0.1 Vectors

Addition and scalar multiplication

Vectors are added component-wise: \[\vb A + \vb B = (A_x + B_x)\,\uv\imath + (A_y + B_y)\,\uv\jmath + (A_z + B_z)\,\uv k .\] Multiplying by a scalar \(s\) multiplies every component: \[s\vb A = sA_x\,\uv\imath + sA_y\,\uv\jmath + sA_z\,\uv k .\]

The dot product

The basis vectors satisfy \[\uv\imath\cdot\uv\imath = \uv\jmath\cdot\uv\jmath = \uv k\cdot\uv k = 1, \qquad \uv\imath\cdot\uv\jmath = \uv\jmath\cdot\uv k = \uv\imath\cdot\uv k = 0 ,\] so, for general vectors, \[\begin{equation} \tag{0.2} \keyeq{\vb A\cdot\vb B = A_x B_x + A_y B_y + A_z B_z} \end{equation}\] The dot product is a scalar, and it is symmetric: \(\vb A\cdot\vb B = \vb B\cdot\vb A\).

Unit vectors

A hat denotes a vector of length one. Any non-zero vector \(\vb A\) defines a unit vector in its direction, \[\keyeq{\displaystyle \uv A = \frac{\vb A}{|\vb A|}}\] so that \(\vb A = |\vb A|\,\uv A\): a length times a direction.

The cross product

The cross product of two vectors is another vector, most easily remembered as a determinant: \[\vb A\times\vb B = \begin{vmatrix} \uv\imath & \uv\jmath & \uv k \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} = (A_y B_z - A_z B_y)\,\uv\imath + (A_z B_x - A_x B_z)\,\uv\jmath + (A_x B_y - A_y B_x)\,\uv k .\] It is antisymmetric: \(\vb A\times\vb B = -\,\vb B\times\vb A\).

0.2 Coordinate systems

Polar coordinates

The area element is \[\keyeq{dA = dx\,dy = r\,dr\,d\theta}\]

Cylindrical coordinates

Add \(z\) to polar coordinates: \((x,y,z)\to(r,\theta,z)\), with \(dV = r\,dr\,d\theta\,dz\).

Spherical coordinates

The projection onto the \(xy\)-plane has length \(r\sin\theta\), so \[\left\{ \begin{aligned} x &= r\sin\theta\cos\phi \\ y &= r\sin\theta\sin\phi \\ z &= r\cos\theta \end{aligned} \right. \qquad\qquad \keyeq{dV = dx\,dy\,dz = r^2\sin\theta\,dr\,d\theta\,d\phi}\]

0.3 Derivatives

Notation

A prime denotes a derivative with respect to \(x\), and a dot a derivative with respect to time \(t\): \[\keyeq{\displaystyle \frac{df}{dx} = f' , \qquad \frac{dx}{dt} = \dot x , \qquad \frac{d^2x}{dt^2} = \ddot x}\]

Standard derivatives

• The power rule: \[\frac{d}{dx}\,x^n = n\,x^{n-1} .\]

• Sine and cosine: \[\frac{d}{dx}\sin x = \cos x , \qquad \frac{d}{dx}\cos x = -\sin x .\]

• The product rule: \[\frac{d}{dx}(fg) = f'g + fg' .\]

• The chain rule: \[\keyeq{\displaystyle \frac{d}{dx}\,f\big(g(x)\big) = \frac{dg}{dx}\,\frac{df}{dg}}\]

Example: \(\tan x\)

Write \(\tan x = \sin x\cdot\dfrac{1}{\cos x}\) and use the product rule, then the chain rule for \(1/\cos x\): \[\frac{d}{dx}\tan x = \frac{\cos x}{\cos x} + \sin x\,\frac{d}{dx}\!\left(\frac{1}{\cos x}\right) = 1 + \sin x\cdot\frac{\sin x}{\cos^2 x} = 1 + \tan^2 x .\] Since \(1 + \tan^2 x = \dfrac{\cos^2 x + \sin^2 x}{\cos^2 x} = \dfrac{1}{\cos^2 x}\), \[\frac{d}{dx}\tan x = 1 + \tan^2 x = \sec^2 x .\] The derivatives of the other trigonometric functions follow in the same way.

Exponentials and logarithms

\[\keyeq{\displaystyle \frac{d}{dx}\,e^x = e^x} \qquad\qquad \keyeq{\displaystyle \frac{d}{dx}\ln x = \frac{1}{x}}\] Together with the rules above, these are all you need. For example, writing \(a^x = e^{x\ln a}\) and using the chain rule, \[\frac{d}{dx}\,a^x = \frac{d}{dx}\,e^{x\ln a} = \ln a\; e^{x\ln a} = a^x\ln a .\]

0.4 Integrals

The area is signed: where \(f(x) < 0\), the rectangles count negatively.

The fundamental theorem of calculus

A function \(F\) is an antiderivative of \(f\) if \(F' = f\). The indefinite integral is the most general antiderivative, \[\int f(x)\,dx = F(x) + C ,\] with an arbitrary constant \(C\) (its derivative is zero). The fundamental theorem of calculus says that integration undoes differentiation: \[\begin{equation} \tag{0.7} \keyeq{\displaystyle \int_a^b f(x)\,dx = \Big[F(x)\Big]_a^b = F(b) - F(a)} \end{equation}\] So every derivative you know, read backwards, gives an integral.

Polynomials

Reading the power rule backwards, \[\keyeq{\displaystyle \int x^n\,dx = \frac{x^{n+1}}{n+1} + C} \qquad (n \neq -1), \qquad\qquad \int \frac{dx}{x} = \ln|x| + C .\] Integrals are linear, so a polynomial is integrated term by term. For example, \[\int_0^2 \left(3x^2 - 2x + 1\right)dx = \Big[x^3 - x^2 + x\Big]_0^2 = (8 - 4 + 2) - 0 = 6 .\]

Sine and cosine

Reading the derivatives of sine and cosine backwards, \[\keyeq{\displaystyle \int \sin x\,dx = -\cos x + C} \qquad\qquad \keyeq{\displaystyle \int \cos x\,dx = \sin x + C}\] For example, over half a period \[\int_0^\pi \sin x\,dx = \Big[-\cos x\Big]_0^\pi = -(-1) - (-1) = 2 ,\] but over a full period the positive and negative areas cancel: \[\int_0^{2\pi} \sin x\,dx = \Big[-\cos x\Big]_0^{2\pi} = -1 - (-1) = 0 .\] When the argument is \(\omega t\), the chain rule brings out a factor of \(\omega\), which the integral must divide back out: \[\int \cos(\omega t)\,dt = \frac{1}{\omega}\sin(\omega t) + C , \qquad \int \sin(\omega t)\,dt = -\frac{1}{\omega}\cos(\omega t) + C .\] Check by differentiating!

Problems

Problems are taken from the sources listed under Sources and Attribution. Numbers refer to the end-of-chapter problems; figures have been redrawn.

OpenStax Calculus Vol. 3, Chapter 2: Vectors in Space

OpenStax Calculus Vol. 1, Chapter 3: Derivatives

Find the derivatives of the following functions.

OpenStax Calculus Vol. 1, Chapter 5: Integration