section 12.3 – velocity and acceleration

Section 12.3 – Velocity and Acceleration

Vector Function

A vector function is a function that takes one or more variables and returns a vector:

Where and are called the component functions.

A vector function is essentially a different notation for a parametric function.

Particle MotionIn AP Calculus AB, particle motion was defined in functions of time versus motion on a horizontal or vertical line.

In AP Calculus BC, particle motion will ALSO be defined in functions of position versus position (along a curve).

How successful you are with Particle Motion is a good predictor of how successful you will be on the AP Test.

Position Vector FunctionWhen a particle moves on the xy-plane, the coordinates of its position can be given as parametric functions:

for

The particle’s position can also be expressed as a position vector function:

The coordinates of the parametric function at time

t…

A vector function is essentially a different notation for a parametric function.

… are equal to the components of the vector

function at time t.

Example 1Let the position vector of a particle moving along a curve is defined by (a) Find and graph the position vector of the particle at

.

-4 -3 -2 -1 1 2 3 4

-3

-2

-1

1

2

3 y

x

4 4 43cos ,2sins

2.121,1.414𝑠( 𝜋4 )

Velocity Vector FunctionThe vector function for position is differentiable at if and have derivatives at .

The derivative of , , is defined as the velocity vector:

A vector function is essentially a different notation for a parametric function.

Example 1 ContinuedLet the position vector of a particle moving along a curve is defined by (b) Find the velocity vector.

-4 -3 -2 -1 1 2 3 4

-3

-2

-1

1

2

3 y

x

ddtv t s t

3sin ,2cost t

𝑠( 𝜋4 ) 3cos , 2sind ddt dtt t

Example 1 ContinuedLet the position vector of a particle moving along a curve is defined by (c) Find and graph the velocity vector of the particle at

.

-4 -3 -2 -1 1 2 3 4

-3

-2

-1

1

2

3 y

x

4 4 43sin ,2cosv

2.121,1,414 𝑠( 𝜋4 )

𝑣 ( 𝜋4 )

If the initial point of the velocity vector is also the

terminal point of the position vector, the velocity vector is

tangent to the curve.

Acceleration Vector Function

The second derivative of , , is defined as the acceleration vector:

A vector function is essentially a different notation for a parametric function.

Example 1 ContinuedLet the position vector of a particle moving along a curve is defined by (b) Find the acceleration vector.

-4 -3 -2 -1 1 2 3 4

-3

-2

-1

1

2

3 y

x

ddta t v t

3cos , 2sint t

𝑠( 𝜋4 ) 3sin , 2cosd ddt dtt t

Arc Length and SpeedConsider a particle moving along a parametric curve. The distance traveled by the particle over the time interval is given by the arc length integral:

On the other hand, speed is defined as the rate of change of distance traveled with respect to time, so by the Second Fundamental Theorem of Calculus:

dsdtSpeed

0

2 2'( ) '( )t

ddt t

x u y u du 2 2'( ) '( )x t y t

This is the magnitude of the velocity vector!

Speed with a Vector Function

The particle’s speed is the magnitude of , denoted :

Speed is a scalar, not a vector.

ReminderFrom AP Calculus AB:

Speed is the absolute value of velocity:

Integrating speed gives total distance traveled:

Example: If , find the speed at and the total distance traveled during .

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CTO

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White Board ChallengeA particle moves along a curve so that and . What is the speed of the particle when .

222( ) 6 ln dyddt dtv t t t

2 2112 sin 2tt t

2 212(2) 12 2 sin 2 2v

24.672

Example 2A particle moves along the graph of , with its x-coordinate changing at the rate of for . If , find

(a) The particle’s position at .

You can use the FTOC on components:

2

12 1 dx

dtx x dt 3

22

11

tdt 2 2

1

11

t

2 21 12 1

1 1.75

Example 2 ContinuedA particle moves along the graph of , with its x-coordinate changing at the rate of for . If , find

(b) The speed of the particle at .

22 dydxdt dtv t

Find the speed equation: Since y is a function of x, we need to use the Chain Rule:

2dy dx dxdt dt dtx

3 32 22t t

x 3 3 3

2 22 2 22t t t

x Substitute this and dx/dt into the speed equation.

3 3 3

2 22 2 22 2 2

2 ___ Substitute t=2

2v 1.75

From (a), we know x=1.75 when t=2

1.152

We know dx/dt.

Example 3A particle moves a long a curve with its position vector given by for . Find the time when the particle is at rest.

0 0,0v t

The particle is at rest when the velocity vector is:

Find the velocity vector: 2 43cos , 5sind t d tdt dtv t

3 52 2 4 4sin , cost t

Solve:3 52 2 4 4sin 0 cos 0t t

0,2 ,4 2t t is when the particle is at rest because it is the only time on the interval when BOTH components are 0.

SummaryIf is the position vector of a particle moving along a smooth curve in the xy-plane, then, at any time t,

1. The particle’s velocity vector is ; if drawn from the position point, it is tangent to the curve.

2. The particle’s speed along the curve is the length of the velocity vector, .

3. The particle’s acceleration vector is , is the derivative of the velocity vector and the second derivative of the position vector.