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Mathematic 1 - Calculus 1 - Summary

1. Vector

1.1 Geometric Vectors

Position Vector

  • A vector that originates from the origin and points to a location in space.

Joining Vectors

  • Geometric relation: OP+PQ=OQPQ=OQOP\overrightarrow{OP} + \overrightarrow{PQ} = \overrightarrow{OQ} \quad \Rightarrow \quad \overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}
  • Component form: PQ=x2,y2x1,y1=x2x1,y2y1\overrightarrow{PQ} = \langle x_2, y_2 \rangle - \langle x_1, y_1 \rangle = \langle x_2 - x_1, y_2 - y_1 \rangle

1.2 Algebraic Vectors

Vector Addition

  • 2D:

    a1,a2+b1,b2=a1+b1,a2+b2\langle a_1, a_2 \rangle + \langle b_1, b_2 \rangle = \langle a_1 + b_1, a_2 + b_2 \rangle
  • 3D:

    a1,a2,a3+b1,b2,b3=a1+b1,a2+b2,a3+b3\langle a_1, a_2, a_3 \rangle + \langle b_1, b_2, b_3 \rangle = \langle a_1 + b_1, a_2 + b_2, a_3 + b_3 \rangle

Scalar Multiplication

  • 2D:

    ka,b=ka,kbk \langle a, b \rangle = \langle ka, kb \rangle
  • 3D:

    ka,b,c=ka,kb,kck \langle a, b, c \rangle = \langle ka, kb, kc \rangle
  • Direction of the vector depends on scalar ( k ):

    • If ( k > 0 ), the vector keeps the same direction.
    • If ( k < 0 ), the vector points in the opposite direction.

Norm (Magnitude)

  • 2D:

    v=a2+b2|\vec{v}| = \sqrt{a^2 + b^2}
  • 3D:

    v=a2+b2+c2|\vec{v}| = \sqrt{a^2 + b^2 + c^2}

💡 Always perform vector operations (e.g., addition, scalar multiplication) before computing the norm.

Unit Vector

  • A unit vector has a magnitude of 1. It is obtained by dividing a vector by its norm: u=vv=xv,yv\vec{u} = \frac{\vec{v}}{|\vec{v}|} = \left\langle \frac{x}{|\vec{v}|}, \frac{y}{|\vec{v}|} \right\rangle

Basis Vectors

  • A 3D vector can be expressed using basis vectors: x,y,z=xi+yj+zk\langle x, y, z \rangle = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}

1.3 Direction Cosines & Angles (3D)

Direction Cosines

  • Direction cosines represent the cosine of the angle between a vector and each coordinate axis:

    cosα=xv,cosβ=yv,cosγ=zv\cos\alpha = \frac{x}{|\vec{v}|}, \quad \cos\beta = \frac{y}{|\vec{v}|}, \quad \cos\gamma = \frac{z}{|\vec{v}|}

Direction Angles

  • The angles themselves are found using the inverse cosine: α=cos1(xv),β=cos1(yv),γ=cos1(zv)\alpha = \cos^{-1} \left( \frac{x}{|\vec{v}|} \right), \quad \beta = \cos^{-1} \left( \frac{y}{|\vec{v}|} \right), \quad \gamma = \cos^{-1} \left( \frac{z}{|\vec{v}|} \right)

Identity

  • The sum of the squares of the direction cosines always equals 1: cos2α+cos2β+cos2γ=1\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1

2. Lines & Planes

2.1 Equations of Lines

Line Through a Point and Direction Vector

  • A line passing through point P=x0,y0,z0P = \langle x_0, y_0, z_0 \rangle with direction vector v=a,b,c\vec{v} = \langle a, b, c \rangle can be written as:

    r=P+tv=x0,y0,z0+ta,b,c\vec{r} = P + t\vec{v} = \langle x_0, y_0, z_0 \rangle + t \langle a, b, c \rangle

Line Through Two Points

  • Given points A=(xa,ya,za)A = (x_a, y_a, z_a) and B=(xb,yb,zb)B = (x_b, y_b, z_b), the direction vector is:

    v=AB=xbxa,ybya,zbza\vec{v} = \overrightarrow{AB} = \langle x_b - x_a, y_b - y_a, z_b - z_a \rangle
  • Line equation becomes:

    r=A+tvorr=B+tv\vec{r} = A + t\vec{v} \quad \text{or} \quad \vec{r} = B + t\vec{v}

Forms of Line Equation

  • Vector Form:

    r=x0,y0,z0+ta,b,c\vec{r} = \langle x_0, y_0, z_0 \rangle + t \langle a, b, c \rangle
  • Parametric Form:

    {x=x0+tay=y0+tbz=z0+tc\begin{cases} x = x_0 + ta \\ y = y_0 + tb \\ z = z_0 + tc \end{cases}
  • Symmetric Form:

    xx0a=yy0b=zz0c\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}

2.2 Equations of Planes

Plane Through a Point and Normal Vector

  • Given a point P=(xP,yP,zP)P = (x_P, y_P, z_P) and normal vector n=a,b,c\vec{n} = \langle a, b, c \rangle, the plane equation is:

    nPr=0a(xxP)+b(yyP)+c(zzP)=0\vec{n} \cdot \overrightarrow{Pr} = 0 \quad \Rightarrow \quad a(x - x_P) + b(y - y_P) + c(z - z_P) = 0
  • Expanded form:

    ax+by+cz+d=0ax + by + cz + d = 0

Plane Through Three Points

  • Given three points A,B,CA, B, C, form two vectors and compute:

    n=AB×AC\vec{n} = \overrightarrow{AB} \times \overrightarrow{AC}
  • Then use the point-normal form with one of the points.

