riaology — where curiosity flows

Integration techniques

Transfer a derivative to the simpler factor.

A time-weighted decaying response combines a polynomial with an exponential. Integration by parts reorganizes this product.

Model relationship

\int_0^bte^{-t}\,dt=\left[-(t+1)e^{-t}\right]_0^b=1-(b+1)e^{-b}

Connected relationship

\int u\,dv=uv-\int v\,du

Assumptions and scope

Use $u=t$ and $dv=e$ ^(−t)dt. Choosing the polynomial for u makes its derivative simpler.

Integration techniques

An identity can expose a simpler integrand.

Full visual lesson ↗

The time-average of a squared oscillation is connected to an integral of sine squared. Reduce the power before integrating.

Model relationship

\sin^2x=\frac{1-\cos2x}{2},\quad\int_0^b\sin^2x\,dx=\frac b2-\frac{\sin2b}{4}

Connected relationship

\sin^2x=\frac{1-\cos2x}{2}

Assumptions and scope

Radian angles; the half-angle identity reduces an even power.

Integration techniques

Turn a radical into a trigonometric identity.

Full visual lesson ↗

A circular cross-section produces square roots such as sqrt( $a^2-x^2$ ). An angle parameter makes that geometry simpler.

Model relationship

x=2\sin\theta,\quad\int_0^b\sqrt{4-x^2}\,dx=\frac b2\sqrt{4-b^2}+2\arcsin(b/2)

Connected relationship

\sqrt{a^2-a^2\sin^2\theta}=a|\cos\theta|

Assumptions and scope

Choose −π/2≤θ≤π/2 so cosθ is nonnegative. The square root is the upper circle of radius 2.

Integration techniques

A hanging cable links exponentials and arc length.

Full visual lesson ↗

A uniformly weighted flexible cable hangs in a catenary rather than a parabola. Hyperbolic functions describe its ideal shape.

Model relationship

y=a\cosh(x/a),\quad y'=\sinh(x/a),\quad\cosh^2u-\sinh^2u=1

Connected relationship

\cosh u=\frac{e^u+e^{-u}}2,\quad\sinh u=\frac{e^u-e^{-u}}2

Assumptions and scope

Ideal catenary shape for a uniform flexible cable under gravity; height reference is a convention.

Integration techniques

Break a rational integrand into recognizable pieces.

Full visual lesson ↗

A rational response with multiple factors can integrate more easily when decomposed into simple reciprocal terms.

Model relationship

\frac1{(x+1)(x+2)}=\frac1{x+1}-\frac1{x+2},\quad\int_0^b=\ln\frac{2(b+1)}{b+2}

Connected relationship

\frac1{(x+1)(x+2)}=\frac1{x+1}-\frac1{x+2}

Assumptions and scope

Distinct linear factors, no poles on the integration interval. Partial fractions requires proper rational form first.

Integration techniques

An infinite tail needs a limiting argument.

Full visual lesson ↗

A curve can stretch forever while its area remains finite. Compare tail exponents and increasingly distant cutoffs.

Model relationship

\int_1^Bx^{-p}\,dx=\begin{cases}\frac{B^{1-p}-1}{1-p}&p\ne1\\\ln B&p=1\end{cases},\quad\int_1^\infty x^{-p}\,dx\text{ converges iff }p>1

Connected relationship

\int_1^\infty f=\lim_{B\to\infty}\int_1^Bf

Assumptions and scope

A finite cutoff is not the improper integral. Convergence requires the limit as B tends to infinity.

Integration techniques

Estimate an integral when no elementary formula is handy.

Full visual lesson ↗

A measured rate can be sampled at discrete times. Trapezoids or parabolic panels turn those observations into an accumulated quantity.

Model relationship

T_n=h\left[\frac{f_0+f_n}{2}+\sum_{i=1}^{n-1}f_i\right],\quad S_n=\frac h3[f_0+4f_1+2f_2+\cdots+f_n]

Connected relationship

h=(b-a)/n

Assumptions and scope

Integrate exp(−x²) on [0,2]. Simpson uses even n. A high-resolution Simpson value provides a numerical reference, not an elementary exact antiderivative.

Integrals in the physical world

Probability is area beneath a density.

Full visual lesson ↗

Waiting times can be modeled by a density curve. The chance of an event before a deadline is accumulated area, not curve height.

Model relationship

f(t)=\lambda e^{-\lambda t}\ (t\ge0),\quad P(T\le b)=\int_0^bf(t)\,dt=1-e^{-\lambda b}

Connected relationship

P(a<T<b)=\int_a^bf(t)\,dt

Assumptions and scope

Exponential waiting-time model, nonnegative time, constant event rate λ.

Integrals in the physical world

Curve length adds tiny diagonal segments.

Full visual lesson ↗

A road or cable follows a curve rather than a straight horizontal line. Its length must include both horizontal and vertical change.

