Calculus 2 Exam 2 Version 2

$ \tan(\theta) $

\frac{\sin(\theta)}{\cos(\theta)}

\sec(\theta)

\frac{1}{\cos(\theta)}

\sin^2(\theta) + \cos^2(\theta)

1

\tan^2(\theta) + 1

\sec^2(\theta)

1 + \cot^2(\theta)

\csc^2(\theta)

\sin(2\theta)

2\sin(\theta)\cos(\theta)

\cos(2\theta)

\cos^2(\theta) - \sin^2(\theta)

\cos^2(\theta)

\frac{1}{2}(1 + \cos(2\theta))

\sin^2(\theta)

\frac{1}{2}(1 - \cos(2\theta))

\sin(0)

0

\cos(0)

1

\tan(0)

0

\sin(\pi/6)

\frac{1}{2}

\cos(\pi/6)

\frac{\sqrt{3}}{2}

\tan(\pi/6)

\frac{1}{\sqrt{3}}

\sin(\pi/4)

\frac{1}{\sqrt{2}}

\cos(\pi/4)

\frac{1}{\sqrt{2}}

\tan(\pi/4)

1

\sin(\pi/3)

\frac{\sqrt{3}}{2}

\cos(\pi/3)

\frac{1}{2}

\tan(\pi/3)

\sqrt{3}

\sin(\pi/2)

1

\cos(\pi/2)

0

\tan(\pi/2)

\text{undefined}

\frac{d}{dt}(\sin(t))

\cos(t)

\frac{d}{dt}(\cos(t))

-\sin(t)

\frac{d}{dt}(\tan(t))

\sec^2(t)

\frac{d}{dt}(\sec(t))

\sec(t)\tan(t)

\frac{d}{dt}(\arctan(t))

\frac{1}{1+t^2}

\frac{d}{dt}(t^k)

kt^{k-1}

\frac{d}{dt}(\ln|t|)

\frac{1}{t}

\frac{d}{dt}(e^{kt})

ke^{kt}

\int \tan(t) \, dt

\ln|\sec(t)| + C

\int \sec(t) \, dt

\ln|\sec(t) + \tan(t)| + C

\int u \, dv

uv - \int v \, du

\text{Substitution for } \sqrt{b^2 - u^2}

u = b \sin(\theta), \quad -\pi/2 \le \theta \le \pi/2

\text{Substitution for } \sqrt{b^2 + u^2}

u = b \tan(\theta), \quad -\pi/2 < \theta < \pi/2

\text{Substitution for } \sqrt{u^2 - b^2}

u = b \sec(\theta), \quad 0 \le \theta < \pi/2 \text{ or } \pi \le \theta < 3\pi/2

\text{Volume by Cross-sections}

V = \int{a}^{b} A(x) \, dx \quad (\text{or } \int{c}^{d} A(y) \, dy)

\text{Volume (Disk/Washer Method, x-axis rotation)}

V = \int_{a}^{b} \pi [ (R(x))^2 - (r(x))^2 ] \, dx

\text{Volume (Shell Method, y-axis rotation)}

V = \int{a}^{b} 2\pi (\text{radius}) (\text{height}) \, dx = \int{a}^{b} 2\pi x h(x) \, dx

\text{Arc Length (Cartesian)}

L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} \, dx

\text{Arc Length (Parametric)}

L = \int_{a}^{b} \sqrt{[x'(t)]^2 + [y'(t)]^2} \, dt

\text{Surface Area (Revolution about x-axis)}

SA = \int_{a}^{b} 2\pi f(x) \sqrt{1 + [f'(x)]^2} \, dx

\text{Surface Area (Parametric, about x-axis)}

SA = \int_{a}^{b} 2\pi y(t) \sqrt{[x'(t)]^2 + [y'(t)]^2} \, dt

\text{Work (Variable Force)}

W = \int_{a}^{b} F(x) \, dx

\text{Work (Pumping Liquid)}

W = \int{c}^{d} (\text{weight density}) (\text{Area of slice}) (\text{distance lifted}) \, dy = \int{c}^{d} \rho g A(y) D(y) \, dy

\text{Hydrostatic Force (on vertical plate)}

F = \int{c}^{d} (\text{weight density}) (\text{depth of strip}) (\text{width of strip}) \, dy = \int{c}^{d} \rho g h(y) w(y) \, dy

\text{Mass (1D Rod, variable density } \rho(x)\text{)}

M = \int_{a}^{b} \rho(x) \, dx

\text{Moment about y-axis (1D Rod)}

My = \int{a}^{b} x \rho(x) \, dx

\text{Moment about x-axis (2D Lamina, constant density } \rho\text{)}

Mx = \int{a}^{b} \frac{1}{2} \rho [ (f{top}(x))^2 - (f{bottom}(x))^2 ] \, dx

\text{Moment about y-axis (2D Lamina, constant density } \rho\text{)}

My = \int{a}^{b} x \rho [ f{top}(x) - f{bottom}(x) ] \, dx

\text{Center of Mass (x-coordinate)}

\bar{x} = \frac{M_y}{M}

\text{Center of Mass (y-coordinate)}

\bar{y} = \frac{M_x}{M}

\text{Polar to Cartesian Conversion (x)}

x = r \cos(\theta)

\text{Polar to Cartesian Conversion (y)}

y = r \sin(\theta)

\text{Cartesian/Polar Relationship}

r^2 = x^2 + y^2

\text{Area in Polar Coordinates}

A = \int_{\alpha}^{\beta} \frac{1}{2} r^2 \, d\theta

\text{Area between Polar Curves}

A = \frac{1}{2} \int{\alpha}^{\beta} [ (r{outer})^2 - (r_{inner})^2 ] \, d\theta

\text{Arc Length in Polar Coordinates}

L = \int_{\alpha}^{\beta} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta

\text{Trapezoidal Rule Formula}

Tn = \frac{\Delta x}{2} [f(x0) + 2f(x1) + 2f(x2) + \dots + 2f(x{n-1}) + f(xn)]

\text{Trapezoidal Rule Error Bound}

|E_T| \le \frac{M(b - a)^3}{12n^2} \quad (\text{where } |f''(x)| \le M \text{ on } [a, b])

\text{Simpson's Rule Formula}

Sn = \frac{\Delta x}{3} [f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + \dots + 2f(x{n-2}) + 4f(x{n-1}) + f(xn)]

\text{Simpson's Rule Error Bound}

|E_S| \le \frac{M(b - a)^5}{180n^4} \quad (\text{where } |f^{(4)}(x)| \le M \text{ on } [a, b])