2.3 Projections

Vector Projection

  • Projection of a\vec{a} onto b\vec{b}:

    projba=(abb2)b\text{proj}_{\vec{b}} \vec{a} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b}

Scalar Projection

  • Component of a\vec{a} along b\vec{b}:

    compba=abb\text{comp}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}

2.4 Shortest Distance

From a Point to a Line

  • Given point PP, point AA on the line, and direction vector v\vec{v}:

    d=PA×vvd = \frac{|\overrightarrow{PA} \times \vec{v}|}{|\vec{v}|}

From a Point to a Plane

  • With normal vector n\vec{n} and vector AP\overrightarrow{AP}:

    d=nAPnd = \left| \frac{\vec{n} \cdot \overrightarrow{AP}}{|\vec{n}|} \right|

Given Plane Equation and Any Point

  • Plane: ax+by+cz+d=0ax + by + cz + d = 0

  • Point: (x1,y1,z1)(x_1, y_1, z_1)

    d=ax1+by1+cz1+da2+b2+c2d = \left| \frac{ax_1 + by_1 + cz_1 + d}{\sqrt{a^2 + b^2 + c^2}} \right|

From Origin to Plane

d=da2+b2+c2d = \left| \frac{d}{\sqrt{a^2 + b^2 + c^2}} \right|

Between Two Parallel Planes

  1. Ensure both planes have the same normal vector.

  2. Find the distance difference from the origin:

    d=d2d1a2+b2+c2d = \frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}}

2.5 Intersection of Two Lines

Step-by-Step

  1. Write both lines in parametric form:

    • L1:x=x1+a1t, y=y1+b1t, z=z1+c1tL_1: x = x_1 + a_1 t,\ y = y_1 + b_1 t,\ z = z_1 + c_1 t
    • L2:x=x2+a2s, y=y2+b2s, z=z2+c2sL_2: x = x_2 + a_2 s,\ y = y_2 + b_2 s,\ z = z_2 + c_2 s
  2. Solve the resulting system of equations to find values of tt and ss.

  3. Substitute back into either line to find the intersection point.

✅ The lines intersect only if all equations result in the same point.


3. Complex Numbers

3.1 Introduction and Operations

Complex Number and Its Conjugate

  • A complex number is defined as:

    z=a+biwith conjugatezˉ=abiz = a + bi \quad \text{with conjugate} \quad \bar{z} = a - bi

Equality of Complex Numbers

  • For z1+z2=a+biz_1 + z_2 = a + bi, then:

    Re(z1)+Re(z2)=a,Im(z1)+Im(z2)=b\text{Re}(z_1) + \text{Re}(z_2) = a, \quad \text{Im}(z_1) + \text{Im}(z_2) = b

Complex Number Operations

  • Addition/Subtraction:

    z±w=(a±c)+(b±d)iz \pm w = (a \pm c) + (b \pm d)i
  • Multiplication:

    zw=(a+bi)(c+di)=(acbd)+(ad+bc)iz \cdot w = (a + bi)(c + di) = (ac - bd) + (ad + bc)i
  • Division:

    a+bic+di=a+bic+dicdicdi\frac{a + bi}{c + di} = \frac{a + bi}{c + di} \cdot \frac{c - di}{c - di}

    Multiply numerator and denominator by the conjugate of the denominator, then simplify into real and imaginary parts.

Properties with Conjugate

  • Addition:

    z+zˉ=2Re(z)z + \bar{z} = 2\text{Re}(z)
  • Multiplication:

    zzˉ=z2=a2+b2z \cdot \bar{z} = |z|^2 = a^2 + b^2

3.2 Geometrical Interpretation – Argand Diagram

Argand Diagram
  • The x-axis represents the real part.
  • The y-axis represents the imaginary part.

3.3 Modulus and Argument

  • The modulus (or absolute value) of a complex number is:

    z=a2+b2|z| = \sqrt{a^2 + b^2}
  • The argument θ\theta is the angle from the positive real axis to the vector:

    θ=tan1(ba),where 0θ<2π\theta = \tan^{-1} \left( \frac{b}{a} \right), \quad \text{where } 0 \leq \theta < 2\pi

💡 Tip: For purely imaginary numbers (i.e., a=0a = 0):

  • If b>0b > 0, then θ=π2\theta = \frac{\pi}{2}
  • If b<0b < 0, then θ=3π2\theta = \frac{3\pi}{2}

3.4 Cartesian, Polar, and Exponential Forms

  • Euler's Formula:

    cosθ+isinθ=eiθ\cos\theta + i\sin\theta = e^{i\theta}
  • Cartesian Form:

    z=a+biz = a + bi
  • Polar Form:

    z=r(cosθ+isinθ)=rcisθz = r(\cos\theta + i\sin\theta) = r\,\text{cis}\,\theta
  • Exponential Form:

    z=reiθz = r e^{i\theta}

    Where r=zr = |z| and θ=arg(z)\theta = \arg(z)

