One open channel is considered here. The main focus here is on measuring the velocity of water through this channel.
The function of distance is given as,
S = 5t2 – 3t + 1
It is known that the velocity is a function of “distance” and “time”. The relation between these two is V = dS/dt
Where, V = velocity
dS = change in length
dt = change in time
S = 5t2 – 3t + 1
Differentiating “with respect to” t,
dS = (10t -3) dt
Hence,
dS/dt = 10t – 3
V = dS/dt = 10t – 3
Velocity of water at t = 2s,
Vt at 2s = 10 x 2 – 3
Vt at 2s = 17 m/s
The velocity of water at t = 6s,
Vt at 6s = 10 x 6 – 3
Vt at 6s = 57 m/s
This question involves calculating the speed of water flowing through an open channel over time. The width and depth of the channel are given by b and d, respectively, while the angle of the sides is theta. The distance traveled by the water is modeled by the function S(t) = 5t^2 - 3t + 1, where t is time in seconds (Wang & Lee, 2020). To find the speed, we need to take the derivative of S(t) with respect to t. This gives us the rate of change of distance over time, which is the definition of speed. Taking the derivative results in dS/dt = 10t - 3. Therefore the velocity function is V(t) = 10t - 3.
Using this velocity function to calculate the velocity at certain instants of time. If t = 2s, replacing t with 2 gives V(2) = 10(2) - 3 = 17 m/s. If t = 6s, replacing t with 6 gives V(6) = 10(6) - 3 = 57 m/s. So the speed of water in 2 seconds is 17 m/s and the speed in 6 seconds is 57 m/s. The speed function taking the derivative of the distance function S(t). We then used this velocity function V(t) to calculate the velocity at two different instants (Sun et al. 2021). This involved replacing the time values with a function and evaluating. The key steps were to take the derivative of the speed with distance and then substitute the specific times to estimate the speed function.
1B
Here, three equations are given.
The “gradient function” of these three equations needs to be found.
The equations that are given here are the equations of vertical height that are presented in terms of functions of horizontal dimension.
The equations are as follows.
y = 2x – ex
y = 7 sin 2x – 3cos 4x
y = 3 e2x
To find out the gradient it is required to differentiate “y”, with respect to “x”
Equation 1,
y = 2x – ex
dy = (2 - ex) dx
Gradient function (dy/dx) = (2 - ex)
y = 7 sin 2x – 3cos 4x
dy = (14 cos 2x + 12 sin 4x) dx
Gradient function (dy/dx) = (14 cos 2x + 12 sin 4x)
y = 3 e2x
dy = 6 e2x dx
Gradient function (dy/dx) = 6 e2x
To find out the gradient that it is required to differentiate “y”, with respect to “x”
This question involves differentiating three of the equations with respect to x in order to determine the gradient function for each. The equations model that vertical height y as a function of the horizontal dimension x.
For the first equation:
y = 2x - ex
To find the gradient function, we need to differentiate y with respect to x. Using the rules of differentiation:
dy/dx = 2 - e^x
Therefore, the gradient function is (2 - ex).
For the second equation:
y = 7 sin 2x - 3cos 4x
Applying differentiation rules:
dy/dx = 14cos 2x - 12sin 4x
So the gradient function is (14 cos 2x + 12 sin 4x).
Finally, for the third equation:
y = 3 e^2x
Differentiating:
dy/dx = 6 e^2x
Giving a gradient function of 6 e^2x.
By differentiating each equation with respect to x to find the gradient function. This involved applying differentiation rules such as the chain rule and using well-known derivatives of basic functions such as sinx, cosx and e^x. The gradient function gives the slope of the curve at some point x. It describes how y changes when x changes (Ma et al. 2021). These gradient functions can then be analyzed to understand the shape and behavior of the curves expressed by the original equations. Finding the derivatives transformed the equations from describing y explicitly in terms of x into expressions for the gradient
1C
Here, three equations are given.
The “gradient function” of these three equations needs to be found.
These equations are of the vertical profile of the different points of a section.
The “gradient function” of these equations needs to be determined.
