riaology
Calculus I / Formula & reasoning reference

A compact map.
Keep the meaning.

Find a relationship, check its assumptions, and connect it to the model. Each entry includes a distinction worth remembering.

Back to the course ↗

35 lesson references found.

Foundations and limits

Functions describe a changing system.

Full visual lesson ↗

A cart's position depends on time. Choose an input and watch its position respond; the graph records that relationship rather than the shape of its path.

Model relationship

s(t)=at2,s:[0,8]→Rs(t)=at^2,\quad s:[0,8]\to\mathbb R

Connected relationship

f:D→R,s(t)=at2f:D\to\mathbb R,\quad s(t)=at^2

Assumptions and scope

A cart starts at the origin with position at². The coefficient a is half the constant physical acceleration.

Foundations and limits

Undo a measurement with an inverse.

Full visual lesson ↗

A sensor reports a squared displacement. To recover displacement, you must know which sign of the original input is allowed.

Model relationship

f(x)=x2 (x≥0),f−1(y)=yf(x)=x^2\ (x\ge0),\quad f^{-1}(y)=\sqrt y

Connected relationship

f−1(f(x))=x(x≥0)f^{-1}(f(x))=x\quad(x\ge0)

Assumptions and scope

Restrict the square function to nonnegative inputs so it is one-to-one.

Foundations and limits

Follow a value toward a missing point.

Full visual lesson ↗

A measurement formula can have a hole at one input while its nearby predictions approach a perfectly finite value.

Model relationship

f(x)=x2−4x−2=x+2 (x≠2),lim⁡x→2f(x)=4f(x)=\frac{x^2-4}{x-2}=x+2\ (x\ne2),\quad\lim_{x\to2}f(x)=4

Connected relationship

x2−4=(x−2)(x+2)x^2-4=(x-2)(x+2)

Assumptions and scope

The original function is undefined at x=2.x=2. At h=0h=0 the readout shows its limit, not a newly defined function value.

Foundations and limits

A limit and a value must agree.

Full visual lesson ↗

A sensor calibration should not produce an isolated jump in output. Move the assigned value and see whether it joins the surrounding response.

Model relationship

f(x)={x+2x≠2cx=2,continuous at 2 iff c=4f(x)=\begin{cases}x+2&x\ne2\\c&x=2\end{cases},\quad\text{continuous at 2 iff }c=4

Connected relationship

lim⁡x→af(x)=f(a)\lim_{x\to a}f(x)=f(a)

Assumptions and scope

The left and right limits are both 4. Only the single assigned value changes.

Foundations and limits

A continuous response must cross intermediate levels.

Full visual lesson ↗

If a continuous temperature or sensor response changes from below a target to above it, some intermediate input reaches that target.

Model relationship

f(x)=x3+x,f(−1)=−2, f(2)=10,f(c)=L for some c∈(−1,2)f(x)=x^3+x,\quad f(-1)=-2,\ f(2)=10,\quad f(c)=L\text{ for some }c\in(-1,2)

Connected relationship

f(a)<L<f(b)⇒∃c∈(a,b):f(c)=Lf(a)<L<f(b)\Rightarrow\exists c\in(a,b):f(c)=L

Assumptions and scope

The function is continuous on [−1,2]. Controls keep L strictly between the endpoint values; a bisection calculation locates the guaranteed crossing.

Foundations and limits

A tiny angle connects arc length and height.

Full visual lesson ↗

A small rotation changes a point's vertical coordinate almost in proportion to the rotation angle. The unit chosen for that angle matters.

Model relationship

lim⁡h→0sin⁡hh=1\lim_{h\to0}\frac{\sin h}{h}=1

Connected relationship

lim⁡x→0sin⁡(kx)x=k\lim_{x\to0}\frac{\sin(kx)}x=k

Assumptions and scope

Angles are in radians. At zero the curve displays the continuous extension of the quotient.

Foundations and limits

A response can level off without ever reaching its limit.

Full visual lesson ↗

Increasing the input can produce diminishing changes. A rational response illustrates how a horizontal asymptote emerges.

Model relationship

f(x)=3x+1x+2,lim⁡x→∞f(x)=3f(x)=\frac{3x+1}{x+2},\quad\lim_{x\to\infty}f(x)=3

Connected relationship

3x+1x+2=3+1/x1+2/x\frac{3x+1}{x+2}=\frac{3+1/x}{1+2/x}

Assumptions and scope

A rational response with a horizontal asymptote; denominator is positive over this control range.