3.5 Operations in Polar/Exponential Form

Multiplication

  • Multiply moduli and add arguments:

    z1z2=r1r2ei(θ1+θ2)=r1r2cis(θ1+θ2)z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)} = r_1 r_2\, \text{cis}(\theta_1 + \theta_2)

Division

  • Divide moduli and subtract arguments:

    z1z2=r1r2ei(θ1θ2)=r1r2cis(θ1θ2)\frac{z_1}{z_2} = \frac{r_1}{r_2} e^{i(\theta_1 - \theta_2)} = \frac{r_1}{r_2}\, \text{cis}(\theta_1 - \theta_2)

3.6 De Moivre’s Theorem

  • For powers of complex numbers:

    zn=(reiθ)n=rneinθz^n = \left( r e^{i\theta} \right)^n = r^n e^{i n \theta}
  • Using polar form:

    zn=rn(cosnθ+isinnθ)z^n = r^n (\cos n\theta + i\sin n\theta)

🔁 Multiply the argument by nn and raise the modulus to the nthn^{\text{th}} power.

Reducing Large Arguments (Radian Mod)

  • If the angle becomes too large, subtract 2π2\pi as many times as needed: θreduced=θmod2π\theta_{\text{reduced}} = \theta \mod 2\pi

3.7 Roots of Unity (Not in Exam)

  • The nthn^{\text{th}} roots of 1 are:

    zk=cos(2πkn)+isin(2πkn),for k=0,1,,n1z_k = \cos\left( \frac{2\pi k}{n} \right) + i\sin\left( \frac{2\pi k}{n} \right), \quad \text{for } k = 0, 1, \dots, n-1
  • Example for z3=1z^3 = 1:

    • z1=cos(0)+isin(0)=1z_1 = \cos(0) + i\sin(0) = 1
    • z2=cos(2π3)+isin(2π3)z_2 = \cos\left( \frac{2\pi}{3} \right) + i\sin\left( \frac{2\pi}{3} \right)
    • z3=cos(4π3)+isin(4π3)z_3 = \cos\left( \frac{4\pi}{3} \right) + i\sin\left( \frac{4\pi}{3} \right)

4. Related Rates, Differentials, and Newton’s Method

  • If y=f(x)y = f(x), the rate of change of yy with respect to xx is defined as:

    dydx=f(x)\frac{dy}{dx} = f'(x)
  • dydx\frac{dy}{dx} is Leibniz's notation for the derivative, and it represents the rate at which yy changes with xx.

Chain Rule

  • If y=f(x)y = f(x) and x=f(t)x = f(t), then:

    dydt=dydxdxdt\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}
  • This rule connects rates of change between dependent variables.

  • Use the chain rule to relate changing quantities.
  • Establish an equation that relates all variables.
  • Differentiate both sides with respect to time.
  • Ratios often appear and can simplify or cancel out variables.

4.2 Differentials & Linear Approximation

Differentials

  • Given y=f(x)y = f(x):

    dydx=f(x)dy=f(x)dx\frac{dy}{dx} = f'(x) \quad \Rightarrow \quad dy = f'(x)\,dx
  • Differentials are used to estimate small changes in yy based on small changes in xx.

  • Used for approximation:

    dyf(x1)(x2x1)dy \approx f'(x_1)(x_2 - x_1)

    Where dydy approximates f(x2)f(x1)f(x_2) - f(x_1).

Linear Approximation

  • The linear approximation formula at x=ax = a is:

    L(x)=f(a)+f(a)(xa)L(x) = f(a) + f'(a)(x - a)
  • This estimates f(x)f(x) using a tangent line near x=ax = a.

4.3 Newton’s Method

  • An iterative method for approximating roots of a function.

  • Formula:

    xn+1=xnf(xn)f(xn)x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
  • Process:

    1. Start with an initial guess x0x_0
    2. Use the formula to compute better approximations
    3. Repeat until convergence
  • Works well when:

    • f(x)f(x) is differentiable near the root
    • f(x)f'(x) is not zero near the root

5. Inverse & Transcendental Functions

5.1 Inverse Functions

Definition

  • A function ff has an inverse f1f^{-1} if it is one-to-one.

  • The graph of f1(x)f^{-1}(x) is the reflection of f(x)f(x) about the line y=xy = x.

  • Point relationship:

    If f(a)=b, then f1(b)=a\text{If } f(a) = b, \text{ then } f^{-1}(b) = a (x=a,y=b)    (x=b,y=a)(x = a, y = b) \iff (x = b, y = a)

Inverse Function Theorem

  • If f(b)=af(b) = a, then:

    (f1)(a)=1f(b)(f^{-1})'(a) = \frac{1}{f'(b)}

5.2 Exponential Functions

Laws of Exponents

  • ax+y=axaya^{x+y} = a^x \cdot a^y
  • axy=axaya^{x-y} = \frac{a^x}{a^y}
  • (ax)y=axy(a^x)^y = a^{xy}
  • (ab)x=axbx(ab)^x = a^x b^x

