Courtesy

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• Courtesy
• OF Perry
• Hubert

PARAMETRIC AND POLAR CURVES (and vectors)
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PARAMETRIC CURVES

• This is the graph of a parametric equation

(Its crazy)
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Relationships Between Parametric and Cartesian
Coordinates

• x x(t) , y y(t)
• To convert from parametric to Cartesian you gotta
  solve for t in one of the given equations and
  plug your new t into the remaining equation.
• Example c(t) (2t 4, 3 t2)
• so. x 2t 4 y 3 t2
• In this example it is more convenient to solve
  for t using the x equation. See
  moreooOoOooOoOoOoooOooOoO?

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Example continued. ?

• x 2t 4
• t (x 4)/2
• y 3 t2
• y 3 (x 4)/22
• 7 2x ¼x2
• So.
• y ¼x2 2x 7

PEAR-ametric
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Slopes of Tangent Lines of Parametric Equations

• The slope of the tan line of a parametric
  equation is still found using dy/dx, BUT since
  you are not starting with a function y in terms
  of x, you need to make one using the previous
  method!
• Its just normal from therelook?

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Another Example

• The graph x(t) sin t , y(t) sin 2t
• (0 t 2p) crosses itself at the origin. Find
  the tangent lines at this point

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Solution

• y 2x
• y -2x
• Remember to take the derivative of the y equation
  and the x equation to find dy/dt and dx/dt. This
  allows you to solve for dy/dx using algebra
  skills.
• dy/dx (dy/dt) / (dx/dt)

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Vectors velocity, acceleration and other fun
stuff

• Velocity and speed can be found using
  derivatives, in a way similar to finding the
  tangent line because velocity is the derivative
  of the position function with respect to time.
• ds/dt v(dx/dt2 dy/dt2)
• Prepare for an example problem D

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Because it is here.

• Find the velocity vector and speed for the
  parametric curve x(t) et, y(t) arctan t
• at t 1

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Solution!!

• v(t) lt e, ½ gt
• Speed ds/dt v e2 ¼

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POLAR CURVES
Coordinates (r, ?)
This is the polar equation r 4cos2?
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Relationships BetweenPolar and Cartesian
Coordinates

• x r cos ?
• y r sin ?
• r2 x2 y2
• tan ? y/x

Polar bear curve
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Slope of Tangent Lines of Polar Equations

• Using those relationships you can find the slopes
  of tangent lines again ?
• Simply convert from polar coordinates to
  Cartesian coordinates and take the derivative!
• Example Finding the derivative of
• r sec2?. We know that x r cos ?
• and y r sin ?
• So substituting we have x sec2? cos ?
• and y sec2? sin ?
• dy/dx becomes
• sec ? ( sec2 ? tan2 ?) / sec ? tan ?
• And simplifies to 1sin2 ? / cos ? sin ?

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Area Bounded by a Polar Curve

• You can find the area between a polar curve and
  the origin OR between to polar curves themselves
  using this magical equation
• Area ½ ? r22 r12 d?
• Find the intersection by setting the two
  equations r1 and r2 equal to each other.
• Then use the points of intersection as your
  limits of integration!

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Other useful stuff to know

• You can also find the area between parametric
  curves by using this handy formula
• Area ? y(t) x(t) dt
• The limits of integration would be from t0 to t1.

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Other useful stuff to know Part II

• Arc Length
• Rectangular
• s ?v(1 dy/dx2) dx
• Parametric
• s ?v(dx/dt2 dy/dt2) dt
• Remember to use x and t for your limits of
  integration for rectangular and parametric
  equations respectively!!
• THE END!