The Rosetta Mission

The Rosetta Mission

Rosetta is a robotic space probe built and launched by the EuropeanSpace Agency to perform a detailed study of comet 67P/Churyumov -Gerasimenko. On August 6th, 2014, it approached the comet to adistance of about 100 km (62 mi) and reduced its relative velocity to 1m/s (3.3 ft/s), thus becoming the first spacecraft to rendezvous with acomet with the intent to orbit. Following further manoeuvres, it willenter orbit after approaching to 30 km (19 mi) five weeks later.

The Rosetta Mission

More info:Mission website: http://rosetta.jpl.nasa.gov/Wikipedia site.Flight path GIF.Interactive flight path.

http://rosetta.jpl.nasa.gov/

http://en.wikipedia.org/wiki/Rosetta_(spacecraft)

http://i.imgur.com/TUkKuhf.gif

http://sci.esa.int/where_is_rosetta/

The Rosetta Mission

QuestionHow do you calculate the flight path of the spacecraft?

Two fundamental formulas:1 F = m · a,

2 F =G ·m ·M

r2.

If X is a Solar system object, let PX (t) be its position (in space) withrespect to time, and let MX be its mass. Let R be the Rosettaspacecraft, E = Earth, Mo = Moon, M = Mars, S = Sun, etc. Then:

MR ·d2PR(t)

dt2= (PE (t)− PR(t)) ·

G ·MR ·ME

|PE (t)− PR(t)|2+ (PMo(t)− PR(t)) ·

G ·MR ·MMo

|PMo(t)− PR(t)|2

+ (PS(t)− PR(t)) ·G ·MR ·MS

|PS(t)− PR(t)|2+ · · ·

The Rosetta Mission



2 F =G ·m ·M

r2.


MR ·d2PR(t)

dt2= (PE (t)− PR(t)) ·

G ·MR ·ME

|PE (t)− PR(t)|2+ (PMo(t)− PR(t)) ·

G ·MR ·MMo

|PMo(t)− PR(t)|2


|PS(t)− PR(t)|2+ · · ·

The Rosetta Mission



2 F =G ·m ·M

r2.


MR ·d2PR(t)

dt2= (PE (t)− PR(t)) ·

G ·MR ·ME

|PE (t)− PR(t)|2+ (PMo(t)− PR(t)) ·

G ·MR ·MMo

|PMo(t)− PR(t)|2


|PS(t)− PR(t)|2+ · · ·

The Rosetta Mission



2 F =G ·m ·M

r2.


MR ·d2PR(t)

dt2= (PE (t)− PR(t)) ·

G ·MR ·ME

|PE (t)− PR(t)|2+ (PMo(t)− PR(t)) ·

G ·MR ·MMo

|PMo(t)− PR(t)|2


|PS(t)− PR(t)|2+ · · ·

MATH 1131Q - Calculus 1.

Álvaro Lozano-Robledo

Department of MathematicsUniversity of Connecticut

Day 7

Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 6 / 29

Derivatives(Rates of Change)

How do we calculate the equation of a tangent line?

First, recall the equation of a line:

0

y = mx + k

1 Line with slope m through (a,b):

y − b = m · (x − a).

2 Line through (a,b) and (c,d):

y − b =

�d − b

c − a

�

· (x − a).



First, recall the equation of a line:

0

y = mx + k

1 Line with slope m through (a,b):

y − b = m · (x − a).

2 Line through (a,b) and (c,d):

y − b =

�d − b

c − a

�

· (x − a).



Let us calculate the equation of the tangent line to a graph y = f (x) ata point P = (a, f (a)) on the graph...

We just need the slope of thetangent line at the point P.

0



Let us calculate the equation of the tangent line to a graph y = f (x) ata point P = (a, f (a)) on the graph... We just need the slope of thetangent line at the point P.

0



Let us calculate the equation of the tangent line to a graph y = f (x) ata point P = (a, f (a)) on the graph... We just need the slope of thetangent line at the point P.

0



Let us calculate the equation of the tangent line to a graph y = f (x) ata point P = (a, f (a)) on the graph... We just need the slope of thetangent line at the point P. The slope is given by

m = limx→a

f (x)− f (a)

x − a= lim

h→0

f (a + h)− f (a)

h.