Foundations and limits

Choose an input tolerance that guarantees an output tolerance.

Full visual lesson ↗

A calibration engineer specifies an allowed output error. The slope determines how tightly the input must be controlled.

Model relationship

f(x)=mx+1,L=ma+1,∣f(x)−L∣=∣m∣∣x−a∣<ε if ∣x−a∣<δ=ε∣m∣f(x)=mx+1,\quad L=ma+1,\quad |f(x)-L|=|m||x-a|<\varepsilon\ \text{if }|x-a|<\delta=\frac\varepsilon{|m|}

Connected relationship

0<∣x−a∣<δ⇒∣f(x)−L∣<ε0<|x-a|<\delta\Rightarrow|f(x)-L|<\varepsilon

Assumptions and scope

Linear function with positive slope m. The strict inequalities apply inside the δ interval, not at its boundary.

Derivatives and connected rates

Watch a secant become a tangent.

Full visual lesson ↗

Two position observations give average velocity. Bringing their times closer reveals instantaneous velocity.

Model relationship

s(t)=t2,s(t+h)−s(t)h=2t+h→s′(t)=2ts(t)=t^2,\quad\frac{s(t+h)-s(t)}h=2t+h\to s'(t)=2t

Connected relationship

s′(t)=lim⁡h→0s(t+h)−s(t)hs'(t)=\lim_{h\to0}\frac{s(t+h)-s(t)}h

Assumptions and scope

A cart position s=t2\mathrm{s}=t^2 m. h is positive so the second secant point is later in time.

Derivatives and connected rates

Use derivative rules to measure sensitivity.

Full visual lesson ↗

A nonlinear response can become much more sensitive as its input grows. Compare its value with its local slope.

Model relationship

f(x)=xn,f′(x)=nxn−1f(x)=x^n,\quad f'(x)=nx^{n-1}

Connected relationship

ddx(af+bg+c)=af′+bg′\frac{d}{dx}(af+bg+c)=af'+bg'

Assumptions and scope

Positive inputs and integer powers. The derivative curve is shown alongside the function.

Derivatives and connected rates

Changing one dimension can change another.

Full visual lesson ↗

A fixed-perimeter rectangle grows one side while shrinking the other. Its area depends on both changes.

Model relationship

A(x)=x(5−x),A′=1(5−x)+x(−1)=5−2xA(x)=x(5-x),\quad A'=1(5-x)+x(-1)=5-2x

Connected relationship

(fg)′=f′g+fg′(fg)'=f'g+fg'

Assumptions and scope

Rectangle sides x and 5−x5-x; total perimeter is fixed at 10 m. Both factors vary.

Derivatives and connected rates

A ratio responds to numerator and denominator changes.

Full visual lesson ↗

A rectangle's aspect ratio can change sharply as one side narrows, even when total perimeter remains fixed.

Model relationship

R(x)=x5−x,R′=(5−x)(1)−x(−1)(5−x)2=5(5−x)2R(x)=\frac{x}{5-x},\quad R'=\frac{(5-x)(1)-x(-1)}{(5-x)^2}=\frac5{(5-x)^2}

Connected relationship

(fg)′=f′g−fg′g2\left(\frac fg\right)'=\frac{f'g-fg'}{g^2}

Assumptions and scope

Rectangle sides x and 5−x5-x. The denominator is positive over the control range.

Derivatives and connected rates

An oscillation's slope has its own phase.

Full visual lesson ↗

A vibrating mass stops momentarily at a turning point, yet moves fastest near the center. Follow displacement and velocity together.

Model relationship

s(t)=2sin⁡(ωt),v(t)=2ωcos⁡(ωt)s(t)=2\sin(\omega t),\quad v(t)=2\omega\cos(\omega t)

Connected relationship

s′′=−ω2ss''=-\omega^2s

Assumptions and scope

Sinusoidal displacement with amplitude 2 m; radians and time in seconds.

Derivatives and connected rates

A rate can pass through two transformations.

Full visual lesson ↗

A sensor first rescales time-dependent input, then cubes the response. Follow each stage to understand its final rate.

Model relationship

y(t)=(1+0.2t)3,dydt=3(1+0.2t)2(0.2)y(t)=(1+0.2t)^3,\quad\frac{dy}{dt}=3(1+0.2t)^2(0.2)

Connected relationship

dydt=dydududt\frac{dy}{dt}=\frac{dy}{du}\frac{du}{dt}

Assumptions and scope

A dimensionless sensor response to a linearly changing dimensionless input.