Derivative

  • ddx(ex)=ex\frac{d}{dx}(e^x) = e^x
  • ddx(ef(x))=ef(x)f(x)\frac{d}{dx}(e^{f(x)}) = e^{f(x)} \cdot f'(x)

Integral

  • exdx=ex+C\int e^x dx = e^x + C
  • ef(x)dx=ef(x)f(x)+C\int e^{f(x)} dx = \frac{e^{f(x)}}{f'(x)} + C

5.3 Logarithmic Functions

Laws of Logarithms

  • loge(xy)=logex+logey\log_e(xy) = \log_e x + \log_e y
  • loge(xy)=logexlogey\log_e\left(\frac{x}{y}\right) = \log_e x - \log_e y
  • loge(xr)=rlogex\log_e(x^r) = r \log_e x

⚠️ Note: (logex)rrlogex(\log_e x)^r \neq r \log_e x

Derivative

  • ddxlnx=1x\frac{d}{dx} \ln x = \frac{1}{x}
  • ddxln(f(x))=f(x)f(x)\frac{d}{dx} \ln(f(x)) = \frac{f'(x)}{f(x)}

Integral

  • 1xdx=lnx+C\int \frac{1}{x} dx = \ln|x| + C
  • f(x)f(x)dx=lnf(x)+C\int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + C

5.4 Inverse Trigonometric Functions

Inverse Sine

  • ddx(sin1x)=11x2\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1 - x^2}}
  • 11x2dx=sin1(x)+C\int \frac{1}{\sqrt{1 - x^2}} dx = \sin^{-1}(x) + C

Inverse Cosine

  • ddx(cos1x)=11x2\frac{d}{dx}(\cos^{-1}x) = -\frac{1}{\sqrt{1 - x^2}}

Inverse Tangent

  • ddx(tan1x)=11+x2\frac{d}{dx}(\tan^{-1}x) = \frac{1}{1 + x^2}
  • 11+x2dx=tan1(x)+C\int \frac{1}{1 + x^2} dx = \tan^{-1}(x) + C

Solving Using Triangle

  • Use a triangle to define angle relationships.
  • Use geometry to rewrite expressions in simpler terms.

5.5 Hyperbolic Functions

Definition of Hyperbolic Functions

  • sinhx=exex2\sinh x = \frac{e^x - e^{-x}}{2}
  • coshx=ex+ex2\cosh x = \frac{e^x + e^{-x}}{2}
  • tanhx=sinhxcoshx=exexex+ex\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}
  • \cschx=1sinhx,\sechx=1coshx,cothx=coshxsinhx\csch x = \frac{1}{\sinh x}, \quad \sech x = \frac{1}{\cosh x}, \quad \coth x = \frac{\cosh x}{\sinh x}

💡 sinh(log(ex))=exex2\sinh(\log(e^x)) = \frac{e^x - e^{-x}}{2}
Think of exe^x as a variable — so you can still use sinh\sinh even if it’s not explicitly written as exe^x

Hyperbolic Identities

  • sinh(x)=sinhx,cosh(x)=coshx\sinh(-x) = -\sinh x, \quad \cosh(-x) = \cosh x
  • cosh2xsinh2x=1\cosh^2 x - \sinh^2 x = 1
  • 1tanh2x=\sech2x1 - \tanh^2 x = \sech^2 x
  • sinh(x+y)=sinhxcoshy+coshxsinhy\sinh(x + y) = \sinh x \cosh y + \cosh x \sinh y
  • cosh(x+y)=coshxcoshy+sinhxsinhy\cosh(x + y) = \cosh x \cosh y + \sinh x \sinh y
  • tanh(x+y)=tanhx+tanhy1+tanhxtanhy\tanh(x + y) = \frac{\tanh x + \tanh y}{1 + \tanh x \tanh y}
  • tanh(xy)=tanhxtanhy1tanhxtanhy\tanh(x - y) = \frac{\tanh x - \tanh y}{1 - \tanh x \tanh y}

Derivatives of Hyperbolic Functions

  • ddxsinhx=coshx\frac{d}{dx} \sinh x = \cosh x
  • ddxcoshx=sinhx\frac{d}{dx} \cosh x = \sinh x
  • ddxtanhx=\sech2x\frac{d}{dx} \tanh x = \sech^2 x
  • ddxcothx=\csch2x\frac{d}{dx} \coth x = -\csch^2 x
  • ddx\sechx=\sechxtanhx\frac{d}{dx} \sech x = -\sech x \tanh x
  • ddx\cschx=\cschxcothx\frac{d}{dx} \csch x = -\csch x \coth x

Integrals of Hyperbolic Functions

  • sinhxdx=coshx+C\int \sinh x \, dx = \cosh x + C
  • coshxdx=sinhx+C\int \cosh x \, dx = \sinh x + C
  • \sech2xdx=tanhx+C\int \sech^2 x \, dx = \tanh x + C
  • \csch2xdx=cothx+C\int \csch^2 x \, dx = -\coth x + C
  • \sechxtanhxdx=\sechx+C\int \sech x \tanh x \, dx = -\sech x + C
  • \cschxcothxdx=\cschx+C\int \csch x \coth x \, dx = -\csch x + C

