Parametric Equations, Polar Coordinates, and Vector-Valued Functions BC only

Nine topics on curves that are not graphs of functions. A parameter supplies both coordinates, so the slope is a quotient of derivatives and the second derivative is divided by dx/dt a second time. A vector packages the same motion, so its constant of integration is a vector too, and speed is the magnitude of velocity rather than velocity itself. In polar form dr/dθ is not a slope, area carries a one-half and a square, and the bounds are wherever the curve finishes tracing the region once.

AB exam n/aBC exam 10-15%9 topics

Topics

9.1

Defining and Differentiating Parametric EquationsBC only

Second Derivatives of Parametric EquationsBC only

Finding Arc Lengths of Curves Given by Parametric EquationsBC only

Defining and Differentiating Vector-Valued FunctionsBC only

Integrating Vector-Valued FunctionsBC only

Solving Motion Problems Using Parametric and Vector-Valued FunctionsBC only

Defining Polar Coordinates and Differentiating in Polar FormBC only

Find the Area of a Polar Region or the Area Bounded by a Single Polar CurveBC only

Finding the Area of the Region Bounded by Two Polar CurvesBC only

Lesson

Diagnostic

after diagnostic

Key forms For every problem in this unit

Slope

(dy/dt) OVER (dx/dt). Never the other way up

Where it fails

dx/dt = 0 while dy/dt is NOT is a VERTICAL tangent, not an error. BOTH zero is indeterminate: cusp, corner or either tangent, so look further

Horizontal tangent

dy/dt = 0 while dx/dt is NOT

Second derivative

d/dt of (dy/dx), THEN divide by dx/dt again

The trap

it is NOT (d²y/dt²) over (d²x/dt²)

Arc length

√((dx/dt)² + (dy/dt)²) dt, over t-bounds

No 1 here

the 1 in Unit 8 WAS (dx/dx)²

Orientation

the parameter picks a DIRECTION and a start

Retracing

check whether the interval covers the curve ONCE

Derivative

differentiate EACH component, separately

Velocity

a VECTOR: the derivative of position

Speed

a SCALAR: the magnitude of velocity

Magnitude

√(sum of squares), never the sum

Acceleration

the derivative of VELOCITY, component by component

Antiderivative

one constant PER COMPONENT, so C is a vector

Position

initial position PLUS the integral of velocity

Displacement

a VECTOR: the integral of velocity

Distance travelled

a SCALAR: the integral of SPEED

Three answers

position, displacement, distance are all different

Converting

x = r cosθ, y = r sinθ, with r a function of θ

Slope

(dy/dθ) OVER (dx/dθ). NOT dr/dθ

Both need it

x and y each take the PRODUCT RULE

Area, one curve

½ ∫ r² dθ. The half AND the square

Square first

square r INSIDE the integral, never after

Bounds

where the region is traced ONCE. Solve r = 0

One petal

is NOT 0 to 2π. Find its own two angles

Area, two curves

½ ∫ (R² − r²) dθ

Not this

NEVER ½ ∫ (R − r)² dθ

First step

set the two r's equal to find the crossing angles

Unit 9 tools

Challenge bank

1 / 60

60 open-ended problems.

Read the question, work it out, then flip the card to compare your reasoning to the worked solution. Mark each card so you can return to the ones that still bite.

Question

Tap card to reveal explanation

Worked solution

Tap card to return to question

Cumulative assessment

Test the unit.

Twenty mixed items drawn from across all 9 topics, with guaranteed misconception-code coverage. Identifies which misconceptions still bite when you cannot see which topic the question came from. AB students are served only the AB items; BC students get the whole unit.

Begin assessment →

Check what stuck.

Units 1 through 9, drawn evenly so earlier units get the same share as this one. Twenty questions or a full 45-question section, your choice. Even coverage means this is a retention check rather than a score estimate.

20 or 45questions

96topics

128codes covered

Begin assessment →