Example

Calculate the equation of the tangent line to y = x2 at (2,4).

.



Let us calculate the equation of the tangent line to a graph y = f (x) ata point P = (a, f (a)) on the graph... We just need the slope of thetangent line at the point P. The slope is given by

m = limx→a

f (x)− f (a)

x − a= lim

h→0

f (a + h)− f (a)

h.

Example

Calculate the equation of the tangent line to y = x2 at (2,4).

.


This slide left intentionally blank


Example

Calculate the equation of the tangent line to y =x

x + 1at (1,1/2).

m = limx→a

f (x)− f (a)

x − a= lim

h→0

f (a + h)− f (a)

h.

−3 −2 −1 1 2 3 4 5 6

−1

1

2

3

0

f

.

Example


x + 1at (1,1/2).

m = limx→a

f (x)− f (a)

x − a= lim

h→0

f (a + h)− f (a)

h.

−3 −2 −1 1 2 3 4 5 6

−1

1

2

3

0

f

.

Example


x + 1at (1,1/2).

m = limx→a

f (x)− f (a)

x − a= lim

h→0

f (a + h)− f (a)

h.

−3 −2 −1 1 2 3 4 5 6

−1

1

2

3

0

f

.

How do we calculate the velocity of an object?

average velocity =displacement

time

If P(t) indicates the position of the object at time t (in the real line), and we want tocalculate the average velocity between time t = t0 and time t = t1, then


time=

P(t1)− P(t0)

t1 − t0

If we want to calculate the average velocity between time t = a and t = a + h (forsome small h > 0), then


time=

P(a + h)− P(a)

(a + h)− a=

P(a + h)− P(a)

h

DefinitionIf the position of an object X is given by P(t), then we define the instantaneousvelocity of X at time t = a as

V (a) = limh→0

P(a + h)− P(a)

h.



time



time=

P(t1)− P(t0)

t1 − t0



time=

P(a + h)− P(a)

(a + h)− a=

P(a + h)− P(a)

h


V (a) = limh→0

P(a + h)− P(a)

h.



time



time=

P(t1)− P(t0)

t1 − t0



time=

P(a + h)− P(a)

(a + h)− a=

P(a + h)− P(a)

h


V (a) = limh→0

P(a + h)− P(a)

h.



time



time=

P(t1)− P(t0)

t1 − t0



time=

P(a + h)− P(a)

(a + h)− a=

P(a + h)− P(a)

h


V (a) = limh→0

P(a + h)− P(a)

h.

ExampleAfter having his hand cut off, and hearing some upsetting news, LukeSkywalker decides to jump through the main shaft in the Cloud City ofBespin.

If we assume the acceleration due to gravity in Bespin is8m/s2, what is Luke’s velocity after 11 seconds?

Fact: P(t) = a2 t2

ExampleAfter having his hand cut off, and hearing some upsetting news, LukeSkywalker decides to jump through the main shaft in the Cloud City ofBespin. If we assume the acceleration due to gravity in Bespin is8m/s2, what is Luke’s velocity after 11 seconds?

Fact: P(t) = a2 t2

ExampleAfter having his hand cut off, and hearing some upsetting news, LukeSkywalker decides to jump through the main shaft in the Cloud City ofBespin. If we assume the acceleration due to gravity in Bespin is8m/s2, what is Luke’s velocity after 11 seconds?

Fact: P(t) = a2 t2

The Derivative of a Function

The slope of a tangent line, or the velocity of an object are examples ofderivatives of functions.

DefinitionThe derivative of a function f at a number a, denoted by f ′(a), isdefined by

f ′(a) = limh→0

f (a + h)− f (a)

hif this limit exists and it is finite.

Alternatively,

f ′(a) = limx→a

f (x)− f (a)

x − a= lim

h→0

f (a + h)− f (a)

h.


Example

Let us calculate the derivative of f (x) = x2 + 3x + 2 at an arbitrarypoint x = a.


Rates of ChangeIf the quantity y is a function of x , and follow a rule y = f (x), then

f (x2)− f (x1)

x2 − x1

represents the average rate of change of y with respect to the variable x inthe interval [x1, x2].