Inverse Hyperbolic Functions

  • sinh1x=ln(x+x2+1),xR\sinh^{-1} x = \ln \left( x + \sqrt{x^2 + 1} \right), \quad x \in \mathbb{R}
  • cosh1x=ln(x+x21),x1\cosh^{-1} x = \ln \left( x + \sqrt{x^2 - 1} \right), \quad x \ge 1
  • tanh1x=12ln(1+x1x),1<x<1\tanh^{-1} x = \frac{1}{2} \ln \left( \frac{1 + x}{1 - x} \right), \quad -1 < x < 1
  • coth1x=12ln(x+1x1),x>1\coth^{-1} x = \frac{1}{2} \ln \left( \frac{x + 1}{x - 1} \right), \quad |x| > 1
  • \sech1x=ln(1+1x2x),0<x1\sech^{-1} x = \ln \left( \frac{1 + \sqrt{1 - x^2}}{x} \right), \quad 0 < x \le 1
  • \csch1x=ln(1x+1+x2x),x0\csch^{-1} x = \ln \left( \frac{1}{x} + \frac{\sqrt{1 + x^2}}{|x|} \right), \quad x \ne 0

Derivatives of Inverse Hyperbolic Functions

  • ddxsinh1x=11+x2\frac{d}{dx} \sinh^{-1} x = \frac{1}{\sqrt{1 + x^2}}
  • ddxcosh1x=1x21\frac{d}{dx} \cosh^{-1} x = \frac{1}{\sqrt{x^2 - 1}}
  • ddxtanh1x=11x2\frac{d}{dx} \tanh^{-1} x = \frac{1}{1 - x^2}
  • ddxcoth1x=11x2\frac{d}{dx} \coth^{-1} x = \frac{1}{1 - x^2}
  • ddx\sech1x=1x1x2\frac{d}{dx} \sech^{-1} x = \frac{-1}{x \sqrt{1 - x^2}}
  • ddx\csch1x=1x1+x2\frac{d}{dx} \csch^{-1} x = \frac{-1}{|x| \sqrt{1 + x^2}}

6. Techniques of Integration

6.1 Integration by Substitution

  • Observe the function and choose an expression to substitute as uu

    • The chosen part should be easily differentiable and cancel with the remaining part of the integrand.
  • When u=g(x)u = g(x), we differentiate:

    dudx=g(x)dx=dug(x)\frac{du}{dx} = g'(x) \quad \Rightarrow \quad dx = \frac{du}{g'(x)}
  • Replace dxdx and the corresponding expression in the integrand with dudu and uu.

  • For definite integrals, convert limits of xx into limits in terms of uu:

    abf(g(x))g(x) dx=u(a)u(b)f(u) du\int_a^b f(g(x))g'(x)\ dx = \int_{u(a)}^{u(b)} f(u)\ du

6.2 Integration by Parts

  • Formula:

    udvdx dx=uvdudxv dx    u dv=uvv du\int u\frac{dv}{dx}\ dx = uv - \int \frac{du}{dx}v\ dx \quad \iff \quad \int u\ dv = uv - \int v\ du
  • Use the LIATE Rule to decide which part of the integrand should be uu:

    • LL: Logarithmic functions (lnx\ln x)
    • II: Inverse trigonometric functions (sin1x\sin^{-1}x, tan1x\tan^{-1}x, ...)
    • AA: Algebraic functions (xx, x2x^2, xnx^n)
    • TT: Trigonometric functions (sinx\sin x, cosx\cos x, ...)
    • EE: Exponential functions (exe^x)

⚠️ The LIATE rule is a guideline, not a strict rule — exceptions may apply depending on the problem.

Integration by Parts for Inverse Trig Functions

  • When integrating an inverse trigonometric function (like tan1x\tan^{-1}x), use the trick:

    tan1x dx=tan1x1 dx\int \tan^{-1}x\ dx = \int \tan^{-1}x \cdot 1\ dx
  • Then apply integration by parts with:

    • u=tan1xu = \tan^{-1}x
    • dv=dxdv = dx

6.3 Integration by Partial Fractions

  • Use algebra to decompose a rational function into simpler fractions that are easier to integrate.

Case 1: Two linear factors

  • For an expression like:

    px+q(ax+b)(cx+d)=Aax+b+Bcx+d\frac{px + q}{(ax + b)(cx + d)} = \frac{A}{ax + b} + \frac{B}{cx + d}

Case 2: Repeated linear factors

  • For a repeated linear factor like (ax+b)2(ax + b)^2:

    px+q(ax+b)2(cx+d)=Aax+b+B(ax+b)2+Ccx+d\frac{px + q}{(ax + b)^2(cx + d)} = \frac{A}{ax + b} + \frac{B}{(ax + b)^2} + \frac{C}{cx + d}
  • For (ax+b)n(ax + b)^n, general form:

    P(x)(ax+b)n=A1ax+b+A2(ax+b)2++An(ax+b)n\frac{P(x)}{(ax + b)^n} = \frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \cdots + \frac{A_n}{(ax + b)^n}

Case 3: Linear and Irreducible Quadratic

  • If the denominator has both a linear factor and a non-factorizable quadratic (cx2+d)(cx^2 + d):

    px+q(ax+b)(cx2+d)=Aax+b+Bx+Ccx2+d\frac{px + q}{(ax + b)(cx^2 + d)} = \frac{A}{ax + b} + \frac{Bx + C}{cx^2 + d}
  • After decomposition, integrate each term using basic rules or known formulas.


