Math 114 – Rimmer

13.2 Integrals of Vector Functions (Space Curves) and

Projectile Motion

Definite integral of a vector function

( ) ( ) ( ) ( ), ,t f t g t h t=r

( ) ( ) ( ) ( ) , ,

b b b b

a a a a

t dt f t dt g t dt h t dt=∫ ∫ ∫ ∫r

- taken component-wise

provided the vector function

is continuous

( )( )2

2

1

1 sin t t t t t dtπ+ − +∫ i j k ( )2 2 2

2

1 1 1

, 1 , sint dt t t dt t t dtπ= −∫ ∫ ∫

22 32

1 13

tt dt

=

∫

8 1

3 3= −

7

3=

2

1

1t t dt−∫1u t= −

du dt=1t u⇒ = +

( )1

0

1u udu= +∫

1 0t u= ⇒ =

2 1t u= ⇒ =

( )1

3/2 1/2

0

u u dt= +∫1

5/2 3/2

0

2 2

5 3u u

= +

2 2

5 3= +

6 10

15 15= +

16

15=

Math 114 – Rimmer

13.2 Integrals of Vector Functions (Space Curves) and

Projectile Motion

( )2

1

sint t dtπ∫ u t= ( )sindv tπ=

du dt= ( )1

cosv tππ

−=

( ) ( )2 2

1 1

1cos cos

tt t dtπ π

π π

− − = −

∫

udv uv vdu= −∫ ∫

( )2

1

1cos t dtπ

π+ ∫

( ) ( )2

2

1

1cos sin

tt tπ π

π π

− = +

( ) ( ) ( ) ( )2 2

2 1 1 1cos 2 sin 2 cos sinπ π π π

π π π π

− − = + − +

( ) ( )2 1

1 0 1 0π π

− − = + − − +

3

π

−=

( )( )2

2

1

7 16 31 sin , ,

3 15t t t t t dtπ

π

−+ − + =∫ i j k

Math 114 – Rimmer

13.2 Integrals of Vector Functions (Space Curves) and

Projectile Motion

A projectile is launched, we want to find the parametric equations for the path.

the projectile has mass m

( )

assume that gravity is the only force acting on the

projectile after it is launched neglect air resistance

( ) mg m⇒ − =j a

( )g= −a j

( )Force due to gravity : mg= −F j

Newton's Second Law : m=F a

( )t dt= ∫v a g dt= −∫ j1

gt= − +j C

( )t dt= ∫r v ( )1gt dt= − +∫ j C2

1 2

1

2gt t= − + +j C C

0r

0v

( ) 00 =v v

( ) 1t gt= − +v j C

1 0⇒ =C v

2

0 2

1

2gt t= − + +j v C

( ) 0t gt= − +v j v( ) 00 =r r

2 0⇒ =C r ( ) 2

0 0

1

2t gt t= − + +r j v r

Math 114 – Rimmer

13.2 Integrals of Vector Functions (Space Curves) and

Projectile Motion( )g= −a j

( ) 0t gt= − +v j v

( ) 2

0 0

1

2t gt t= − + +r j v r

Often you are given an initial height, an initial speed,

and the angle at which the projectile is launched.θ

0r

0v

h =

θ

0 0 0cos , sinθ θ=v v v

0 0, h=r

initial speed

( ) 0 00, cos , sint gt θ θ= − +v v v

( )g= −a j ( ) 0,t g= −a

( ) 0 0cos , sint gtθ θ= −v v v

( ) 2

0 0

10, cos , sin 0,

2t gt t hθ θ= − + +r v v

( ) ( ) ( ) 2

0 0

1cos , sin

2t t h t gtθ θ= + −r v v

Math 114 – Rimmer

13.2 Integrals of Vector Functions (Space Curves) and

Projectile Motion

A baseball is hit 3 feet above the ground level at 100 feet per second

and at an angle of 45 with respect to the ground. Find the maximum

height reached by the baseball. Will it clear a 10 foot fence

�

located

300 feet from home plate?

3h =

0 100=v

45θ = °

( ) ( ) ( ) 21100cos 45 ,3 100sin 45 32

2t t t= ° + ° −

( ) 250 2 , 3 50 2 16t t t t= + −r

32g =

( ) 50 2,50 2 32t t= −v

The maximum height occurs when the component of the velocity vector is zero.j

50 2 32 0t⇒ − =25 2

2.21 sec.16

t⇒ = ≈

2

25 2 25 2The maximum height 3 50 2 16

16 16

= + −

( )625 225003

16 16

= + −

1298 81.125 ft.

16= =

( ) ( ) ( ) 2

0 0

1cos , sin

2t t h t gtθ θ= + −r v v

The ball is 300 ft. from home plate when the component of the position vector is 300.i

50 2 300t⇒ =

63 2 4.24 sec.

2t = = ≈

Plug this time into the component of the

position vector to find the height of the ball.

j

( ) ( )2

The height of the ball at the fence 3 50 2 3 2 16 3 2= + − 3 300 288= + − 15 ft.=

Yes the ball clears the 10 ft. fence for a home run!

Math 114 – Rimmer

13.2 Integrals of Vector Functions (Space Curves) and

Projectile Motion

Math 114 – Rimmer

13.2 Integrals of Vector Functions (Space Curves) and

Projectile Motion

Math 114 – Rimmer

13.2 Integrals of Vector Functions (Space Curves) and

Projectile Motion

( )g= −a j

( ) 0t gt= − +v j v

( ) 2

0 0

1

2t gt t= − + +r j v r

( ) 0,t g= −a

( ) 0 0cos , sint gtθ θ= −v v v

( ) ( ) ( ) 2

0 0

1cos , sin

2t t h t gtθ θ= + −r v v

0If launch angle , initial speed , and initial height is known:hθ v

General formulas that are always true:

0 0If 0, we have the following "shortcut" formulas : here h v= = v

( )02 sin 2Flight time

v

g

θ=

( )2

0sin 2

Max. ht.2

v

g

θ =

( ) ( )2

0sin 2

Rangev

g

θ=

Math 114 – Rimmer

13.2 Integrals of Vector Functions (Space Curves) and

Projectile Motion

( )g= −a j ( ) 0t gt= − +v j v ( ) 2

0 0

1

2t gt t= − + +r j v r