10.1 Parametric Functions. In chapter 1, we talked about parametric equations. Parametric equations...

10.1 Parametric Functions

In chapter 1, we talked about parametric equations.Parametric equations can be used to describe motion that is not a function.

x f t y g t

If f and g have derivatives at t, then the parametrized curve also has a derivative at t.

dx dy

f t g tdt dt

The formula for finding the slope of a parametrized curve is:

dy

dy dtdxdxdt

This makes sense if we think about canceling dt.

Since the derivatives of each parametric equation are:

dx dy

f t g tdt dt

Example 2 (page 514): The parametric curve below is given by the equations 2 3 and 1.5 1x t t y t t t Find the values of t for which the line tangent to this curve is

a) vertical

b) horizontal

21 3

1 2

dy t

dx t

When is this equal to 0?

1

2t

1

4x

3

8y

Example 2 (page 514): The parametric curve below is given by the equations

a) vertical

b) horizontal

Find the values of t for which the line tangent to this curve is

21 3

1 2

dy t

dx t

When is this equal to 0?

1

3t 3 1

3x

2

3 3y

3 1

3x

2

3 3y

2 3 and 1.5 1x t t y t t t

The formula for finding the slope of a parametrized curve is:

dy

dy dtdxdxdt

Remember since this is still a parametric function

that will be given in terms of tdx

dy

In the next slides, we will be using y in place of

for simplicity.dx

dy

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

dydtdxdt

2

2

d y

dx d

ydx

1. Find the first derivative (dy/dx).2. Find the derivative of dy/dx with respect to t.

3. Divide by dx/dt.

Example 2 (page 514):2

2 32

Find as a function of if and .d y

t x t t y t tdx

Example 2 (page 514):2

2 32

Find as a function of if and .d y

t x t t y t tdx

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

dy

dy dtydxdxdt

21 3

1 2

t

t

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

21 3

1 2

dy d t

dt dt t

2

2

2 6 6

1 2

t t

t

Quotient Rule

3. Divide by dx/dt.

2

2

d y

dx

dydtdxdt

2

2

2 6 6

1 2

1 2

t t

t

t

2

3

2 6 6

1 2

t t

t

2 2( ) ( )L dx dy

The length of a segment of a parametric curve can be approximated using the Pythagorean theorem:

L

dx

dy

2 2( ) ( )L dx dy

dt

dt

2 2dx dy

dt dt

2 22

1( ) ( )

( )dx dy

dt dt

dt

Multiply by dt/dt

After some algebra

As dt gets smaller, the approximation gets better. As dt goes to zero, we can determine the exact length of the curve.

This equation for the length of a parametrized curve is similar to our previous “length of curve” equation.

(Notice the use of the Pythagorean Theorem.)

2 2b

a

dy dxL dt

dt dt

If we wanted to add up many segments over an interval of t, we can add an infinite amount of infinitely small segments to get…An Integral:

Example 2 (page 514): The parametric curve below is given by the equations 2 3 and 1.5 1x t t y t t t Find the length of this curve over the given interval

2 21

1.5

dx dyL dt

dt dt

21 2 2

1.51 2 1 3L t t dt

5.69

Likewise, the equations for the surface area of a parametrized curve are similar to our previous “surface area” equations:

Revolution about the -axis 0x y 2 2

2b

a

dx dyS y dt

dt dt

Revolution about the -axis 0y x

2 2

2b

a

dx dyS x dt

dt dt

Revolution about the -axis 0x y 2 2

2b

a

dx dyS y dt

dt dt

By the way, this is an actual curve whose equations are:

sin 2

2cos 5

x t t t

y t t t