Model relationship

y=\frac23x^{3/2},\quad L=\int_0^b\sqrt{1+(y')^2}\,dx=\frac23[(1+b)^{3/2}-1]

Connected relationship

ds=\sqrt{dx^2+dy^2}

Assumptions and scope

Coordinates are numeric meter values: $y=\frac23(1\,\mathrm m)[x/(1\,\mathrm m)]^{3/2}$ . The arc-length density includes horizontal and vertical movement.

Integrals in the physical world

Rotation sweeps circumference along a slant length.

Full visual lesson ↗

A cone's side area comes from rings with circumference set by radius and slant thickness set by the curve.

Model relationship

y=x,\quad S=\int_0^b2\pi y\sqrt{1+(y')^2}\,dx=\pi\sqrt2\,b^2

Connected relationship

dS=2\pi(\text{radius})\,ds

Assumptions and scope

Rotate $y=x$ , 0≤x≤b, about the x axis. This is the lateral area of a cone, excluding the base.

Integrals in the physical world

Pressure builds force one submerged strip at a time.

Full visual lesson ↗

A deeper gate experiences more pressure and a force that acts below its center. Change gate height, width, and top depth to see all three effects.

Model relationship

p(y)=\rho g(d+y),\quad dF=p(y)b\,dy,\quad F=\rho gb\left(dH+\frac{H^2}{2}\right)

Connected relationship

y_{CP}=\frac{\int_0^H y(d+y)\,dy}{\int_0^H(d+y)\,dy}

Assumptions and scope

Water ρ=1000 kg/m³, $g=9.81 \mathrm{m}/\mathrm{s}$ ², vertical rectangular gate. Atmospheric pressure cancels because the other side is exposed to the same atmosphere.

Integrals in the physical world

The balance point is a weighted average.

Full visual lesson ↗

A rod can be heavier at one end even with a uniform shape. Its balance point depends on mass distribution.

Model relationship

\lambda(x)=1+kx,\quad M=\int_0^b\lambda\,dx=b+\frac{kb^2}2,\quad\bar x=\frac{\int_0^bx\lambda\,dx}{M}

Connected relationship

\bar x=\frac{\int x\,dm}{\int dm}

Assumptions and scope

A thin rod along the x axis with line density in kg/m; the constant density scale is 1 kg/m.

Differential equations and models

A differential equation specifies how a state changes.

Full visual lesson ↗

A population rate law links growth to the current amount. Instead of prescribing the curve, the equation prescribes its slope.

Model relationship

y'=0.3y,\quad y(0)=y_0,\quad y(t)=y_0e^{0.3t}

Connected relationship

y'=f(t,y),\quad y(t_0)=y_0

Assumptions and scope

A first-order initial-value problem with constant proportional growth. The derivative specifies a rate, not the value itself.

Differential equations and models

A hot cup approaches the room temperature.

Full visual lesson ↗

Cooling depends on the temperature gap. Change the initial temperature and rate constant to see a family of physical solutions.

Model relationship

T'=-k(T-20),\quad T(t)=20+(T_0-20)e^{-kt}

Connected relationship

\ln|T-20|=-kt+C,\quad\tau=1/k

Assumptions and scope

Lumped uniform-temperature object, ambient 20°C, constant k; evaporation and heat generation neglected.

Differential equations and models

Approximate a trajectory by following short tangents.

Full visual lesson ↗

An Euler solver predicts the next temperature using the present cooling rate. Compare its line segments with the exact curve.

Model relationship

T_{n+1}=T_n+\Delta t[-0.1(T_n-20)],\quad T_0=80

Connected relationship

y_{n+1}=y_n+h f(t_n,y_n)

Assumptions and scope

Euler approximates the same cooling IVP. A final shorter step lands exactly on the chosen endpoint.

Differential equations and models

Growth changes when resources become limited.

Full visual lesson ↗

An exponential model grows forever, but a carrying capacity makes growth slow and can pull an oversized population downward.

Model relationship

P'=rP(1-P/100),\quad P(t)=\frac{100}{1+(100/P_0-1)e^{-rt}}

Connected relationship

P'=0\text{ at }P=0\text{ or }P=K

Assumptions and scope

Carrying capacity 100; positive initial population may start below or above it. r is constant.

Differential equations and models

An integrating factor turns a sum into a product derivative.

Full visual lesson ↗

A system with constant input and proportional loss approaches a balance where input equals loss.

Model relationship

y'+0.2y=b,\quad y(0)=1,\quad y=5b+(1-5b)e^{-0.2t}

Connected relationship

\mu=e^{\int a(t)dt},\quad(\mu y)'=\mu b

Assumptions and scope

Constant-coefficient linear first-order equation; units are consistent with time in minutes.

Differential equations and models

A tank can keep its volume while changing its contents.