Derivatives and connected rates

A constraint determines the allowed slope.

Full visual lesson ↗

A particle constrained to a circular guide cannot change its horizontal position independently of its height.

Model relationship

x2+y2=9,2x+2ydydx=0,dydx=−xyx^2+y^2=9,\quad2x+2y\frac{dy}{dx}=0,\quad\frac{dy}{dx}=-\frac xy

Connected relationship

ddx(y2)=2ydydx\frac d{dx}(y^2)=2y\frac{dy}{dx}

Assumptions and scope

Upper semicircle of radius 3 m. Endpoints with vertical tangents are excluded from the control range.

Derivatives and connected rates

An angle changes at a nonlinear rate.

Full visual lesson ↗

A viewing angle responds less and less to the same extra horizontal displacement as the geometry spreads out.

Model relationship

θ(x)=arctan⁡(x/2),θ′(x)=1/21+(x/2)2\theta(x)=\arctan(x/2),\quad\theta'(x)=\frac{1/2}{1+(x/2)^2}

Connected relationship

ddxarcsin⁡x=11−x2,ddxarctan⁡x=11+x2\frac d{dx}\arcsin x=\frac1{\sqrt{1-x^2}},\quad\frac d{dx}\arctan x=\frac1{1+x^2}

Assumptions and scope

Viewing geometry has a fixed adjacent side of length 2 m; angles in radians.

Derivatives and connected rates

A proportional growth rate makes an exponential.

Full visual lesson ↗

When growth is proportional to the amount already present, equal time intervals multiply rather than add the same quantity.

Model relationship

P(t)=2ekt,P′=kP,ddxln⁡x=1xP(t)=2e^{kt},\quad P'=kP,\quad\frac{d}{dx}\ln x=\frac1x

Connected relationship

(sinh⁡x)′=cosh⁡x,(cosh⁡x)′=sinh⁡x,(ln⁡x)′=1/x(\sinh x)'=\cosh x,\quad(\cosh x)'=\sinh x,\quad(\ln x)'=1/x

Assumptions and scope

Continuous exponential growth with fixed k and no carrying-capacity limitation.

Derivatives and connected rates

Position, velocity, and acceleration tell different stories.

Full visual lesson ↗

A cart can move backward while accelerating forward. The signs of its velocity and acceleration determine whether its speed rises or falls.

Model relationship

s=t3−3t2,v=3t2−6t,a=6t−6s=t^3-3t^2,\quad v=3t^2-6t,\quad a=6t-6

Connected relationship

v=s′,a=v′=s′′,speed=∣v∣v=s',\quad a=v'=s'',\quad\text{speed}=|v|

Assumptions and scope

A prescribed one-dimensional trajectory; signed velocity differs from speed.

Derivatives and connected rates

The same inflow can raise a cone's level at different rates.

Full visual lesson ↗

A conical tank widens as it fills. Change its depth and flow to see why level rise slows even with a constant inlet.

Model relationship

r=h/2,V=π12h3,dVdt=π4h2dhdtr=h/2,\quad\mathcal V=\frac\pi{12}h^3,\quad\frac{d\mathcal V}{dt}=\frac\pi4h^2\frac{dh}{dt}

Connected relationship

r=h2,V=π3r2h=π12h3r=\frac h2,\quad\mathcal V=\frac\pi3r^2h=\frac\pi{12}h^3

Assumptions and scope

An inverted cone has height 4 m and top radius 2 m. Prescribed inflow, no outflow, no evaporation.

Using derivatives

A tangent is a local prediction tool.

Full visual lesson ↗

A small measurement change can be translated into an approximate output change using sensitivity at a reference point.

Model relationship

f(x)=x,L(x)=2+14(x−4)f(x)=\sqrt x,\quad L(x)=2+\frac14(x-4)

Connected relationship

f(a+h)≈f(a)+f′(a)hf(a+h)\approx f(a)+f'(a)h

Assumptions and scope

Linearization near x=4.x=4. All controls keep x positive.

Using derivatives

Find critical points, then inspect the interval.

Full visual lesson ↗

Choosing a maximum response requires more than finding where a derivative is zero. Endpoints may win.

