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
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 is a VERTICAL tangent, not an error
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