Full visual lesson ↗

Clean water can become salty while equal inlet and outlet flows keep the level fixed. The mass balance must track solute separately from total liquid.

Model relationship

M'=q c_{in}-q\frac M{100},\quad M(0)=0,\quad M=100c_{in}(1-e^{-qt/100})

Connected relationship

\text{solute accumulation}=\text{solute in}-\text{solute out}

Assumptions and scope

Well-mixed 100 L tank, equal inlet/outlet volume flow, constant density/volume, initially no solute.

Sequences and infinite series

A sequence approaches a value term by term.

Full visual lesson ↗

Repeated approximations can settle toward a target. The index labels discrete observations rather than continuous time.

Model relationship

a_n=\frac n{n+1},\quad\lim_{n\to\infty}a_n=1

Connected relationship

1-a_n=\frac1{n+1}\to0

Assumptions and scope

The input is an integer. Connecting segments are a visual guide, not additional sequence terms.

Sequences and infinite series

Partial sums determine a series' value.

Full visual lesson ↗

A repeating fractional contribution can add to a finite total. Follow the partial sums instead of only the individual contributions.

Model relationship

S_n=\sum_{j=0}^{n-1}r^j=\frac{1-r^n}{1-r},\quad S_\infty=\frac1{1-r}\ (|r|<1)

Connected relationship

(1-r)S_n=1-r^n

Assumptions and scope

Geometric series starts with 1; controls keep |r|<1. Negative ratios alternate.

Sequences and infinite series

An infinite sum can be compared with an infinite area.

Full visual lesson ↗

A decreasing positive sequence of contributions can be bracketed by areas. The same construction also bounds the remainder.

Model relationship

\sum_{n=1}^\infty n^{-p}\text{ converges iff }p>1,\quad\int_{N+1}^\infty f\le R_N\le\int_N^\infty f

Connected relationship

\int_{N+1}^\infty f\le\sum_{n=N+1}^\infty f(n)\le\int_N^\infty f

Assumptions and scope

Positive continuous decreasing x^(−p) on x≥1. Tail bounds apply only in the convergent case p>1.

Sequences and infinite series

A known convergent sum can control a smaller one.

Full visual lesson ↗

A complicated positive term may be bounded by a simpler power-law term. Use that inequality to decide convergence.

Model relationship

0<\frac1{n^2+n}<\frac1{n^2},\quad\sum_{n=1}^N\frac1{n(n+1)}=1-\frac1{N+1}

Connected relationship

\lim_{n\to\infty}\frac{a_n}{b_n}=L\in(0,\infty)

Assumptions and scope

Positive terms permit direct comparison. This series also telescopes because 1/[n( $n+1$ )]= $1/n-1$ /( $n+1$ ).

Sequences and infinite series

A factorial can outrun an exponential.

Full visual lesson ↗

Terms involving factorials can initially grow, yet eventually shrink rapidly. Inspect how one term compares with the next.

Model relationship

u_n=\frac{a^n}{n!},\quad\left|\frac{u_{n+1}}{u_n}\right|=\frac{|a|}{n+1}\to0

Connected relationship

L=\lim\left|\frac{a_{n+1}}{a_n}\right|

Assumptions and scope

Positive a and factorial denominators. The ratio limit is zero, so the series converges absolutely for every fixed a.

Sequences and infinite series

A root test exposes an exponential factor.

Full visual lesson ↗

An nth root can separate exponential growth from slower polynomial effects. Its limiting value decides most cases.

Model relationship

u_n=\frac{a^n}{n^2},\quad\sqrt[n]{|u_n|}=\frac{|a|}{n^{2/n}}\to|a|

Connected relationship

n^{2/n}=e^{2\ln n/n}\to1

Assumptions and scope

Positive a. Root limit <1 proves absolute convergence; >1 proves divergence; =1 is inconclusive, and this particular $a=1$ series converges by the p-series test.

Sequences and infinite series

Cancellation can produce conditional convergence.

Full visual lesson ↗

Positive and negative contributions can settle even when the sum of their magnitudes grows without bound.

Model relationship

\sum_{n=1}^\infty\frac{(-1)^{n+1}}n=\ln2,\quad|R_N|\le\frac1{N+1}

Connected relationship

|S-S_N|\le b_{N+1}

Assumptions and scope

Term magnitudes decrease to zero. The harmonic absolute-value series diverges, so convergence is conditional.

Sequences and infinite series

An infinite polynomial has a convergence interval.

Full visual lesson ↗

A finite polynomial can be evaluated anywhere, but an infinite series may represent a function only inside a specific radius.

Model relationship

\sum_{n=0}^\infty x^n=\frac1{1-x}\ (|x|<1),\quad R=1

Connected relationship

\sum c_n(x-a)^n,\quad|x-a|<R

Assumptions and scope

Finite polynomials exist for every x, but the infinite geometric identity is valid only inside |x|<1. Both endpoints diverge.