We usually write:

∆x = x2 − x1 for the increment in x ,

∆y = y2 − y1 = f (x2)− f (x1) for the increment in y ,

so thatf (x2)− f (x1)

x2 − x1=

∆y

∆x

DefinitionThe instantaneous rate of change of a function f (x) at x = x1 is given by

f ′(x1) = limx2→x1

f (x2)− f (x1)

x2 − x1= lim

∆x→0

∆y

∆x.



f (x2)− f (x1)

x2 − x1

represents the average rate of change of y with respect to the variable x inthe interval [x1, x2]. We usually write:




x2 − x1=

∆y

∆x


f ′(x1) = limx2→x1

f (x2)− f (x1)

x2 − x1= lim

∆x→0

∆y

∆x.



f (x2)− f (x1)

x2 − x1

represents the average rate of change of y with respect to the variable x inthe interval [x1, x2]. We usually write:




x2 − x1=

∆y

∆x


f ′(x1) = limx2→x1

f (x2)− f (x1)

x2 − x1= lim

∆x→0

∆y

∆x.


Rates of Change

Note:

f ′(a) = limh→0

f (a + h)− f (a)

h

so, if we use h = 1, we obtain an approximation of f ′(a):

f ′(a) ∼=f (a + 1)− f (a)

1= f (a + 1)− f (a).

This means that f ′(a) is approximately the increment in y value afteran increment of one unit in x value after a, i.e., the increment in ywhen going from a→ a + 1.


Rates of Change

Note:

f ′(a) = limh→0

f (a + h)− f (a)

hso, if we use h = 1, we obtain an approximation of f ′(a):

f ′(a) ∼=f (a + 1)− f (a)

1= f (a + 1)− f (a).

This means that f ′(a) is approximately the increment in y value afteran increment of one unit in x value after a, i.e., the increment in ywhen going from a→ a + 1.


Example (Ebola cases in West Africa)

Let us define E(t) (with t in days) as the number of Ebola casesreported in West Africa since March 1st, 2014. Then, E ′(t) is the rateof change in the number of cases.

At each time t = a, the value E ′(a) is approximately the number ofnew cases the next day (i.e., day a + 1 of the epidemic).

Example (Ebola cases in West Africa)

Let us define E(t) (with t in days) as the number of Ebola casesreported in West Africa since March 1st, 2014. Then, E ′(t) is the rateof change in the number of cases.At each time t = a, the value E ′(a) is approximately the number ofnew cases the next day (i.e., day a + 1 of the epidemic).

The Derivative as a Function


f ′(a) = limh→0

f (a + h)− f (a)


We define a new function, the derivative of f :

DefinitionLet f (x) be a function. We define the derivative of f , denoted by f ′, by

f ′(x) = limh→0

f (x + h)− f (x)

h.

The domain of f ′(x) are those values x where the limit exists and it isfinite.




f ′(a) = limh→0

f (a + h)− f (a)




f ′(x) = limh→0

f (x + h)− f (x)

h.





f ′(a) = limh→0

f (a + h)− f (a)




f ′(x) = limh→0

f (x + h)− f (x)

h.



Example

Let us calculate the derivative of the function f (x) = x2 + 3x + 2.

We just saw that for any a, we have

f ′(a) = 2a + 3

Thus,f ′(x) = 2x + 3.


Example



f ′(a) = 2a + 3

Thus,f ′(x) = 2x + 3.


Example



f ′(a) = 2a + 3

Thus,f ′(x) = 2x + 3.


It is interesting to compare the graph of f (x) and the graph of f ′(x):

−4 −3 −2 −1 1

−1

1

2

3

4

5

0

f

f (x) = x2 + 3x + 2

−4 −3 −2 −1 1

−1

1

2

3

4

5

0

f

f ′(x) = 2x + 3.


It is interesting to compare the graph of f (x) and the graph of f ′(x):

−4 −3 −2 −1 1

−1

1

2

3

4

5

0

f

f (x) = x2 + 3x + 2

−4 −3 −2 −1 1

−1

1

2

3

4

5

0

f

f ′(x) = 2x + 3.


Documents

The Rosetta Mission