← Unit 9

Unit 9 Cheat Sheet

Every formula in Parametric, Polar & Vector-Valued Functions — and, more importantly, when to use which. The pieces that look alike (displacement vs. distance, \(dr/d\theta\) vs. the tangent slope, one-curve vs. two-curve area) are exactly where points slip away, so each block ends with the decision that picks the right tool — and the hardest skill of all, finding the limits, gets its own panel at the bottom.

Parametric Calculus 9.1 – 9.3

A curve traced by \(x(t),y(t)\). Slopes and lengths come from the \(t\)-derivatives.

first derivative (slope)

second derivative (concavity)

arc length on [a, b]

Second derivative is a ratio too: differentiate the slope \(\dfrac{dy}{dx}\) with respect to \(t\), then divide by \(\dfrac{dx}{dt}\) again — never \(\dfrac{d^2y/dt^2}{d^2x/dt^2}\).

Horizontal or vertical tangent?

horizontal\(\dfrac{dy}{dt}=0\) and \(\dfrac{dx}{dt}\neq 0\)

vertical\(\dfrac{dx}{dt}=0\) and \(\dfrac{dy}{dt}\neq 0\)

trap

Slope is \(\dfrac{dy/dt}{dx/dt}\) — don't differentiate \(y\) by \(x\) directly.
If both derivatives are \(0\), the tangent is undefined there (a cusp), not automatically horizontal.

Vector-Valued Functions 9.4 – 9.6

Position \(\vec r(t)=\langle x(t),y(t)\rangle\). Differentiate for motion, integrate for accumulation.

velocity & acceleration

speed (a scalar)

displacement (vector) & distance (scalar)

position from velocity

Displacement or distance?

net change\(\displaystyle\int \vec v\,dt\) — integrate components, keep signs

total traveled\(\displaystyle\int |\vec v|\,dt\) — integrate the speed

Speeding up or slowing down?

\(\vec v\cdot\vec a\gt 0\)speeding up

\(\vec v\cdot\vec a\lt 0\)slowing down

trap

Speed combines both components: \(\sqrt{v_x^2+v_y^2}\) — never one component, never their plain sum.
Distance puts the magnitude inside the integral; displacement takes the magnitude outside.

Polar Graphing 9.7

A point is \((r,\theta)\): distance \(r\) from the origin at angle \(\theta\).

polar → rectangular

rectangular → polar

tangent slope (in the xy-plane)

\(\dfrac{dr}{d\theta}\) is not the slope. It's the rate the distance from the origin changes. The tangent slope needs the full formula above (from \(x=r\cos\theta,\ y=r\sin\theta\)).

circle

\(r=a\cos\theta\) or \(a\sin\theta\)

diameter \(a\); closes over \([0,\pi]\)

rose

\(r=a\cos(k\theta)\)

\(k\) odd \(\Rightarrow k\) petals; \(k\) even \(\Rightarrow 2k\)

cardioid

\(r=a\pm a\cos\theta\)

ratio \(a/b=1\); a single dimple to the pole

limaçon

\(r=a\pm b\cos\theta\)

\(a/b\lt 1\) inner loop; \(\ge 2\) convex

lemniscate

\(r^2=a^2\cos 2\theta\)

figure-eight; two loops, \(r=0\) at \(\theta=\pm\tfrac{\pi}{4}\)

spiral

\(r=a\theta\)

Archimedean; \(r\) grows with \(\theta\), never closes

trap

Negative \(r\) plots in the opposite direction — it's what draws inner loops and the extra rose petals.
Quadrant check when going to polar: \(\tan^{-1}(y/x)\) only gives a reference angle.

Polar Area 9.8 – 9.9

Sum of thin circular sectors. The square and the \(\tfrac12\) come from the sector.

area swept by one curve

between two curves (outer − inner)

Subtract the squares, not the radii: the integrand is \(r_{\text{out}}^2-r_{\text{in}}^2\), since \((a-b)^2\neq a^2-b^2\).

One region or two curves?

one curve\(\tfrac12\int r^2\,d\theta\) between the two angles that bound the region

inside–outside\(\tfrac12\int(r_{\text{out}}^2-r_{\text{in}}^2)\,d\theta\) over their overlap

inside bothsplit at the intersection; integrate whichever curve is closer to the pole on each piece

trap

Never blindly use \(0\) to \(2\pi\). Wrong limits double-count or sweep the wrong region.
Don't drop the \(\tfrac12\) — it halves the sector area.

Symmetry integrate less

Spot a symmetric region and you can integrate over half (or a quarter) and multiply — fewer limits to misplace.

Test the equation for…

x-axisunchanged when \(\theta\to-\theta\) — e.g. \(r=a\cos\theta\)

y-axisunchanged when \(\theta\to\pi-\theta\) — e.g. \(r=a\sin\theta\)

the poleunchanged when \(r\to-r\) (or \(\theta\to\theta+\pi\))

Payoff: a petal symmetric about its axis ⇒ integrate \(0\) to the half-angle and double. Roses, cardioids, and lemniscates almost always hand you one.

Before you start every problem

A 20-second checklist that catches most lost points.

Always

sketch first→ see the curve, the loops, and where they cross

radians→ keep the calculator in radian mode for every polar/trig step

set up exactly→ on calc-active parts, write the exact integral, then evaluate

Units tell: a speed integrated over time is a length; \(\tfrac12\int r^2\,d\theta\) is an area. If the units don't match what's asked, the setup is wrong.

Vectors are componentwise: every derivative and integral acts on \(x\) and \(y\) separately — there is no shortcut around doing both.

Finding the limits of integration the whole-unit skill

Almost every lost point in this unit is a limits problem, not an algebra problem. Match the situation to its rule.

one loop / petal

solve \(r=0\)

consecutive angles where \(r=0\) bound a single petal (e.g. \(\cos k\theta=0\)).

whole closed curve

\(0\) to \(2\pi\) — but check

a circle \(r=a\cos\theta\) closes in \(\pi\); using \(2\pi\) traces it twice.

two curves meet

set \(r_1=r_2\)

the solutions are your limits — plus check the pole separately, since curves can pass through it at different \(\theta\).

Sketch, then bound

parametric / vectorthe interval is given in \(t\); for "farthest left / turning point" find where \(\dfrac{dx}{dt}\) (or \(v_x\)) changes sign

polar singletrace the region once — the start and end angles are the limits, not the full circle

polar two-curveintersection angles split the picture; on each piece, the inner curve is whichever has smaller \(r\) there

The reliable move: sketch first, mark every intersection and pole-crossing, shade the region, and read the limits off the picture. A correct setup with a sketch earns the credit even on a calculator-active part.

Unit 9 is one idea in three costumes: describe a curve by a parameter — time \(t\) for motion, the angle \(\theta\) for polar — and every question becomes ordinary calculus on those pieces. Slope is still a ratio of rates, length is still \(\int\)(speed), area is still a sum of slivers. Pick the right parameter and the machinery you already know does the rest.

Test yourself on the exam set →

Parametric Vectors Polar Graphing Polar Area