Multivariable Calculus Comprehensive Cheat Sheet
1. Partial Derivatives
1.1 Definition of Partial Derivatives
The partial derivative of a function \( f(x, y) \) with respect to \( x \) is the derivative of \( f \) with \( y \) held constant, denoted as \( \frac{\partial f}{\partial x} \). Similarly, the partial derivative with respect to \( y \) is denoted as \( \frac{\partial f}{\partial y} \).
Example:
Find the partial derivatives of \( f(x, y) = x^2y + y^3 \).
- Partial derivative with respect to \(x\): \( \frac{\partial f}{\partial x} = 2xy \).
- Partial derivative with respect to \(y\): \( \frac{\partial f}{\partial y} = x^2 + 3y^2 \).
- Result: \( \frac{\partial f}{\partial x} = 2xy \), \( \frac{\partial f}{\partial y} = x^2 + 3y^2 \).
2. Gradient and Directional Derivatives
2.1 Gradient Vector
The gradient of a scalar function \( f(x, y, z) \) is a vector field given by:
\[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \]
The gradient points in the direction of the steepest ascent of the function.
Example:
Compute the gradient of \( f(x, y, z) = x^2 + y^2 + z^2 \).
- Compute the partial derivatives: \( \frac{\partial f}{\partial x} = 2x \), \( \frac{\partial f}{\partial y} = 2y \), \( \frac{\partial f}{\partial z} = 2z \).
- Result: \( \nabla f = (2x, 2y, 2z) \).
2.2 Directional Derivative
The directional derivative of \( f \) in the direction of a unit vector \( \mathbf{u} = \langle a, b, c \rangle \) is given by:
\[ D_{\mathbf{u}}f = \nabla f \cdot \mathbf{u} = \frac{\partial f}{\partial x}a + \frac{\partial f}{\partial y}b + \frac{\partial f}{\partial z}c \]
Example:
Find the directional derivative of \( f(x, y, z) = x^2 + y^2 + z^2 \) at the point \( (1, 1, 1) \) in the direction of the vector \( \mathbf{v} = \langle 1, 2, 2 \rangle \).
- Normalize \( \mathbf{v} \) to find the unit vector \( \mathbf{u} = \frac{1}{3}\langle 1, 2, 2 \rangle \).
- Compute the gradient \( \nabla f = \langle 2x, 2y, 2z \rangle = \langle 2, 2, 2 \rangle \) at \( (1, 1, 1) \).
- Compute the directional derivative: \( D_{\mathbf{u}}f = \nabla f \cdot \mathbf{u} = \frac{2}{3}(1 + 2 + 2) = \frac{10}{3} \).
- Result: The directional derivative is \( \frac{10}{3} \).
3. Multiple Integrals
3.1 Double Integrals
A double integral is used to integrate a function over a two-dimensional region. It is denoted as:
\[ \iint_R f(x, y) \, dA \]
where \( dA \) represents the differential area element. The limits of integration correspond to the region \( R \) over which the function is integrated.
Example:
Compute \( \iint_R (x + y) \, dA \) where \( R \) is the region bounded by \( x = 0 \), \( y = 0 \), and \( x + y = 1 \).
- Set up the integral: \( \iint_R (x + y) \, dA = \int_0^1 \int_0^{1-y} (x + y) \, dx \, dy \).
- Integrate with respect to \(x\): \( \int_0^1 \left[ \frac{x^2}{2} + yx \right]_0^{1-y} \, dy \).
- Simplify and integrate with respect to \(y\).
- Result: \( \frac{1}{6} \).
3.2 Triple Integrals
A triple integral extends the concept of a double integral to three dimensions. It is used to integrate a function over a three-dimensional region:
\[ \iiint_V f(x, y, z) \, dV \]
where \( dV \) represents the differential volume element.
Example:
Compute \( \iiint_V z \, dV \) where \( V \) is the region bounded by \( x^2 + y^2 \leq 1 \), \( z = 0 \), and \( z = 1 - x^2 - y^2 \).
- Set up the integral in cylindrical coordinates: \( \iiint_V z \, r \, dz \, dr \, d\theta \).
- Determine the limits: \( z \) from 0 to \( 1 - r^2 \), \( r \) from 0 to 1, \( \theta \) from 0 to \( 2\pi \).
- Integrate with respect to \(z\), then \(r\), and finally \( \theta \).
- Result: \( \frac{\pi}{8} \).
4. Surface Integrals
4.1 Surface Area
The surface area of a surface parameterized by \( \mathbf{r}(u, v) = \langle x(u, v), y(u, v), z(u, v) \rangle \) is given by:
\[ A = \iint_D \left|\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}\right| \, du \, dv \]
Example:
Find the surface area of the portion of the plane \( z = 1 + x + y \) that lies above the region \( x^2 + y^2 \leq 1 \).
- Parameterize the surface: \( \mathbf{r}(u, v) = \langle u, v, 1 + u + v \rangle \).
- Compute the cross product \( \mathbf{r}_u \times \mathbf{r}_v \) and find its magnitude.
- Set up the integral over the region \( x^2 + y^2 \leq 1 \).