Sequences and infinite series

Match more derivatives to improve a local approximation.

Full visual lesson ↗

An exponential response can be approximated near zero by a line, a quadratic, or a higher-degree polynomial. Compare error as you change degree.

Model relationship

P_N(x)=\sum_{n=0}^N\frac{x^n}{n!},\quad e^x=\sum_{n=0}^\infty\frac{x^n}{n!}

Connected relationship

P_N(x)=\sum_{n=0}^N\frac{f^{(n)}(a)}{n!}(x-a)^n

Assumptions and scope

Maclaurin expansion of exp(x). Finite polynomial accuracy depends on degree and distance from the center.

Curves and spatial geometry

A path can use a parameter instead of a single graph.

Full visual lesson ↗

An orbit can pass through vertical tangents and loop back in x. A parameter records both coordinates without forcing y to be single-valued in x.

Model relationship

x=a\cos t,\quad y=\sin t,\quad\frac{dy}{dx}=\frac{dy/dt}{dx/dt},\quad\frac{ds}{dt}=\sqrt{(x')^2+(y')^2}

Connected relationship

L=\int\sqrt{(x')^2+(y')^2}\,dt

Assumptions and scope

Ellipse with vertical semiaxis 1 m; t is a parameter in radians, not necessarily physical time. At $t=0$ the tangent is vertical.

Curves and spatial geometry

A swept angle builds area from thin sectors.

Full visual lesson ↗

A cardioid is easier to describe by distance and direction from a pole. Sweep its angle and watch the enclosed region grow.

Model relationship

r(\theta)=1+\cos\theta,\quad A=\frac12\int_0^b(1+\cos\theta)^2\,d\theta=\frac{3b}{4}+\sin b+\frac{\sin2b}{8}

Connected relationship

x=r\cos\theta,\quad y=r\sin\theta,\quad dA=\frac12r^2d\theta

Assumptions and scope

Cardioid traced once on 0≤θ≤2π; swept region has nonnegative r and no repeated full turns.

Curves and spatial geometry

Components describe direction and magnitude.

Full visual lesson ↗

A displacement has both length and direction. Change its components and follow the arrow and its magnitude.

Model relationship

\mathbf v=\langle v_x,v_y\rangle,\quad|\mathbf v|=\sqrt{v_x^2+v_y^2}

Connected relationship

\hat{\mathbf v}=\mathbf v/|\mathbf v|\quad(\mathbf v\ne\mathbf0)

Assumptions and scope

A two-dimensional displacement vector. Length is nonnegative; zero vector has no direction.

Curves and spatial geometry

A point and direction generate a spatial line.

Full visual lesson ↗

A particle path can be described by a starting point plus a scaled direction vector. Its parameter controls position along the line.

Model relationship

\mathbf r(t)=\langle1,2,0\rangle+t\langle2,-1,1\rangle

Connected relationship

x=1+2t,\quad y=2-t,\quad z=t

Assumptions and scope

A line in three-dimensional space; graph shows its xy projection. Full coordinates include $z=t$ .

Curves and spatial geometry

Only the along-motion force component does work.

Full visual lesson ↗

Pushing sideways can exert a force without doing work along the displacement. Rotate the force to see the signed result.

Model relationship

W=\mathbf F\cdot\mathbf d=|\mathbf F||\mathbf d|\cos\theta

Connected relationship

\mathbf a\cdot\mathbf b=\sum_i a_i b_i

Assumptions and scope

Force magnitude 10 N and displacement magnitude 2 m. Degrees are converted to radians inside cosine.

Curves and spatial geometry

A perpendicular component creates torque.

Full visual lesson ↗

A wrench is most effective when the force is perpendicular to its handle. Compare that with force along the handle.

Model relationship

\boldsymbol\tau=\mathbf r\times\mathbf F,\quad|\boldsymbol\tau|=|\mathbf r||\mathbf F|\sin\theta

Connected relationship

\mathbf a\times\mathbf b=-\mathbf b\times\mathbf a

Assumptions and scope

Lever length 0.5 m and force magnitude 20 N. Direction follows the right-hand rule, perpendicular to their plane.

Curves and spatial geometry

A normal vector fixes a plane's orientation.

Full visual lesson ↗

A plane can be described by points whose displacement from a reference point is perpendicular to one normal vector.

Model relationship

\Pi:x+2y+2z=6,\quad d(P,\Pi)=\frac{|x_P+2y_P+2z_P-6|}{\sqrt{1^2+2^2+2^2}}

Connected relationship

\mathbf n\cdot(\mathbf r-\mathbf r_0)=0

Assumptions and scope

Point P=(1,1,z); normal vector is (1,2,2). Diagram shows the plane slice at the chosen z, not the full 3D plane.