7. Definite Integrals & Numerical Integration

7.1 The Definite Integral

  • A definite integral is evaluated over a specific interval:

    abf(x) dx=[F(x)]ab=F(b)F(a)\int_a^b f(x)\ dx = [F(x)]_a^b = F(b) - F(a)

Substitution with Limits

  • When using substitution in definite integrals:

    u=f(x),dudx=f(x)dx=duf(x)u = f(x), \quad \frac{du}{dx} = f'(x) \quad \Rightarrow \quad dx = \frac{du}{f'(x)}
  • Convert limits:

    x1x2f(x) dx=u1=f(x1)u2=f(x2)uf(x) du\int_{x_1}^{x_2} f(x)\ dx = \int_{u_1 = f(x_1)}^{u_2 = f(x_2)} \frac{u}{f'(x)}\ du

Integration by Parts with Limits

  • Formula with limits applied:

    abudvdx dx=[uv]ababdudxv dx\int_a^b u \frac{dv}{dx}\ dx = [uv]_a^b - \int_a^b \frac{du}{dx}v\ dx

7.2 The Trapezoidal Rule

  • Approximate the area under a curve by dividing it into trapezoids:

    abf(x) dxh2[y0+yn+2(y1+y2++yn1)]\int_a^b f(x)\ dx \approx \frac{h}{2} \left[ y_0 + y_n + 2(y_1 + y_2 + \cdots + y_{n-1}) \right]
  • Step size (height of each trapezoid):

    h=banh = \frac{b - a}{n}
  • Nodes:

    x0=a, x1=a+h, , xn=bx_0 = a,\ x_1 = a + h,\ \ldots,\ x_n = b
  • Make a table of xx and corresponding y=f(x)y = f(x) values to compute.

7.3 Simpson’s Rule

  • More accurate than the trapezoidal rule, uses parabolas:

    abf(x) dxh3[y0+yn+4(y1+y3++yn1)+2(y2+y4++yn2)]\int_a^b f(x)\ dx \approx \frac{h}{3} \left[ y_0 + y_n + 4(y_1 + y_3 + \cdots + y_{n-1}) + 2(y_2 + y_4 + \cdots + y_{n-2}) \right]
  • Equivalent in simple terms:

    abf(x) dxh3[First + Last+4(Odds)+2(Evens)]\int_a^b f(x)\ dx \approx \frac{h}{3} \left[ \text{First + Last} + 4(\text{Odds}) + 2(\text{Evens}) \right]
  • Step size:

    h=ban,where n is evenh = \frac{b - a}{n}, \quad \text{where $n$ is even}
  • Make a table of xix_i and yi=f(xi)y_i = f(x_i) values to compute.

7.4 Trapezoidal vs Simpson’s Rule

  • Trapezoidal Rule approximates area using straight lines (trapezoids).
  • Simpson’s Rule approximates area using quadratic curves (parabolas).
  • Simpson’s Rule is usually more accurate if f(x)f(x) is smooth and nn is even.

8. First-Order Linear Differential Equations

8.1 Introduction to Differential Equations

  • A differential equation is an equation involving a function and its derivatives.

Order of a Differential Equation

  • Determined by the highest derivative in the equation:

    dydx1st order,d2ydx22nd order,dnydxnn-th order\frac{dy}{dx} \Rightarrow \text{1st order}, \quad \frac{d^2y}{dx^2} \Rightarrow \text{2nd order}, \quad \frac{d^n y}{dx^n} \Rightarrow \text{n-th order}

Degree of a Differential Equation

  • The degree is the exponent of the highest order derivative (if the equation is polynomial in derivatives):

    (dydx)21st order, 2nd degree,(d4ydx4)14th order, 1st degree\left( \frac{dy}{dx} \right)^2 \Rightarrow \text{1st order, 2nd degree}, \quad \left( \frac{d^4 y}{dx^4} \right)^1 \Rightarrow \text{4th order, 1st degree}

Linear vs Nonlinear Differential Equations

  • A differential equation is linear if:
    • All yy terms and derivatives are of degree 1
    • Derivatives of yy are not multiplied by each other

8.2 Standard Form

  • A first-order linear differential equation is written as:

    dydx+P(x)y=Q(x)\frac{dy}{dx} + P(x)y = Q(x)
  • If Q(x)=0Q(x) = 0, the equation is homogeneous.

  • If Q(x)0Q(x) \ne 0, the equation is non-homogeneous.

8.3 The Method of Separation of Variables

General Method

  • Rearrange to separate xx and yy:

    f(y)dydx=g(x)f(y)\frac{dy}{dx} = g(x)
  • Integrate both sides:

    f(y)dydxdx=g(x) dx\int f(y) \frac{dy}{dx} dx = \int g(x)\ dx f(y) dy=g(x) dx\Rightarrow \int f(y)\ dy = \int g(x)\ dx
  • General solution:

    F(y)=G(x)+CF(y) = G(x) + C
  • If initial values are given, substitute to find CC for the particular solution.