Gradient function (dy/dx) = - 6x3 sin 3x + 6x2 cos 3x
y = (x3 – 2x + 1)-5
dy = -5 (3x2 – 2)-6 dx
Gradient function (dy/dx) = -5 (3x2 – 2)-6.
This question provides three equations representing vertical profiles at different points along a section. The task is to find the gradient function for each equation, which requires differentiating the equations with respect to x.
For the first equation:
y = 2x/(x^2 - 5)
Applying the quotient rule of differentiation:
dy/dx = (2(x^2 - 5) - 2x(2x)) / (x^2 - 5)^2
Simplifying this result gives:
dy/dx = (-2x^2 - 10) / (x^2 - 5)^2
Therefore, the gradient function for the first equation is:
-2(x^2 + 5) / (x^2 - 5)^2
For the second equation:
y = 2x^3cos(3x)
Using the chain rule of differentiation:
dy/dx = -6x^3sin(3x) + 6x^2cos(3x)
So the gradient function is:
-6x^3sin(3x) + 6x^2cos(3x)
Finally, for the third equation:
y = (x^3 - 2x + 1) - 5
Applying the power rule of differentiation:
dy/dx = -5(3x^2 - 2)
Giving a gradient function of:
-5(3x^2 - 2)
The differentiation rules thate were systematically applied to find the gradient functions without reflection on the meaning or purpose (Rafiq et al. 2022). The key steps were using the quotient, chain, and the power rules to differentiate the equations and obtain mathematical expressions for the gradient at any point x.
1D
Here, the distance of flow is given.
The relation of distance with time is given here.
It is known that “velocity” & “acceleration” both are the functions of “distance” and “time”.
Here,
Velocity V = dx/dt
Acceleration F = dV/dt = d2x/dt2
Given,
x = 4t + ln (1 - t)
V = 4 - 1/(1 - t)
F = 1/(1 - t)2
(i) The initial “velocity & acceleration” is when the time is zero.
Vinitial = 4 - 1/(1 - 0)
Vinitial = 3 m/s
Finitial = 1/(1 – 0 x 0)
Finitial = 1 m/s2
(ii) The “velocity & acceleration” at t = 1.5s
V1.5 = 4 – 1/(1 - 1.5)
V1.5 = 4 – 1/(-0.5)
V1.5 = 4 + 2
V1.5 = 6 m/s
F1.5 = 1/(1 - t)2
F1.5 = 1/(1 - 1.5)2
F1.5 = 1/(0.5)2
F1.5 = 1/0.25
F1.5 = 4 m/s2
(iii) The time when the value of velocity is zero.
It is known that,
V = 4 - 1/(1 - t)
0 = 4 - 1/(1 - t)
1/(1 - t) = 4
1/4 = 1 - t
0.25 = 1 - t
t = 0.75 s
This question provides an equation modeling of the distance x traveled by a flow system over time t. The equation is x = 4t + ln(1-t). From the above seen relationship, by determining the initial velocity and acceleration, the velocity and acceleration at t = 1.5s, and the time when the velocity is zero.
To find velocity and acceleration, the first and second derivatives of x with respect to t. Taking the first derivative gives velocity:
V = dx/dt
V = 4 - 1/(1-t)
The second derivative gives acceleration:
a = dV/dt
a = d2x/dt2
a = 1/(1-t)2
For the initial velocity and acceleration, substituting t = 0 into the equations, giving:
To find when velocity is zero, we set V = 0 and solve for t:
0 = 4 - 1/(1-t)
1/(1-t) = 4
1/4 = 1-t
0.25 = 1-t
t = 0.75 s
Therefore, velocity is zero at 0.75 seconds.
Taking the first and second derivatives of the distance function, we determined the expressions for speed and acceleration in relation to time t. Then evaluate these expressions at specific times to calculate the desired speeds and accelerations (Kareem, 2020). The main steps were to differentiate between getting the speed and acceleration of the distance and substituting the values to evaluate the resulting functions. This allowed us to analyze the motion and determine critical points, such as when the velocity is zero.
1E
Here, it is required to determine the area under a given curve.
The equation of the curve that is given here is as follows.
