^2+y^2} \, ds[/latex] , over the line segment from [latex](1,1)[/latex] to [latex](3,0)[/latex].

[latex]\vec{F}(x,y) = \langle P(x, y) , Q(x, y) \rangle[/latex] or [latex]\vec{F}(x,y,z)= \langle P(x, y,z) , Q(x, y,z) , R(x,y,z)\rangle[/latex]

Looking to answer the question of how we can compute the work done by the river of moving the kayak along route [latex]C[/latex], we can calculate the work [latex]W[/latex] done by force field [latex]\vec{F}[/latex] along the curve [latex]C[/latex] as the following equation[latex]W = \int_C \vec{F} \cdot dr= \int_a^b F(r(t)) r'(t)\,dt[/latex]

- Fill in [latex] P(x, y)[/latex] and [latex]Q(x, y)[/latex] (i.e., the first and second compartments of the vector field function).
- Adjust xmin, xmax, ymin and max and they are the lower and upper bounds for the [latex]x[/latex]-axis and [latex]y[/latex]-axis.
- Input the function for curve C [latex]y=f(x)[/latex] (i.e., the trajectory that the object travels along).
- Adjust [latex]a[/latex] and [latex]b[/latex] (i.e., the [latex]x[/latex]-coordinates of the start and end points of curve C).
- The result of the work is shown.

Check your understanding by solving the following question[footnote]Gilbert Strang, Edwin “Jed” Herman, OpenStax, Calculus Volume 3, Calculus Volumes 1, 2, and 3 are licensed under an Attribution-NonCommercial-Sharealike 4.0 International License (CC BY-NC-SA).[/footnote]:

- Find the work done by vector field [latex] \vec{F}(x,y)=y\vec{i}+2x\vec{j} [/latex] in moving an object along path [latex]C[/latex], which joins points (1,0) and (0,1).

- Fill in [latex]P(x, y,z)[/latex], [latex]Q(x, y,z)[/latex] and [latex]R(x, y,z) [/latex](i.e., three compartments of the vector field function).
- Input the surface function.
- The graph depicted shows the flux.

Check your understanding by solving the following question[footnote]Gilbert Strang, Edwin “Jed” Herman, OpenStax, Calculus Volume 3, Calculus Volumes 1, 2, and 3 are licensed under an Attribution-NonCommercial-Sharealike 4.0 International License (CC BY-NC-SA).[/footnote]:

- Consider the radial field [latex]\vec{F}(x,y,z)= \frac{\langle x,y,z \rangle}{(x^2+y^2+z^2)}[/latex] and sphere [latex]S[/latex] centred at the origin with radius 1. Find the total outward flux across [latex]S[/latex].

- If [latex] \vec{F}=\langle P(x,y), Q(x,y) \rangle [/latex] is a vector field in 2D, and [latex]P_x [/latex] and [latex]Q_y[/latex] both exist, then the divergence of [latex]\vec{F}[/latex] is defined by [latex] div(F) =P_x+Q_y = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}[/latex].

- If [latex] \vec{F}=\langle P(x,y,z), Q(x,y,z), R(x,y,z)\rangle [/latex] is a vector field in 3D and [latex] P_x [/latex], [latex]Q_y[/latex], and [latex]R_z[/latex] all exist, then the divergence of [latex]\vec{F}[/latex] is defined by [latex] div(F) =P_x+Q_y+R_z = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} [/latex].

- If [latex] \vec{F}=\langle P, Q \rangle [/latex] is a vector field in 2D, and [latex] P_x [/latex] and [latex] Q_y[/latex] both exist, then the curl of [latex]\vec{F}[/latex] is defined by [latex] Curl(\vec{F}) = (Q_x - P_y)\vec{k} = \langle 0, 0, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \rangle [/latex].

- If [latex] \vec{F}=\langle P, Q ,R\rangle [/latex] is a vector field in 3D, and [latex] P_x, Q_y, and R_z[/latex] all exist, then the curl of [latex]\vec{F}[/latex] is defined by [latex] Curl(\vec{F}) = (R_y-Q_z)\vec{i} + (P_z-R_x)\vec{j} + (Q_x - P_y)\vec{k} = \langle \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}. \rangle [/latex]

- Fill in a field function.
- Choose a path.
- The graph depicted shows the divergence and curl.

Check your understanding by solving the following question[footnote]Gilbert Strang, Edwin “Jed” Herman, OpenStax, Calculus Volume 3, Calculus Volumes 1, 2, and 3 are licensed under an Attribution-NonCommercial-Sharealike 4.0 International License (CC BY-NC-SA).[/footnote]:

- Find the divergence and Curl of [latex] D_u f[/latex] of the function: [latex]f(x,y,z) = x(cos(y))\vec{i} +xy^2\vec{j}[/latex]

**The Concept:**In this section, we’ll share the key concepts you’ll need to know for the unit topic.**The Plot:**In this section, we’ll provide step-by-step instructions on engaging with a 3D plot related to the unit topic. Following the instructions, you should be able to manipulate the 3D graph to understand the key concepts for the unit.**Self-Checking Questions:**In this section, you’ll find questions to test your understanding of the unit concepts. Answers to some of the questions will be provided; however, some questions will only be able to be found through using the 3D plot in the unit.

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