Applications

Population Growth and Decay

  • General model:

    P(t)=P0ektP(t) = P_0 e^{kt}

Newton’s Law of Cooling

  • Temperature model:

    T(t)=P+T0ektT(t) = P + T_0 e^{kt}

8.4 Integrating Factors

General Form

  • Arrange into standard linear form:

    dydx+p(x)y=Q(x)\frac{dy}{dx} + p(x)y = Q(x)

Step 1: Find Integrating Factor

  • The integrating factor is:

    v(x)=ep(x) dxv(x) = e^{\int p(x)\ dx}
  • Derivative of v(x)v(x):

    v(x)=v(x)p(x)v'(x) = v(x)p(x)

Step 2: Multiply Equation by Integrating Factor

  • Multiply both sides by v(x)v(x):

    v(x)dydx+v(x)p(x)y=v(x)Q(x)v(x) \frac{dy}{dx} + v(x)p(x)y = v(x)Q(x)
  • Recognize the left side as the derivative of a product:

    ddx[v(x)y]=v(x)Q(x)\frac{d}{dx} [v(x)y] = v(x)Q(x)

Step 3: Integrate Both Sides

  • Integrate:

    v(x)y=v(x)Q(x) dx+Cv(x)y = \int v(x)Q(x)\ dx + C
  • Solve for yy:

    y=1v(x)(v(x)Q(x) dx+C)y = \frac{1}{v(x)} \left( \int v(x)Q(x)\ dx + C \right)
  • If initial condition is given, substitute to find CC.

8.5 When to Use Which Method?

  • Given:

    dydx+p(x)y=q(x)\frac{dy}{dx} + p(x)y = q(x)
  • If q(x)=0q(x) = 0: homogeneous → use separation of variables.

  • If q(x)0q(x) \ne 0: non-homogeneous → use integrating factors (separation won't work).


9. Second-Order Differential Equations

9.1 Standard Form

  • A second-order linear differential equation has the form:

    P(x)y+Q(x)y+R(x)y=f(x)P(x)y'' + Q(x)y' + R(x)y = f(x)
  • For constant coefficients, this becomes:

    ay+by+cy=f(x)ay'' + by' + cy = f(x)
  • If f(x)=0f(x) = 0, the equation is homogeneous.

  • If f(x)0f(x) \ne 0, the equation is non-homogeneous.

9.2 Characteristic Equation and General Solution

  • Standard homogeneous form:

    ay+by+cy=0ay'' + by' + cy = 0
  • Associated characteristic equation:

    am2+bm+c=0am^2 + bm + c = 0
  • Solve for roots m1m_1 and m2m_2 (via quadratic formula, factoring, etc.)

  • General solution depends on the type of roots:

Case 1: Real and Distinct Roots (m1m2m_1 \ne m_2)

  • Solution:

    y(t)=c1em1t+c2em2ty(t) = c_1 e^{m_1 t} + c_2 e^{m_2 t}

Case 2: Real and Equal Roots (m1=m2m_1 = m_2)

  • Solution:

    y(t)=(c1+c2t)em1ty(t) = (c_1 + c_2 t)e^{m_1 t}

Case 3: Complex Roots (m=p±qim = p \pm qi)

  • Solution:

    y(t)=ept(c1cosqt+c2sinqt)y(t) = e^{pt}(c_1 \cos qt + c_2 \sin qt)
  • Use initial conditions to solve for constants c1c_1, c2c_2 if provided.

9.3 Spring-Mass System

No Friction

  • Forces:

    {F=maF=mxFs=kx\begin{cases} F = ma \Rightarrow F = mx'' \\ F_s = -kx \end{cases}
  • Resulting DE:

    mx+kx=0x+ω2x=0,ω2=kmmx'' + kx = 0 \Rightarrow x'' + \omega^2 x = 0, \quad \omega^2 = \frac{k}{m}
  • Characteristic equation:

    m2+ω2=0m=±iωm^2 + \omega^2 = 0 \Rightarrow m = \pm i\omega
  • General solution:

    x(t)=c1cosωt+c2sinωtx(t) = c_1 \cos \omega t + c_2 \sin \omega t

With Friction

  • Additional force: damping force FD=cvF_D = -cv

  • Forces:

    {F=maF=mxFs=kxFD=cx\begin{cases} F = ma \Rightarrow F = mx'' \\ F_s = -kx \\ F_D = -cx' \end{cases}
  • Resulting DE:

    mx+cx+kx=0mx'' + cx' + kx = 0
  • Characteristic equation:

    mr2+cr+k=0mr^2 + cr + k = 0
  • Roots:

    r=c±c24km2mr = \frac{-c \pm \sqrt{c^2 - 4km}}{2m}

Types of Motion Based on Roots

Overdamped Motion (c2>4kmc^2 > 4km)

  • Roots are real and distinct: r1<0r_1 < 0, r2<0r_2 < 0

    x(t)=c1er1t+c2er2tx(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t}

Critically Damped Motion (c2=4kmc^2 = 4km)

  • Roots are real and equal: r1=r2<0r_1 = r_2 < 0

    x(t)=(c1+c2t)er1tx(t) = (c_1 + c_2 t)e^{r_1 t}

Underdamped Motion (c2<4kmc^2 < 4km)

  • Roots are complex: r=p±qir = p \pm qi with p<0p < 0

    x(t)=ept(c1cosqt+c2sinqt)x(t) = e^{pt}(c_1 \cos qt + c_2 \sin qt)

10. Non-Homogeneous Second-Order Linear Differential Equations

10.1 Standard Form

  • General form of the equation:

    ax+bx+cx=f(t)ax'' + bx' + cx = f(t)
  • If f(t)=0f(t) = 0, it's homogeneous.