- Result: The surface area is \( 2\sqrt{2} \pi \).
5. Line Integrals
5.1 Line Integrals of Scalar Fields
A line integral of a scalar field along a curve \( C \) is given by:
\[ \int_C f(x, y, z) \, ds \]
where \( ds \) is the differential arc length along the curve \( C \).
Example:
Evaluate \( \int_C (xy) \, ds \) where \( C \) is the line segment from \( (0, 0) \) to \( (1, 1) \).
- Parametrize the curve: \( x(t) = t \), \( y(t) = t \), \( 0 \leq t \leq 1 \).
- Compute \( ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt = \sqrt{2} \, dt \).
- Set up the integral: \( \int_0^1 (t^2) \sqrt{2} \, dt = \frac{\sqrt{2}}{3} \).
- Result: \( \frac{\sqrt{2}}{3} \).
5.2 Line Integrals of Vector Fields
For a vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} \), the line integral along a curve \( C \) is:
\[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_C P \, dx + Q \, dy + R \, dz \]
Example:
Evaluate \( \int_C \mathbf{F} \cdot d\mathbf{r} \) for \( \mathbf{F} = (y, -x, z) \) along the curve \( C \) given by \( x = \cos t \), \( y = \sin t \), \( z = t \), \( 0 \leq t \leq 2\pi \).
- Compute \( \mathbf{F} \cdot d\mathbf{r} = \sin t(-\sin t) + (-\cos t)(\cos t) + t \, dt = -1 + t \, dt \).
- Set up the integral: \( \int_0^{2\pi} (-1 + t) \, dt \).
- Result: The integral evaluates to \( -2\pi^2 \).
6. Vector Calculus
6.1 Divergence and Curl
The divergence of a vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} \) is:
\[ \nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \]
The curl of \( \mathbf{F} \) is:
\[ \nabla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \]
Example:
Compute the divergence and curl of \( \mathbf{F}(x, y, z) = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} \).
- Divergence: \( \nabla \cdot \mathbf{F} = 1 + 1 + 1 = 3 \).
- Curl: \( \nabla \times \mathbf{F} = \mathbf{0} \) (since all mixed partial derivatives are zero).
- Result: Divergence = 3, Curl = \( \mathbf{0} \).
7. Green's Theorem
Green's Theorem relates a line integral around a simple closed curve \( C \) to a double integral over the plane region \( R \) bounded by \( C \):
\[ \oint_C (P \, dx + Q \, dy) = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \]
Example:
Use Green's Theorem to evaluate \( \oint_C (x \, dy - y \, dx) \) where \( C \) is the circle \( x^2 + y^2 = 1 \).
- Set up the double integral: \( \iint_R \left( \frac{\partial (-y)}{\partial x} - \frac{\partial x}{\partial y} \right) \, dA = \iint_R (-1 - 1) \, dA \).
- Evaluate the area integral over the unit disk.
- Result: The integral evaluates to \( -2\pi \).
8. Stokes' Theorem
Stokes' Theorem relates a surface integral of a curl over a surface \( S \) to a line integral over the boundary curve \( C \) of \( S \):
\[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} \]
Example:
Use Stokes' Theorem to evaluate \( \oint_C \mathbf{F} \cdot d\mathbf{r} \) for \( \mathbf{F} = (y, -x, z) \) where \( C \) is the boundary of the disk \( x^2 + y^2 \leq 1 \), \( z = 0 \).
- Compute \( \nabla \times \mathbf{F} = (1, 1, -2) \).
- Set up the surface integral over the disk.
- Result: The integral evaluates to \( 2\pi \).
9. Divergence Theorem
The Divergence Theorem relates a flux integral over a closed surface \( S \) to a triple integral over the volume \( V \) enclosed by \( S \):
\[ \iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_V (\nabla \cdot \mathbf{F}) \, dV \]
Example:
Use the Divergence Theorem to compute the flux of \( \mathbf{F} = (x, y, z) \) across the surface of the unit sphere \( x^2 + y^2 + z^2 = 1 \).
- Compute the divergence \( \nabla \cdot \mathbf{F} = 3 \).
- Set up the triple integral over the volume of the sphere.
- Result: The flux is \( 4\pi \).
10. Lagrange Multipliers
Lagrange multipliers are used to find the local maxima and minima of a function subject to equality constraints. For a function \( f(x, y, z) \) with a constraint \( g(x, y, z) = 0 \), the method involves solving:
\[ \nabla f = \lambda \nabla g \]
where \( \lambda \) is the Lagrange multiplier.
Example:
Find the maximum and minimum values of \( f(x, y) = x^2 + y^2 \) subject to the constraint \( x^2 + y^2 = 1 \).
- Compute the gradients \( \nabla f = (2x, 2y) \) and \( \nabla g = (2x, 2y) \).
- Set up the equation \( 2x = \lambda 2x \), \( 2y = \lambda 2y \).
- Solve for \( \lambda \) and find the critical points.
- Result: Maximum at \( (1, 0) \) or \( (0, 1) \), minimum at \( (-1, 0) \) or \( (0, -1) \).