Parametric and Polar Curves and
Vectors
Gerald Cheng
Kevin Chiou
Jay Dasigi
Parametric vs. Cartesian




Parametric equations show the motion and curves
represented by graphs that usually cannot be a function in
Cartesian (or rectangular) graphs.
Instead of relating the function to x in Cartesian equations,
parametrics relate the x and y axis to “t,” usually
representing time.
Parametric equations can be converted to Cartesian by
isolating the “t” variable, then substituting into the other
equation.
Similarly, Cartesian equations can be converted to
parametrics by setting x or y equal to t.
Examples


#1 Express the parametric as a Cartesian
equation of (ln(t) , 2-t).
#2 Express the Cartesian equation as a
parametric: 4x-y^2=5.
Graphs of Parametric Curves


Parametric graphs can depict the velocity and
acceleration of particular graphs.
They can also represent curves that usually
cannot be made into a function by regular
Cartesian equations.
Graphs of Parametric Curves
cont’d
Graphs of Polar Curves




Unlike parametric and Cartesian equations,
polar curves cannot be graphed on rectangular
graph plots, but instead need a polar plot.
Different from the rectangular plots, polar plots
are plotted along a circular path from the origin.
Instead of the x and y axis, this has a radius
and degrees “axis.”
Points are expressed in the
form (r, θ)
Graphs of Polar Curves cont’d
Graphs of Polar Curves cont’d

To find the coordinates of a polar curve from a
Cartesian point, the following equations are
needed.
–
–
r^2=x^2 + y^2
tanθ= (y/x)
Graphs of Polar Curves cont’d

To find the coordinates of a Cartesian point from a
polar point, the following equations are needed.
- x= rcosθ
- y= rsinθ
Examples


#3 Convert (-1, 3^(1/2)) to polar coordinates.
#4 Convert (4, 3π/4) to Cartesian coordinates.
Slope of a Tangent Line-Parametric


To find the slope of a line tangent to a
parametric curve (dy/dx) you must calculate
(dy/dt)/(dx/dt)
To find the second derivative of a parametric
curve(d^2y/dx^2) you must calculate
(d(dy/dx)/dt)/(dx/dt)
Examples


#5 Find the first and second derivatives and
the tangent line of x=t and y=t^(1/2) at t=1/4.
#6 For 0<t<13, an object travels along an
elliptical path given by the parametric
equations x=3cost and y=4sint. At the point
where t=13, the object leaves the path and
travels along the line tangent to the path at that
point. What is the slope of the line on which the
object travels?
Vectors





Vector notation: (x(t),y(t)), <x(t),y(y)>, xi+yi
Take the derivative of both functions and leave in same
notation
Vectors are also used in finding velocity and
acceleration when given position functions
When given a position function of a particle, such as
s=(x(t),y(t)), the velocity vector is (x’(t),y’(t)) and the
acceleration vector is (x’’(t),y’’(t))
The particles speed=((x’(t))^2+(y’(t))^2)^(1/2)
Examples

#7 A particle’s position varies with the function
s(t)=(t^2 –4, t/2). Find a) the particle’s general
velocity vector, b) the particle’s velocity vector
at t=1, c) the particle’s speed at t=1, and d) the
particle’s acceleration vector.
Slope of a Tangent Line-Polar


To find the equation of the tangent line of a
polar curve (dy/dx) you must calculate
(dy/dθ)/(dx/dθ)
To do this it is necessary to express the polar
function as a function of both x and y. For
example, the function r=sin2θ must be
expressed as x=rsin2θcosθ and y=rsin2θsinθ.
Examples

#8 Find the equation of the line tangent to
r=sin2θ at θ=π/4.
Area Bounded by a Polar Curve


When finding the area bounded by a polar
curve, you are really finding the area of a
bounded sector of a curve.
The area of the sector is:
Examples


#9 Find the area bounded by the curve
r=2(1+cosθ) from 0 to 2π.
#10 Find the area of the inside loop of the
function r=2cosθ+1.