when you integrate 4x^5 + 3
you get 4x^6/6 + 3x + c
when you integrate 2
you get 2x + c
Exams is on calculus and matrix only
calculus
differentiation is gradient machine
give a point and you will get the gradient from gradient machine
differentiate polynomial, sin x, cos x, e^x , ln x
real life application RATE OF CHANGE is differentiation
VELOCITY = change in distance/ time
ACCELERATION = change in velocity /time
Integration
polynomial
sin x
cos x
area under curve
definite integral means point given so you can find the value of c
GOOD LUCK
Monday, April 27, 2009
Thursday, April 23, 2009
Integration Questios
Integrate the following
1. dy/dx = 5x^3 + 2x^2 + 5
2. dy/dx = 6x + 1
3. dy/dx = 2
4. dy/dx = 8x^5 - 5x^3 + 4x
5. dy/dx = 7x^2 + 3x + 4
1. dy/dx = 5x^3 + 2x^2 + 5
2. dy/dx = 6x + 1
3. dy/dx = 2
4. dy/dx = 8x^5 - 5x^3 + 4x
5. dy/dx = 7x^2 + 3x + 4
Tuesday, April 21, 2009
Integration +c being forgotten
Too many forgetting the + c in integration
question2
dy/dx=2x(x-3)
dy/dx=2x^2-6x
y=2/3x^3-3x^2
What about the + c
y=2/3x^3-3x^2 + c
To find the value of c use the point provided (3,6) x = 3 and y = 6
so c can be found
What about a question with v= and distance is required integrate yes but do not forget the +c
b)TO find the total distance travelled u need to integrate the (V) velocity.
v = 5t -3t^2 + 2
s= 5/2t^2 - t^3 + 2t
it must be
s= 5/2t^2 - t^3 + 2t + c
question2
dy/dx=2x(x-3)
dy/dx=2x^2-6x
y=2/3x^3-3x^2
What about the + c
y=2/3x^3-3x^2 + c
To find the value of c use the point provided (3,6) x = 3 and y = 6
so c can be found
What about a question with v= and distance is required integrate yes but do not forget the +c
b)TO find the total distance travelled u need to integrate the (V) velocity.
v = 5t -3t^2 + 2
s= 5/2t^2 - t^3 + 2t
it must be
s= 5/2t^2 - t^3 + 2t + c
Monday, April 20, 2009
Errors
1. y = 6x^3 + 4x^2 -5x
when you differentiate how can the y remain y
dy/dx = 18x^2 + 8x -5
2. y^2 = x^3 + 16 is in the syllabus differentiating with 2 unknowns
some of you missed this class
differentiate term by term as normal and if the current term is a y term differentiate w.r.t y and then multiply by dy/dx
2y dy/dx = 3x^2
3. 1/y = 4x + 4
no downstairs
y^-1 = 4x + 4
differentiate term by term
-y^-2 dy/dx = 4
when you differentiate how can the y remain y
dy/dx = 18x^2 + 8x -5
2. y^2 = x^3 + 16 is in the syllabus differentiating with 2 unknowns
some of you missed this class
differentiate term by term as normal and if the current term is a y term differentiate w.r.t y and then multiply by dy/dx
2y dy/dx = 3x^2
3. 1/y = 4x + 4
no downstairs
y^-1 = 4x + 4
differentiate term by term
-y^-2 dy/dx = 4
Sunday, April 19, 2009
Common errors being made
- y = 2x + 8/x^2 find dy/dx no downstairs y = 2x + 8x^-2 BUT when differentiating dy/dx = 2 -16x^-3 NOTE -2 -1 = -3
- y = ln (2x + 3) use sub for m BUT dy/dm CANNOT be the same as dy/dm instead dy/dx must be= dy/dm * dm/dx
Friday, April 17, 2009
Project (20%)
These are all the projects I received except for Brent and Azad (no ID on project but marked)
107000184
20
108002539
20
109000055
20
109000336
20
109003667
20
109003729
20
109004010
20
109004027
20
110000051
20
110000060
20
110000080
20
110000117
20
110000164
20
110000179
20
110000215
20
110000226
20
110000228
20
110000239
20
110000243
20
110000283
20
110000284
20
110000303
20
110000314
20
110000316
20
109004186
19
110000129
19
110000139
19
110000234
19
110000250
19
109000083
18
109002131
18
109002140
18
109002469
18
109004071
18
110000235
18
110000241
18
110000250
18
110000266
18
110000288
18
110000300
18
110000328
18
109002069
17
109002277
17
110000320
17
100000026
16
110000162
16
110000223
16
110000286
16
109000323
15
109002393
15
109004801
15
109004820
15
110000111
15
110000229
15
110000257
15
110000264
15
110000265
15
110000280
15
110000282
15
110000319
15
110000294
14
110000304
14
109002449
13
110000259
13
110000271
13
110000197
12
110000112
10
110000236
8
110000254
7
110000067
5
110000102
5
110000219
5
109003257
14
110000022
20
110000304
4
109002035
20
107000184
20
108002539
20
109000055
20
109000336
20
109003667
20
109003729
20
109004010
20
109004027
20
110000051
20
110000060
20
110000080
20
110000117
20
110000164
20
110000179
20
110000215
20
110000226
20
110000228
20
110000239
20
110000243
20
110000283
20
110000284
20
110000303
20
110000314
20
110000316
20
109004186
19
110000129
19
110000139
19
110000234
19
110000250
19
109000083
18
109002131
18
109002140
18
109002469
18
109004071
18
110000235
18
110000241
18
110000250
18
110000266
18
110000288
18
110000300
18
110000328
18
109002069
17
109002277
17
110000320
17
100000026
16
110000162
16
110000223
16
110000286
16
109000323
15
109002393
15
109004801
15
109004820
15
110000111
15
110000229
15
110000257
15
110000264
15
110000265
15
110000280
15
110000282
15
110000319
15
110000294
14
110000304
14
109002449
13
110000259
13
110000271
13
110000197
12
110000112
10
110000236
8
110000254
7
110000067
5
110000102
5
110000219
5
109003257
14
110000022
20
110000304
4
109002035
20
Sunday, April 12, 2009
Revision
- Find the trap area from (1,4) to (2,1) and the line y = 7 - 3x and the curve y = 4/x^2
- Find the equation of the curve which passes through the point (3,6) and for which dy/dx = 2x(x -3). Hint open brackets first then integrate.
