Introduction to Mathematics Modelling There is huge part of compromise in mathematical modelling. Most of interacting systems in the real life are very complicated to model in their entirety. The first thing is to identify the important parts of the system which will be included in the model, the rest will be excluded. The second thing is to concerns the amount of mathematical manipulation which is valuable. Since mathematics has potential to prove general results, these results depend on the form of equations. Model equations can also be handle by using computers but it may never lead to elegant results. Mathematical modelling can be used for different objectives. Following are some examples: ▷ Development of scientific understanding. …show more content…
The Terrifier coaster is modelled by .for eight seconds. 1.1) which coaster is bigger? Area of Traumatiser roller coaster: Area of Traumatiser roller coaster: Above calculation shows that Traumatiser roller coaster is bigger. 1.2) Use factor theorem to confirm when the Terrifier coaster is at ground level. Factor Theorem: + 1.3) Find maximum and minimum heights using first and second derivatives with appropriate explanations. Traumatiser roller coaster: When t = 1.92, is positive means at this time, height is minimum = -12.09 When t = 6.08, is negative means at this time, height is maximum = 60.08 The Terrifier coaster: When t = 6.12, is positive means at this time, height is minimum = -4.06 When t = 3.21, is negative means at this time, height is maximum = 8.21 1.4) Draw graphs by hand – title, axes labelled, in pencil, appropriate scale, maximum and minimum labelled and include tables of values for …show more content…
It is concluded from section 1.1 and 1.4 that Terrifier coaster has larger underground as it travel underground for 4 seconds (from 0 to 2 and 5 to 7 seconds) while traumatizer coaster travel underground for only 2 seconds (from 1 to 3 seconds). Task 2: Find the maximum speed (5 marks) The velocity of a lorry, between motorway junctions, is modelled by For 0 ≤ t ≤ 30 Where t is the time in seconds after it sets off from the first junction. Find the maximum speed. Solution: Speed is maximum at We know that If we put t=15 in We get the maximum speed = 22.5. Task 3: Find the minimum area (10 marks) A paint manufacturer wants to design a cylindrical can to hold 5 liters of paint (5000).Find the minimum area of material that can be used to make the can and the dimensions of the can. Solution: If r (cm) is radius and h (cm) is height, then volume V = πr2h So if V = 5000 then πr2h = 5000 And the metal used M = 2πr2 + 2πrh M = 2πr2 + 2πrh = 2πr2 + 2πr × ………. (Using ) =2πr2 + 10000r-1 So, =4πr −
4. A) The functions for the problem would look like p=800*0.95*t and 0=50*1.15*t. P represents the pine
... : The difference in slope is positively correlated with a lower temperature. This slope becomes apparent
So using this formula but with the data we collected from our first attempt, this is what it would look like; Tan(60°) x 23m = 39m. As you can tell this answer collected from our first attempt is very well incorrect, but at the time, our group did not know this.
Y = sales of firm, X = average height of employees, α = intercept of the regression line,
60 What is Angle T? When there is more than 500 mils difference between the gun target line and the observer target line.
as the “r-value” and “r” can be any value between -1 and +1. It can be
6. A small stream flowing at a rate of 8 liters per second has a vertical drop of 1.5 m. What is the maximum power that you can obtain from this stream? (1 liter of water has a mass of 1 kg).
the length of the slope can be used to calculate the speed of the car
find the rate I have to find the mass change in 1 hour, and I will
- Temperature was measured after and exact time i.e. 1 minute, 2 minutes, 3 minutes.
Since the ball was not launched horizontally, initial velocity was taken into consideration. The ball was launched at increasing angles of 10° from a starting angle of 15°. The range, R, was calculated by using the initial velocity and the initial angle of the launch, because the horizontal range of the projectile is defined as the horizontal distance traveled by the projectile when it returns to the height at which it was originally launched. As a result, this gives the equation R = (v02sin2θ/g). However, the maximum value of sin2θ = 1 or when θ = 45°. In other words, the range will be at its maximum when the angle of projection is θ=45°. As the experiment progressed, the ranges for the ball being launched at an angle greater than 45° would decrease. Both the measured and calculated ranges were compared, and all values should be similar. With limited data, such as initial velocity or displacement, it is possible to determine the time of flight or range of the projectile motion of an
be the height of the ramp which in turn would affect the angle of the
This graph shows the result that I expect to get, I expect to see a