Calculus of P-metric

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    The Calculus of Parametric Equations

    As we have done, we will do again. Parametric curves havetangent lines, rates of change, area under and above, local andabsolute maxima and minima, horizontal and vertical tangents,etc.

    So we will find derivatives and integrals and interpret theirmeanings.

    I.O.W. SAME CONCEPTSDIFFERENTFUNCTIONS

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    How would one go about differentiating a pair ofparametric equations?...It is not too badbut check out

    how we get to a process that is not too bad.

    Let and be differentiable parametric equations.

    If we eliminate the parameter we receive an equation of y or x in

    terms of one or the other (for convenience we'll assume y in terms of x).

    x f t y g t

    y F x

    , such that is differentiable.

    The trick is to rewrite with the parametric equations

    and then differentiate to find . Let's do it!!!

    F x

    y F x

    dydx

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    2 312 1 and 3 x t t y t t t

    a.) Graph the above on the interval by hand.

    b.) Find the equation of the line tangent to the curve att= 4.

    c.) Find each of the following and discuss their meaning

    0 4t

    2 2 2

    t t t

    dy dydxdt dt dx

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    Finding Vertical and Horizontal Tangents

    0 and 0 means...

    0 and 0 means...

    dx dydt dt

    dy dxdt dt

    Note: if 0, then the resulting indeterminate slopedy dx dydt dt dx

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    What do we need to find the intervals on which thetangent lines are decreasing? I.O.W.where thegraph of the curve is ___________________ .

    Concave down

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    To find the second derivative of a parametrized curve, we find thederivative of the first derivative:

    dydt

    dxdt

    2

    2

    d y

    dx d y

    dx

    1. Find the first derivative ( dy/dx ).

    2. Find the derivative of dy/dx with respect to t .

    3. Divide by dx/dt .

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    For the graph of the parametric curve described

    above, find the interval(s) where the curve is concavedown.

    2 312 1 and 3 x t t y t t t

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    How can we use to find the locations of any

    relative extrema on our curve?

    2

    2

    d ydx

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    Geeit sure would be nice if we could use ourGC to find some of this information we just did

    by hand!!!

    Lets use our GC to check our answers to some of theproblems we completed

    We will try both in graphing mode and from the homescreen!

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    Notice on our graph that the curve intersects itself!Dont worry, I will not ask you to find where this is (nor

    will the BC exam)however you will be asked thefollowing:

    A curve defined by x(t) = and y(t) = intersects itself at the point (5,0). Find the equationsof the two tangent lines at (5,0)

    Two in the what now?!?!? Regraphyour curve until you are convincedthere are two tangent lines.

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    To solve such a beast:

    1.) Find the values of t (yes, the word is plural, butwhy?) where the curve crosses (5,0). Use theparametric equation that is easiest to solve!!!

    2.) Evaluate dy/dx at the above t-values and roll fromhere!

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    We have covered much of the calculus interpretation ofparametric curvesnow a few more examples so you arein great shape!!!

    Given 2 sin and 2 cos and the fact that the

    graph of the curve described by these equations intersects itself

    at 0, 2 , find the following:

    x t t y t

    , , , 2.35, 2.35,1 , 3.1,6.3,1100

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    a.) the equations of the two tangent lines that occurwhere the graph intersects itself

    b.) the point(s) where the curve has a horizontaltangent on the interval

    c.) using a GC, the values of t and the point(s) wherethe graph has a vertical tangent on the same intervalas in b.

    t

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    Find the slope of the tangent line to the curvedescribed by the following equations at t = 1 .

    ln cos csct x t t t y t e t

    GC to the rescue!!!

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    Given the curve described by the equation below,find the equation of the tangent line at t=1.

    11 1 x t y

    t

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    Find all the points of horizontal and verticaltangency that occur on the graph described by theparametric equations below:

    cos 2sin 2 x y

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    The Advantage of Parametric Equations asWitnessed in Things that are Launched

    The following is a practical application which some of you will be

    familiar with. It serves as a highlight of the advantage of

    using p-metric eqns. to describe physical events that involve

    two independent components which change w/ respect to

    time.you will see much more of this thought process when we

    delve into vectors YOU WILL ALSO SEE THIS ON THE

    NEXT TEST!

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    A ball is thrown with an initial velocity of 88 ft/sec. atan angle of 40 degrees to the ground. The ball isreleased at an initial height of 6 ft above the ground.Find the following (assume the acceleration due togravity is -32 ft/sec.):

    a.) A pair of parametric equations that describes the position of

    the ball at any time t (horizontal and vertical components).

    b.) The maximum height of the ball.

    c.) The range of the ball (maximum horizontal distance)

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    Imagine a roller coaster that travels one loop on itstracks modeled after a parametric curve. The arc

    length of the p-metric curve is the distance theroller coaster travels once around the tracks.

    As you calculate arc length MAKE SURE YOU ARECALCULATING ONE LOOP!!!

    For example If you calculate the arc length of acircle by setting your limits so that you are travelling

    the circle a second time, then you will be doublecounting portions of the curve!!!!...which wouldpotentially bring disaster upon the rest of theworldexcept on youbut the rest of the world is outof luck.

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    Lets Generate the Formula!

    1.) With our original arc length formula.

    2.) Conceptually

    2 2

    ' 'b

    a

    Arc Length S x t y t dt

    2 2

    dx dy

    OR dt dt dt

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    Find the length of the curve defined by theparametric equations below on the given interval.

    2 312 1 0 4.73

    x t t y t t t t

    Does the curve repeat itself on the above interval? Check with GC or graph by hand.

    * Note that dx/dt and dy/dt do not equal zero at thesame time on the above interval

    Therefore our curve is considered smooth.

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    Find the length of the curve described by thefollowing parametric equations on the given interval.

    2 31 4 1 0 x t y t t

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    Find the length of the curve described by thefollowing parametric equations on the given interval.

    5

    3

    1 1 2

    10 6t

    x t y t t

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    Find the area of the surface generated by revolvingthe region bounded by the x-axis and the curve and

    on the interval described below about the x axis.

    2 312 3 03

    x t y t t t

    Ummmmyeahdefinitely use your GC!!!

    ABOUT THE Y-AXIS?

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    Before we say goodbye to parametric curves, lets findthe area under a parametric curve!

    WE KNOW: b

    a

    A ydx

    ' 'dx

    x x t x t dx x t dt dt

    y y t

    ' and curve is travelled once!b

    a

    A y t x t dt a t b

    Makes sense with units!

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    a.) Graph the region described by the following in GC

    3cos sin 0

    0, ,.05 , 3,3,1 , 1,1,1

    x t t y t t t

    b.) Find the area of the region.

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    Lets round the p -metric madness out with some FR!