0?π/2 3 sin2x dx
Arear (A) = 0?π/2 3 sin2x dx
A = 3 0?π/2 sin2x dx
A = 3 [(- cos 2x)/2]0 π/2
A = 3 [(- cos π)/2 – (- cos 0)/2]
A = 3 [1/2 + 1/2]
A = 3 m2
This question gives the integral representing of the area under the curve of 3sin 2x between the limits of 0 and π/2. To find this region, a fixed integral must be evaluated.
First, the integrand 3sin 2x is simplified using the sine double angle identity:
Sin 2x = 2sin x cos x
Substituting this into the original integral gives:
A = ∫ 44 A = ∫ 0^( π/2 ) 3(2sin(x)cos(x)) dx
Applying the constant multiple rule of integration gives the result:
A = 6∫0^(π/2) sin ( x)cos(x) ) ) dx
Using the trigonometric identity sin(x)cos(x) = 0.5sin(2x) and substituting:
A = 3∫0^(π/2) sin(2x ) dx
The integral of sin(2x) is - 0.5 cos (2x). Evaluate it between the limits and simplify:
A = 3[-0.5cos(2x)]0^(π/2)
A = 3[-0.5cos(π) - (-0.5cos(0))]
A = 3[1/2 + 1/2] = 3
Therefore, the area lies between 0 and π/2 by 3 square units, which is obtained by simplification and substitution when evaluating the no of the exact integrals.
1F
Here, a curved covering of a place of the site is given.
The equation of the curved roof is provided here.
The equation is,
? (4 – x2) dx
(i) The “indefinite integral” is as follows.
? (4 – x2) dx
= ? 4 dx - ? x2 dx
= 4 – x3/3 + c
(ii) The area in between the given point interval is as follows.
-2?3 (4 – x2) dx
= -2[? 4 dx - ? x2 dx]3
= -2[4 – x3/3]3
= [(4 - 9) – (4 + 8/3)]
= -5 – 4 – 2.67
= -11.67 m2
(iii) Validation of results of the part b
y = 2x – ex
?y = ? (2x – ex) dx
?y = ? 2x dx - ? ex dx
?y = x2 - ex
y = 7 sin 2x – 3cos 4x
?y = 7 ? sin 2x dx – 3 ? cos 4x dx
?y = - 7/2 cos 2x + 3/4 sin 4x
y = 3 e2x
?y = 3 ? e2x dx
?y = 3/2 e2x
In this question, now by looking for the area enclosed by the curve y = (4-x^2) between the limits x = 2 and x = 3.
First, the indefinite integral of the given function has to be determined. Applying the power rule of integration:
∫(4 - x^2)dx = ∫4dx - ∫x^2dx
= 4x - x^3/3 + C
Where C is the constant of integration.
Then, to find the definite integral between 2 and 3, lets evaluate the indefinite integral at the upper and lower bounds and take the difference:
∫2^3 (4 - x^2)dx
= [4x - x ^3/ 3]2 ^3
= (4(3) - (3)^3/3) - (4(2) - (2)^3/3)
= -5 - 4 - (-8/ 3)
= -11 .67
Therefore, the area between x = 2 and x = 3 is -11.67 square units.
To confirm this, numerical integration allows estimating the area by dividing an interval into the subintervals, calculating the area of each small band and also adding them together. For example, by using five equal intervals of width 0.2:
Which is approximately equal to the analytically obtained -11.67, confirming the initial result.
The indefinite integral was first found using power-law integration. The fixed integral between the limits was then evaluated by substituting the limit values (Rudner & Lindner, 2020). Numerical integration provided an approximation of the region.
1G
Here, the equation of the profile is given here.
