The Calculus of Parametric Equations

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

So we will find derivatives and integrals and interpret their meanings.

I.O.W. – SAME CONCEPTS…DIFFERENT FUNCTIONS

How would one go about differentiating a pair of parametric equations?...It is not too bad…but 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

dy

dx

We get the following result…

'

dydy dtF x

dxdxdt

Which makes sense algebraically!!!

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 at t= 4.

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

0 4t

2 2 2

t t t

dy dydxdt dt dx

Finding Vertical and Horizontal Tangents

0 and 0 means...

0 and 0 means...

dx dy

dt dt

dy dx

dt dt

Note: if 0, then the resulting indeterminate slopedy dx dy

dt dt dx

2 312 1 and 3

x t t y t t t

a.) Find the value(s) of t at which the curve described above has a horizontal tangent line. Equation of tangent line? Does this match with our graph?

GRAPH IN YOUR GC TO SEE!!! [-2,5,.1],[-2,30,1],[-2,10,1]

b.) Find the value(s) of t at which the curve described above has a vertical tangent line. Equation of tangent line? Does this match with our graph?

c.) Find the equation of the tangent line at t=3

d.) Find the intervals of t during which the tangent lines have positive slope. Does this match with our graph?

What do we need to find the intervals on which the tangent lines are decreasing? I.O.W….where the graph of the curve is ___________________. Concave down

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

dydtdxdt

2

2

d y

dx dy

dx

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

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

3. Divide by dx/dt.

For the graph of the parametric curve described above, find the interval(s) where the curve is concave down.

2 312 1 and 3

x t t y t t t

How can we use to find the locations of any

relative extrema on our curve?

2

2

d y

dx

Gee…it sure would be nice if we could use our GC to find some of this information we just did by hand!!!

Let’s use our GC to check our answers to some of the problems we completed…

We will try both in graphing mode and from the home screen!

Notice on our graph that the curve intersects itself! Don’t worry, I will not ask you to find where this is (nor will the BC exam)…however you will be asked the following:

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

Two in the what now?!?!? – Regraph your curve until you are convinced there are two tangent lines.

To solve such a beast:

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

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

We have covered much of the calculus interpretation of parametric curves…now a few more examples so you are in 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

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

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

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

t

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

ln cos csctx t t t y t e t

GC to the rescue!!!

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

11 1x t y

t

a.) Find the equations of the tangent lines at the point where the curve described by the equations below crosses itself (you may have to find by graphing the curve on your calculator for the homework!!!)

3 2 6

5,5,.1 , 10,10,1 , 10,10,1

x t t t y t t

b.) Where is the graph concave up?

Find all the points of horizontal and vertical tangency that occur on the graph described by the parametric equations below:

cos 2sin 2x y

The Advantage of Parametric Equations as Witnessed 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!

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

Imagine a roller coaster that travels one “loop” on its tracks modeled after a parametric curve. The arc length of the p-metric curve is the distance the roller coaster travels once around the tracks.

As you calculate arc length MAKE SURE YOU ARE CALCULATING ONE LOOP!!!

For example – If you calculate the arc length of a circle by setting your limits so that you are travelling the circle a second time, then you will be double counting portions of the curve!!!!...which would potentially bring disaster upon the rest of the world…except on you…but the rest of the world is out of luck.

Let’s 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 dtdt dt

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

2 312 1 0 4.7

3x 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 the same time on the above interval

Therefore our curve is considered smooth.

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

2 31 4 1 0x t y t t

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

5

3

1 1 2

10 6

tx t y t

t

SURFACE AREA!!!(Curves must

Same idea as with rectangular form!!!

Each cross section is a disk of which we want the SA

2SA rh

Since we are adding up a bunch of areas of these super tiny disks, we will integrate…but what does the finished integrand look like for figures rotated about the x-axis? y-axis?

Find the area of the surface generated by revolving the region bounded by the x-axis and the curve and on the interval described below about the x – axis.

2 312 3 0

3x t y t t t

Ummmm…yeah…definitely use your GC!!!

ABOUT THE Y-AXIS?

Find the surface area of the curve generated by revolving the curve about the x-axis on the given interval.

sin cos 0 2x y

GC fa sho.

Before we say goodbye to parametric curves, let’s find the area under a parametric curve!

WE KNOW: b

a

A ydx

' 'dx

x x t x t dx x t dtdt

y y t

' and curve is travelled once!b

a

A y t x t dt a t b

Makes sense with units!

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.

Let’s round the p-metric madness out with some FR!