- Differentiate e^(3-2x)
- A particle moves in a straight line so that, t seconds after passing through a fixed point O, its velocity, is giveb by v = 5t -3t^2 + 2. The particle comes to instantaneous rest at the point Q. Find
a) the acceleration of the particle atQ
b) the total distance travelled in the time interval t = 0 to t =3.
Velocity
A particle moves in a straight line so that at time t seconds after passing through a fixed point, its velocity, v m/s, is given by v = 6 cos 2t. Find
- the two smallest positive values of t for which the particle is at instanteneous rest
- the distance between the positions of instantaneous rest corresponding to these two values of t
- the greatest magnitude of the accelerstion
Friday, April 10, 2009
Find dy/dx
1. y = 6x^3 + 4x^2 - 5x
2. y^2 = x^3 + 16
3. 1/y = 4x + 4
4. y = 3x^4 + 3(x^2 + 5)
5. y + 6 = 4x + 9y - 3x^2
2. y^2 = x^3 + 16
3. 1/y = 4x + 4
4. y = 3x^4 + 3(x^2 + 5)
5. y + 6 = 4x + 9y - 3x^2
Wednesday, April 8, 2009
Questions 2
- A curve has equation y = 4/(2)^.5, find dy/dx
- A curve is such that dy/dx = 16/x^3 and (1,4) is a point on the curve, find the equation of the curve.
- y = 6theta - sin 2 theta, find dy/dx
Questions 1
- The equation of a curve is y = 2x + 8/x^2
find dy/dx and d^2ydx^2
- Differentiate ln (2x + 3)
- A curve is such that dy/dx = 2x^2 -5. Given that the point (3,8) lies on the curve, find the equation of the curve.
Tuesday, April 7, 2009
Matrices
- ....... is used to represent space in the matrix
14 ................6
5x-3 ............8
What value of x will this matrix be singular?
- What values of p will make this matrix singular
6p + 2 .........8
5 ...................3p
- Can a singular matrix have an inverse, justify your answer?
- Tom bought 6 plums and 5 mangoes for $40. Jane bought 3 similar plums and 4 similar mangoes for $23. Using the matrix method determine the price of 1 plum and the price of 1 mango.
Find the product of the following matricies
Find the product of the following matricies
1.
| 1 5 | | 2 2 |
| 3 8 | . | 4 5 |
2.
| 5 9 | | 2 1 |
| 3 2 | . | 2 7 |
3.
| 4 1 | | 1 2 |
| 0 2 | . | 4 1 |
4.
| 9 7 | | 3 3 |
| 9 8 | . | 3 3 |
5.
| 4 4 | | 5 2 |
| 4 4 | . | 3 5 |
6.
| 1 0 | | 0 5 |
| 0 8 | . | 4 4 |
7.
| 2 3 | | 3 2 |
| 3 6 | . | 3 5 |
8.
| 0 5 | | 0 0 |
| 0 8 | . | 4 5 |
9.
| 2 2 | | 3 3 |
| 2 2 | . | 3 3 |
10.
| 3 6 | | 6 3 |
| 6 3 | . | 3 6 |
1.
| 1 5 | | 2 2 |
| 3 8 | . | 4 5 |
2.
| 5 9 | | 2 1 |
| 3 2 | . | 2 7 |
3.
| 4 1 | | 1 2 |
| 0 2 | . | 4 1 |
4.
| 9 7 | | 3 3 |
| 9 8 | . | 3 3 |
5.
| 4 4 | | 5 2 |
| 4 4 | . | 3 5 |
6.
| 1 0 | | 0 5 |
| 0 8 | . | 4 4 |
7.
| 2 3 | | 3 2 |
| 3 6 | . | 3 5 |
8.
| 0 5 | | 0 0 |
| 0 8 | . | 4 5 |
9.
| 2 2 | | 3 3 |
| 2 2 | . | 3 3 |
10.
| 3 6 | | 6 3 |
| 6 3 | . | 3 6 |
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