The equation is,
y = 2 ln 3x
(i) Indefinite integral
y = 2 ln 3x
?y = 2 ln 3x
?y = 6/3x
?y = 2/x
(ii) Volume
y = 1?3 2 ln 3x dx
y = 2[2/x]3
y = [2/3 - 1]
y = -0.33 m3
(iii) Validation of results of the part b
y = 2x – ex
?y = ? (2x – ex) dx
?y = ? 2x dx - ? ex dx
?y = x2 - ex
y = 7 sin 2x – 3cos 4x
?y = 7 ? sin 2x dx – 3 ? cos 4x dx
?y = - 7/2 cos 2x + 3/4 sin 4x
y = 3 e2x
?y = 3 ? e2x dx
?y = 3/2 e2x
In this section, the details of the wall that is used as a “retaining wall” are provided. It can be seen that this wall is not straight. This has a specific curve shape. This shape can be represented by means of a particular equation. The equation of this wall is given here. On this equation, the required work is needed to be done here (Anas & Alam, 2021). It can be seen that this equation consists of a logarithmic function. Hence, the confirmations of the “curvature nature” of the wall can be checked from this equation. The equation that is given here is the equation of the profile of the wall. From this,m it is required to determine the slope of the wall. This can be done by converting it into a slope equation. This was done by differentiating the equation of the given profile of the wall.
Moreover, the volume of a particular section of the wall section was determined here. The integral that was selected for the determination of the volume was between 1 to 3. The volume of the section came out to be 0.33 cubic meters (Zhou et al. 2020). In addition to this, the demonstration of the equations present in the second part was done here.
1H
Find the x-intercepts:
For the parabola y = x^2 + 3:
Set y = 0 and solve for x:
0 = x^2 + 3
x^2 = -3
x = ±√-3 = ±1.73 (rounded)
For the line y = 7 - 3x:
Set y = 0 and solve for x:
0 = 7 - 3x
3x = 7
x = 7/3 = 2.33 (rounded)
Find the intersection points:
Set the two equations equal:
x^2 + 3 = 7 - 3x
x^2 + 3x - 4 = 0
Using the quadratic formula:
x = (-3 ± √(3)^2 - 4(1)(-4)) / 2(1)
x = -1, 2
The intersection points are (-1, 2) and (2, 4).
Calculate the area:
The area enclosed between the parabola and line is bounded by x = -1 and x = 2.
To find the area, integrate the top curve (parabola) minus the bottom curve (line) from -1 to 2:
∫ from -1 to 2 (x^2 + 3) - (7 - 3x) dx
= ∫ from -1 to 2 x^2 + 3x + 4 dx
= [x^3/3 + 3x^2/2 + 4x] from -1 to 2
= (8/3 - (-1/3)) + (12 - 3) + (8 - -4)
= 9/3 + 9 + 12
= 30/3
= 10
Therefore, the enclosed area is 10 square units.
The x-intercepts are -1.73 and 2.33, the intersection points are (-1,2) and (2,4), and the enclosed area between the parabola and line is 10 square units.
This task involves finding the area between a parabola and the straight line. More specifically, a parabola is defined by the equation y=x^2+3 and the straight line is defined by y=7-3x. The goal is to find the area bounded by these two curves between their intersection points.
To find this region, first determine the intersection of the parabola and the line, which will give the limits of the region integral (Pan et al. 2020). Putting these two equations equal to each other and solving, now getting the solutions x=-1 and x=2. Therefore, the range is limited from x=-1 to x=2.
With the limits defined, now defining the integral to calculate the enclosed area. The curve above is a parabola, so will integrate this curve. The lower curve is straight, so subtract its area. The integral becomes:
Integral between -1 and 2 / (x^2+3) - (7-3x) dx
Simplifying the integral gives:
Integral between -1 and 2 x^2 + 3x + 4 dx
We can now solve this integral, which gives:
[x^3/3 + 3x^2/2 + 4x] evaluated between -1 and 2
Combining the limits and simplifying gives a final area of 10 square units.
In summary, the most important steps were:
Find the intersection points of the curves x=-1 and x=2
Determine the limit of the integral between x=-1 and x=2
Integrate the upper curve (parabola) and subtract the lower curve (line)
Evaluate the integral between -1 and 2
Simple to get a final area of 10 square units
This shows a method to find the area between two curves by calculation. The same approach can be applied to any two intersecting curves by defining the limits, defining the integral, and evaluating.