  • If f(t)0f(t) \ne 0, it's non-homogeneous.

10.2 Solution Steps

  1. Find the Complementary Solution xcx_c by solving the homogeneous equation:

    ax+bx+cx=0ax'' + bx' + cx = 0
    • Solve for roots using the characteristic equation.
    • Refer to: [[Solutions of Second-Order DEs]].
  2. Find a Particular Solution xpx_p of the non-homogeneous equation.

  3. Combine the Solutions:

    x(t)=xc+xpx(t) = x_c + x_p

10.3 The Method of Undetermined Coefficients

We focus on:

  • f(t)=kcos(bt)f(t) = k\cos(bt)
  • f(t)=ksin(bt)f(t) = k\sin(bt)
  • f(t)=kcos(bt)+lsin(bt)f(t) = k\cos(bt) + l\sin(bt)

First Method

  1. Assume:

    xp=Acos(bt)+Bsin(bt)x_p = A\cos(bt) + B\sin(bt)
  2. Compute derivatives xpx_p' and xpx_p''.

  3. Substitute into:

    axp+bxp+cxp=f(t)ax''_p + bx'_p + cx_p = f(t)
  4. Match coefficients of cos(bt)\cos(bt) and sin(bt)\sin(bt) on both sides to find AA and BB.

Second Method (if First Fails)

Use if:

axp+bxp+cxp=0 when substituted into f(t)0ax''_p + bx'_p + cx_p = 0 \text{ when substituted into } f(t) \ne 0
  1. Assume:

    xp=Atcos(bt)+Btsin(bt)x_p = At\cos(bt) + Bt\sin(bt)
  2. Compute xpx_p' and xpx_p'' using product rule.

  3. Substitute into equation and solve for AA, BB.

Note:

  • Try First Method first.
  • If it fails, use the Second Method.

10.4 Forced Oscillations

  • Focus: solving DEs with oscillating forcing function.

General Form

x+ω2x=asin(bt),x(0)=0,x(0)=0x'' + \omega^2 x = a\sin(bt), \quad x(0) = 0,\quad x'(0) = 0
  • Mass starts at rest and equilibrium.

Three Cases

  1. Case 1: ωb\omega \ne b (Periodic Motion)
    Use:

    xp=Acos(bt)+Bsin(bt)x_p = A\cos(bt) + B\sin(bt)
  2. Case 2: ω=b\omega = b (Resonance)
    Use:

    xp=Atcos(bt)+Btsin(bt)x_p = At\cos(bt) + Bt\sin(bt)
  3. Case 3: ωb\omega \approx b (Beat)
    Use same form as Case 1, but will produce beat phenomenon.

General Solution

x(t)=xc+xpx(t) = x_c + x_p
  • Use initial conditions to solve for constants c1c_1, c2c_2.

10.5 Beat Behavior

  • Equation:

    x+ω2x=asin(bt)x'' + \omega^2 x = a \sin(bt)
  • If ωb\omega \approx b, the system exhibits beats — the amplitude varies slowly over time.


11. Sequences & Series

11.1 Introduction to Infinite Sequences and Series

Geometric Sequences

  • The general form of a geometric sequence:

    a, ar, ar2, ar3, , arn, a,\ ar,\ ar^2,\ ar^3,\ \dots,\ ar^n,\ \dots
    • aa is the first term
    • rr is the common ratio

Geometric Series

  • An infinite geometric series is written as:

    n=1an=a1+a2+a3++an+\sum^{\infty}_{n=1} a_n = a_1 + a_2 + a_3 + \dots + a_n + \dots
  • The n-th partial sum of the series is:

    Sn=a1+a2+a3++an=k=1nakS_n = a_1 + a_2 + a_3 + \dots + a_n = \sum^{n}_{k=1} a_k
    • Examples:
      • S1=a1S_1 = a_1
      • S2=a1+a2S_2 = a_1 + a_2
      • S3=a1+a2+a3S_3 = a_1 + a_2 + a_3
      • and so on...

Sum Formulas

  • Sum to nn terms:

    Sn=a(1rn)1r,r1S_n = \frac{a(1 - r^n)}{1 - r}, \quad r \ne 1
  • Sum to infinity (when r<1|r| < 1):

    S=a1rS_\infty = \frac{a}{1 - r}

References


11.2 The Ratio Test and Power Series

The Ratio Test

Given a series an\sum a_n:

limnan+1an=L\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L
  • If L<1L < 1: the series converges
  • If L>1L > 1 or L=L = \infty: the series diverges
  • If L=1L = 1: the test is inconclusive

Example

Given:

an=2n+53na_n = \frac{2^n + 5}{3^n}

Then:

an+1=2n+1+53n+1a_{n+1} = \frac{2^{n+1} + 5}{3^{n+1}}

Substitute into the ratio test formula.

Power Series

Power series form and convergence not elaborated here but related to the ratio test for determining convergence radius.

References