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cheat sheet calculus 3

cheat sheet calculus 3

4 min read 20-10-2024
cheat sheet calculus 3

Calculus 3 Cheat Sheet: A Quick Refresher

Calculus 3, also known as multivariable calculus, dives into the exciting world of functions with multiple variables. This branch of mathematics extends the concepts of calculus to higher dimensions, leading to a deeper understanding of geometry, physics, and other fields. If you're feeling overwhelmed by the vastness of Calculus 3, this cheat sheet will be your guide to crucial formulas, concepts, and problem-solving strategies.

1. Vectors and Vector Operations

What are vectors?

Vectors are mathematical objects possessing both magnitude (length) and direction. They are often represented as arrows in space.

How are vectors added and subtracted?

  • Addition: Add corresponding components of each vector. For example, the sum of vectors u=(u1,u2,u3)\textbf{u} = (u_1, u_2, u_3) and v=(v1,v2,v3)\textbf{v} = (v_1, v_2, v_3) is u+v=(u1+v1,u2+v2,u3+v3)\textbf{u} + \textbf{v} = (u_1 + v_1, u_2 + v_2, u_3 + v_3).
  • Subtraction: Subtract corresponding components. uv=(u1v1,u2v2,u3v3)\textbf{u} - \textbf{v} = (u_1 - v_1, u_2 - v_2, u_3 - v_3).

How do you multiply a vector by a scalar?

Multiply each component of the vector by the scalar. For example, cu=(cu1,cu2,cu3)c\textbf{u} = (cu_1, cu_2, cu_3).

What are the dot product and cross product?

  • Dot Product: Measures the projection of one vector onto another. It's a scalar value calculated as uv=u1v1+u2v2+u3v3\textbf{u} \cdot \textbf{v} = u_1v_1 + u_2v_2 + u_3v_3.
  • Cross Product: Produces a vector orthogonal (perpendicular) to both input vectors. It's calculated as u×v=(u2v3u3v2,u3v1u1v3,u1v2u2v1)\textbf{u} \times \textbf{v} = (u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1).

Example:

Let u=(1,2,3)\textbf{u} = (1, 2, 3) and v=(4,5,6)\textbf{v} = (4, 5, 6).

  • u+v=(1+4,2+5,3+6)=(5,7,9)\textbf{u} + \textbf{v} = (1+4, 2+5, 3+6) = (5, 7, 9)
  • uv=(1)(4)+(2)(5)+(3)(6)=32\textbf{u} \cdot \textbf{v} = (1)(4) + (2)(5) + (3)(6) = 32
  • u×v=(3,6,3)\textbf{u} \times \textbf{v} = (-3, 6, -3)

2. Functions of Multiple Variables

How are functions of multiple variables defined?

They take multiple input variables and produce a single output. For example, f(x,y)=x2+y2f(x, y) = x^2 + y^2 takes two inputs (xx and yy) and outputs their squares summed.

What are level curves and surfaces?

  • Level Curves: Represent sets of points where a function of two variables has a constant output. They are obtained by setting f(x,y)=cf(x, y) = c, where cc is a constant.
  • Level Surfaces: Similar to level curves, they represent sets of points where a function of three variables has a constant output. They are obtained by setting f(x,y,z)=cf(x, y, z) = c.

Example:

The level curves of the function f(x,y)=x2+y2f(x, y) = x^2 + y^2 are circles centered at the origin with radii equal to the square root of the constant value.

3. Partial Derivatives

What are partial derivatives?

Partial derivatives measure the rate of change of a multivariable function with respect to one variable, keeping all other variables constant.

How do you calculate partial derivatives?

Treat all other variables as constants and differentiate the function with respect to the desired variable.

Example:

For f(x,y)=x2+y2f(x, y) = x^2 + y^2:

  • fx=2x\frac{\partial f}{\partial x} = 2x (treat yy as a constant)
  • fy=2y\frac{\partial f}{\partial y} = 2y (treat xx as a constant)

4. Gradient and Directional Derivatives

What is the gradient?

The gradient of a function is a vector containing all its partial derivatives. It points in the direction of the steepest ascent of the function.

What is the directional derivative?

The directional derivative measures the rate of change of a function along a specific direction. It's calculated as the dot product of the gradient and a unit vector pointing in the desired direction.

Example:

For f(x,y)=x2+y2f(x, y) = x^2 + y^2:

  • The gradient is f=(2x,2y)\nabla f = (2x, 2y)
  • The directional derivative in the direction of u=(1,1)\textbf{u} = (1, 1) is fu=(2x,2y)(1,1)=2x+2y\nabla f \cdot \textbf{u} = (2x, 2y) \cdot (1, 1) = 2x + 2y.

5. Optimization in Multiple Variables

What is optimization?

Finding the maximum or minimum values of a function within a given domain.

How do you find critical points?

Critical points occur where the gradient is zero or undefined.

How do you classify critical points?

Use the second partial derivative test to determine whether a critical point is a maximum, minimum, or saddle point.

Example:

To find the critical points of f(x,y)=x2+y22x4yf(x, y) = x^2 + y^2 - 2x - 4y, we find the gradient: f=(2x2,2y4)\nabla f = (2x - 2, 2y - 4). Setting this equal to zero, we get the critical point (1,2)(1, 2). Applying the second partial derivative test, we find it's a local minimum.

6. Multiple Integrals

What are multiple integrals?

Integrals used to calculate volumes, areas, and other quantities in multivariable calculus.

How do you evaluate double integrals?

Iterate through integration with respect to one variable at a time, treating the other variable as a constant.

How do you evaluate triple integrals?

Similar to double integrals, iterate through integration with respect to three variables.

Example:

To calculate the volume of a solid bounded by the planes x=0x = 0, y=0y = 0, z=0z = 0, and x+y+z=1x + y + z = 1, we can use a triple integral: 0101x01xydzdydx=1/6\int_0^1 \int_0^{1-x} \int_0^{1-x-y} dzdydx = 1/6.

7. Line Integrals

What are line integrals?

Integrals taken along a curve. They can be used to calculate work done by a force along a path, for instance.

How do you evaluate line integrals?

Parametrize the curve and express the integral in terms of the parameter.

Example:

To calculate the line integral of the function f(x,y)=x+yf(x, y) = x + y along the curve CC parametrized by r(t)=(t,t2)\textbf{r}(t) = (t, t^2) from t=0t = 0 to t=1t = 1, we evaluate: Cf(x,y)ds=01(t+t2)1+4t2dt\int_C f(x, y) ds = \int_0^1 (t + t^2)\sqrt{1 + 4t^2} dt.

8. Surface Integrals

What are surface integrals?

Integrals taken over a surface. They can be used to calculate flux, which is the rate of flow of a fluid through a surface.

How do you evaluate surface integrals?

Parametrize the surface and use a surface integral formula.

Example:

To calculate the flux of the vector field F=(x,y,z)\textbf{F} = (x, y, z) across the surface SS defined by the equation z=x2+y2z = x^2 + y^2, we need to parametrize the surface and apply the surface integral formula.

Conclusion

This cheat sheet provides a basic overview of crucial concepts in Calculus 3. Remember, practice is key to mastering these concepts. Use online resources like Khan Academy, MIT OpenCourseware, and the textbook examples for further exploration.

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