Model relationship

f(x)=x3−3x,f′=3(x2−1),f′′=6xf(x)=x^3-3x,\quad f'=3(x^2-1),\quad f''=6x

Connected relationship

f′(x)=0gives candidates, not automatic extremaf'(x)=0\quad\text{gives candidates, not automatic extrema}

Assumptions and scope

Closed interval [−2,2]. Critical points and endpoints all matter for absolute extrema.

Using derivatives

An average rate must occur somewhere instantaneously.

Full visual lesson ↗

A trip with a known average velocity must match that velocity at some instant under the theorem's smoothness conditions.

Model relationship

f′(c)=f(b)−f(a)b−a,f=x2, a=0⇒c=b/2f'(c)=\frac{f(b)-f(a)}{b-a},\quad f=x^2,\ a=0\Rightarrow c=b/2

Connected relationship

f(b)−f(a)=f′(c)(b−a)f(b)-f(a)=f'(c)(b-a)

Assumptions and scope

The function is continuous on [0,b] and differentiable on (0,b), so the mean value theorem applies.

Using derivatives

Read shape from first and second derivatives.

Full visual lesson ↗

A response can be increasing while flattening out. Separate slope sign from the direction in which slope changes.

Model relationship

f=x3−x,f′=3x2−1,f′′=6xf=x^3-x,\quad f'=3x^2-1,\quad f''=6x

Connected relationship

f′>0: increasing,f′′>0: concave upf'>0:\text{ increasing},\quad f''>0:\text{ concave up}

Assumptions and scope

Derivative signs diagnose increasing/decreasing behavior; second derivative signs diagnose concavity.

Using derivatives

Check the indeterminate form before changing the quotient.

Full visual lesson ↗

An apparently undefined small-input response can have a finite limit. A valid limit tool should be applied only after checking its hypotheses.

Model relationship

lim⁡x→0ex−1x=0/0lim⁡x→0ex1=1\lim_{x\to0}\frac{e^x-1}{x}\overset{0/0}{=}\lim_{x\to0}\frac{e^x}{1}=1

Connected relationship

lim⁡fg=lim⁡f′g′under the rule’s hypotheses\lim\frac fg=\lim\frac{f'}{g'}\quad\text{under the rule's hypotheses}

Assumptions and scope

The limit is a 0/0 form and the differentiability hypotheses hold near zero. It is not an equality of the quotients for nonzero x.

Using derivatives

A constraint turns a design question into one variable.

Full visual lesson ↗

A river supplies one side of an enclosure. Decide how to distribute a fixed amount of fencing to maximize useful area.

Model relationship

2x+y=30,A=x(30−2x),A′=30−4x=02x+y=30,\quad A=x(30-2x),\quad A'=30-4x=0

Connected relationship

A′′=−4<0A''=-4<0

Assumptions and scope

A river forms one side, so 30 m of fence encloses only three sides. Feasible x is between 0 and 15 m.

Using derivatives

A tangent can guide a root search.

Full visual lesson ↗

A numerical solver predicts where a local tangent hits zero, then repeats the prediction from that new location.

Model relationship

xn+1=xn−f(xn)f′(xn)=12(xn+2xn)x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=\frac12\left(x_n+\frac2{x_n}\right)

Connected relationship

xn+1=xn−f(xn)f′(xn)x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}

Assumptions and scope

Newton iteration for the positive square root of 2. Initial guess is positive so zero-derivative failure is avoided in this example.

Accumulation and integration

An integral grows from many small contributions.

Full visual lesson ↗

A changing velocity contributes a little displacement over each short time interval. Rectangles make the approximation visible.

Model relationship

∑i=1nf(ti∗)Δt→∫02t2 dt=83\sum_{i=1}^nf(t_i^*)\Delta t\to\int_0^2t^2\,dt=\frac83

Connected relationship

Δs≈∑iv(ti∗)Δt\Delta s\approx\sum_i v(t_i^*)\Delta t

Assumptions and scope

Integrating velocity t² m/s over 0–2 s. Three sampling rules; rectangle heights are visible in the diagram.

Accumulation and integration

Differentiation and accumulation undo one another.

Full visual lesson ↗

Accumulated volume or energy changes at the rate entering at its current endpoint. See the accumulated curve and its slope together.