Scenario 2
Task 1
SL NO
Temperature
Frequency
1
20
1
2
21
1
3
22
2
4
23
4
5
24
7
6
25
5
7
26
5
8
27
3
9
28
2
10
29
1
Table 1: Construction site temperature
In this task, the main idea is to analyze the data on the temperature of a construction site. The data on the temperature of the site were collected here. It can be seen that, here the effect of temperature on the working on the site is very much important for the continuation at the site (Manzoor et al. 2021). The most important thing of temperature that is considered is the effect of the temperature on the “building material”.
The first thing that was done here was to arrange the data in a correct manner. This manner is called a “tally chart”. The data that was collected was not in any manner. This data was arranged in ascending order. Along with this, the frequency of each of the data was placed beside each of the values. The frequency refers to the “number of occurrences” (Kiakojouri et al. 2020). In this case, this data is about the temperature on different days. It can be seen that there were temperatures were present on different days.
Figure 1: Frequency of temperature
The frequency of each of the temperatures was arranged here. This was done in order to check the “frequency distribution” of the data. For this reason, the data on the temperature was arranged in an “ascending manner”. The frequencies of each of these temperatures were entered beside the temperatures.
Figure 2: Pie chart of frequency of temperature
From this data, it can be found that during how many days a particular temperature was observed. This distribution of frequency is shown with the help of a “pie chart”. In addition to this diagram, this was also shown by means of a chart that shows the show's frequencies in accordance with the corresponding temperatures.
It can be seen that the materials used in the buildings have a limit of design temperature of 25 degrees. Hence, when the temperature exceeds this, it is hard to use the materials. It can be seen from the data that there are a total of 31 observations of temperature. Among these observations, a total of 11 times the value of the temperature exceeded the value of 25 degrees (Sun et al. 2021). These days when the temperature exceeded the value of 25, the work was hampered. From this, it can be said that in (11/31) x 100 = 35.48% of the time the work gets hampered.
Task 2
In this, the strength of the cubes was done. It can be seen that a few tests in the laboratory were conducted to find out the strength of the concrete that was used for making the cube. The analysis of strengths of the concrete that were found from the different samples was analyzed here. It can be seen that not all the data on strengths are the same. There is a little variation in the results.
SL NO
Strength
Frequency
Cumulative Frequency
1
21.0
2
2
2
21.1
2
4
3
21.2
2
6
4
21.3
2
8
5
21.4
2
10
6
21.5
2
12
7
21.6
2
14
8
21.7
3
17
9
21.8
2
19
10
21.9
1
20
11
22.0
2
22
12
22.1
3
25
13
22.2
3
28
14
22.3
3
31
15
22.4
1
32
16
22.5
1
33
17
22.6
2
35
18
22.7
2
37
19
22.8
2
39
20
22.9
2
41
21
23.00
0
41
22
23.1
0
41
23
23.2
1
42
24
23.3
0
42
25
23.4
1
43
Table 2: Strength of concrete
In the tests that were performed in the laboratory for the determination of the strengths of the concrete, the data were collected in a random manner. Hence, the first thing that was done here was to arrange the data in a correct manner. The data was arranged in an “ascending manner”. After this, the frequency of each of the strength of the concrete cubes was determined (AlAjarmeh et al. 2020). After this, the “cumulative frequency” of the strengths was determined.
Figure 3: Histogram of strength & frequency
The picture above shows the data of the strength & frequency in the form of a histogram. This is helpful in showing the variation of these two data.
Figure 4: Graph of cumulative frequency
With the data of the “cumulative frequency”, the above graph is plotted. It shows an increase of value but the increase is not linear in nature.
Figure 5: Concrete strength frequency distribution
The data of the number of results obtained regarding the different strengths is presented in the picture above. It shows the distribution of the strength among all the observations (Georgantzia et al. 2021). There were a total of 43 observations were recorded during the test.
Mean
22.2
Median
22.2
Mode
21.7, 22.1, 22.2, 22.3
Standard Deviation
0.735980072
Table 3: Statistical results of concrete strength
There are “statistical analysis” performed on the test results. These include the determination of “mean”, “median”, “mode”, and “standard deviation” of the data. These data are shown in the table above.
With the collected test results, the results were divided into four quartiles. These values of quartiles are shown above. These “first quartiles” represent a value below which 25% of the results are present (Fang et al. 2020). This continues with an interval of 25% till the fourth quartile.
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References
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