Model relationship

F(x)=∫0x(1+t2) dt=x+x33,F′(x)=1+x2F(x)=\int_0^x(1+t^2)\,dt=x+\frac{x^3}3,\quad F'(x)=1+x^2

Connected relationship

∫abf(t) dt=G(b)−G(a),G′=f\int_a^bf(t)\,dt=G(b)-G(a),\quad G'=f

Assumptions and scope

A continuous integrand. The accumulated area is signed if endpoints are reversed; control range is nonnegative here.

Accumulation and integration

Initial data selects an antiderivative.

Full visual lesson ↗

Two carts can have the same velocity history while starting at different positions. The integration constant carries that missing information.

Model relationship

v(t)=2t,s(t)=t2+Cv(t)=2t,\quad s(t)=t^2+C

Connected relationship

∫2t dt=t2+C\int2t\,dt=t^2+C

Assumptions and scope

Integrating velocity leaves an arbitrary constant; an initial position fixes it.

Accumulation and integration

Displacement can cancel while distance accumulates.

Full visual lesson ↗

A cart reverses direction. Its signed displacement and total distance answer different questions.

Model relationship

v(t)=t−2,Δs=∫0Tv dt=T22−2T,d=∫0T∣v∣ dtv(t)=t-2,\quad\Delta s=\int_0^Tv\,dt=\frac{T^2}2-2T,\quad d=\int_0^T|v|\,dt

Connected relationship

distance=∫∣v∣ dt≥∣∫v dt∣\text{distance}=\int|v|\,dt\ge\left|\int v\,dt\right|

Assumptions and scope

One-dimensional motion reverses at 2 s. Distance uses speed and is never negative.

Accumulation and integration

Reverse the chain rule with a new variable.

Full visual lesson ↗

A composed rate law can become simpler when its inner quantity is treated as the integration variable.

Model relationship

∫0b2xex2 dx=∫0b2eu du=eb2−1\int_0^b2xe^{x^2}\,dx=\int_0^{b^2}e^u\,du=e^{b^2}-1

Connected relationship

u=g(x),du=g′(x) dxu=g(x),\quad du=g'(x)\,dx

Assumptions and scope

Let u=x2u=x^2 and transform both differential and bounds. This is the chain rule run backward.

Geometry and work

The top curve can switch in the middle.

Full visual lesson ↗

Two response curves can cross. A geometric area between them adds contributions on both sides of the crossing.

Model relationship

A=∫0b∣x−x2∣ dxA=\int_0^b|x-x^2|\,dx

Connected relationship

A=∫ab∣f−g∣ dxA=\int_a^b|f-g|\,dx

Assumptions and scope

Curves x and x² cross at 0 and 1. Geometric area requires switching top and bottom after 1.

Geometry and work

Stack disks and washers to build volume.

Full visual lesson ↗

A hollow pipe demonstrates why an annular cross-section uses squared radii. The same slicing idea works when radius changes along the axis.

Model relationship

V=∫03π(R2−r2) dx=3π(R2−r2)\mathcal V=\int_0^3\pi(R^2-r^2)\,dx=3\pi(R^2-r^2)

Connected relationship

dV=π[R(x)2−r(x)2] dxd\mathcal V=\pi[R(x)^2-r(x)^2]\,dx

Assumptions and scope

A length-3 m hollow cylinder is the simplest washer solid. r is a fixed fraction of R; r=0r=0 recovers disks.

Geometry and work

A strip parallel to the axis becomes a shell.

Full visual lesson ↗

A thin vertical strip can rotate into a hollow cylindrical layer. Its volume comes from circumference, height, and radial thickness.

Model relationship

V=∫0b2πx x dx=2πb33\mathcal V=\int_0^b2\pi x\,x\,dx=\frac{2\pi b^3}3

Connected relationship

dV=2π(radius)(height) dxd\mathcal V=2\pi(\text{radius})(\text{height})\,dx

Assumptions and scope

Rotate the region under y=xy=x on 0≤x≤b about the y axis. Each vertical strip becomes a cylindrical shell.

Geometry and work

A varying force requires an area, not one multiplication.

Full visual lesson ↗

Stretching a spring becomes harder as the displacement grows. Sum the little force–distance contributions.

Model relationship

F(x)=kx,W=∫0aF(x) dx=12ka2F(x)=kx,\quad W=\int_0^aF(x)\,dx=\frac12ka^2

Connected relationship

dW=F(x) dxdW=F(x)\,dx

Assumptions and scope

Quasi-static linear spring, stretched from its unstressed position; no friction or kinetic-energy change.