Drawing a Graph using Integral Equation

In summary, Homework Statement:The student is trying to solve a homework equation for power in watts over time. They have tried plugging in various values for t, but they are not sure how to do the trapezium rule. If they had already solved the equation for power, they would have included that information in their post.
  • #1
Missy
11
1

Homework Statement


Ok guys, its been a while since I have done this one so I don't even know where to begin. Here is the problem I have to complete:

The power (in watts) from an engine is represented by the equation below, where t is the time in seconds.

P = 30t^2.2 + 3t

1) Draw a table showing values of P for the range t=0 to t=9 seconds in intervals of 1 second. (Stuck on how to do this with the above equation)
2) Plot the graph that represents power against time from 0 to 9 seconds for this engine. Ensure the graph is correctly and fully labelled and titled. (Easy part)
3) Show the area on the graph which represents the energy converted between t = 5s and 9s (Easy part)
4) Using the trapezium rule, estimate the energy converted between t=5 and t=9 for intervals of 1 sec (never done this before)
6) Check your answer by integrating the expression between the limits t=5 and t=9. Account for the differences obtained.

Homework Equations


1) How do I solve my table of co-ordinates to plot the graph?
2) How do I apply the trapezium rule once I have plotted the graph?
The rest I can do

The Attempt at a Solution


I have integrated the expression to 9.375t^3.2 +3t^2/2

Please be nice, I am still relatively new to all of this so where you may be fully experience in dealing with calculus, trapezium rules and definite integration I'm still learning and I can't see how to solve this task.
 
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  • #2
Your integral looks correct. However, you need to provide more of an attempt at a solution. Since you say items #2 and #3 are "easy", please post your solutions to those items.

Also, does your textbook describe how to use the trapezoidal rule (or trapezium rule)? The Wikipedia page seems fairly informative:

http://en.wikipedia.org/wiki/Trapezoidal_rule
 
  • #3
PeterDonis said:
Your integral looks correct. However, you need to provide more of an attempt at a solution. Since you say items #2 and #3 are "easy", please post your solutions to those items.

Also, does your textbook describe how to use the trapezoidal rule (or trapezium rule)? The Wikipedia page seems fairly informative:

http://en.wikipedia.org/wiki/Trapezoidal_rule
Hi, I didn't post anything up for 2 and 3 because I don't know how to solve the table to be able to draw the graph and the area between 5 and 9 seconds. Once I get the table I can draw the graph and area on easily. We have to leave our textbooks at work so I don't have it with me.
 
  • #4
Missy said:
I don't know how to solve the table

What have you tried? Have you tried plugging various values for ##t## into the formula?
 
  • #5
PeterDonis said:
What have you tried? Have you tried plugging various values for ##t## into the formula?

Hi, I have tried plugging in the time constants I.e. 0,1,2,3 but that then makes them 1^2.2, 2^2.2, 3^2.2 which produces high values, it didn't seem right?
 
  • #6
Missy said:
Hi, I have tried plugging in the time constants I.e. 0,1,2,3 but that then makes them 1^2.2, 2^2.2, 3^2.2 which produces high values, it didn't seem right?

Just TELL us what values you finally obtained. How can we possibly be of help if you won't give us the information?
 
  • #7
Ray Vickson said:
Just TELL us what values you finally obtained. How can we possibly be of help if you won't give us the information?

Ok, firstly, please could you atleast be civil. I don't appreciate the tone in your message, I would have happily given you the answers had you asked me nicely. I'm here for help with my homework because I don't know how to solve it to get the right answers, I would appreciate it if you reconsidered your approach to me from here forward.

The t is for the time constants 0-9. And has a power of 2.2 in the equation. So 0 to the power of 2.2 is 0
1 to the power of 2.2 is 158.49 (2dp)
2 to the power of 2.2 is 316.98 (2dp)
3 to the power of 2.2 is 475.46 (2dp)
4 to the power of 2.2 is 633.96 (2dp)
5 to the power of 2.2 is 792.45 (2dp)
6 to the power of 2.2 is 950.94 (2dp)
7 to the power of 2.2 is 1109.43 (2dp)
8 to the power of 2.2 is 1267.91 (2dp)
9 to the power of 2.2 is 1426.40 (2dp)
 
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Likes Ray Vickson
  • #8
Missy said:
I don't appreciate the tone in your message, I would have happily given you the answers had you asked me nicely.

Just as a note, if you read the instructions given on how to post homework questions, they will tell you to show everything you've already done, even if you don't understand how to go any further. If you had already made the calculations you gave in post #7, according to the instructions, they should have been part of the "attempt at solution" section of your original post.

Missy said:
I'm here for help with my homework because I don't know how to solve it to get the right answers

Please bear in mind that we are trying to help you, but we can't just give you the answers or tell you how to get them; that's not the purpose of the homework help forum. The purpose is for you to show us what you've done, and for us to try to help you see for yourself how to get the rest of the way to the solution.

Missy said:
The t is for the time constants 0-9. And has a power of 2.2 in the equation.

And a coefficient of 30, and a second term with 3t; so the complete equation, as you posted it in the original post, is ##P = 30 t^{2.2} + 3 t##.

Missy said:
1 to the power of 2.2 is 158.49 (2dp)

Given the above, I don't understand how you got this. 1 to any power is 1; 30 times 1 to the power 2.2 is therefore 30; and 3 times 1 is 3, so plugging ##t = 1## into the above equation for ##P## should give ##P = 33##, shouldn't it?
 
  • #9
Missy said:
Ok, firstly, please could you atleast be civil. I don't appreciate the tone in your message, I would have happily given you the answers had you asked me nicely. I'm here for help with my homework because I don't know how to solve it to get the right answers, I would appreciate it if you reconsidered your approach to me from here forward.

The t is for the time constants 0-9. And has a power of 2.2 in the equation. So 0 to the power of 2.2 is 0
1 to the power of 2.2 is 158.49 (2dp)
2 to the power of 2.2 is 316.98 (2dp)
3 to the power of 2.2 is 475.46 (2dp)
4 to the power of 2.2 is 633.96 (2dp)
5 to the power of 2.2 is 792.45 (2dp)
6 to the power of 2.2 is 950.94 (2dp)
7 to the power of 2.2 is 1109.43 (2dp)
8 to the power of 2.2 is 1267.91 (2dp)
9 to the power of 2.2 is 1426.40 (2dp)

Nothing in my response was uncivil. Besides, what you really need to show are the values of ##P##, not of ##t^{2.2}##, because ##P## is what you have been asked to tabulate and plot.
 
  • #10
Ok sorry, it was a misunderstanding of context. I have found out that my powers have calculated incorrectly because I was using a poor and unbranded calculator. This is why I was having trouble solving the formula. I will re attempt it in the next 48 hours and let you know if I have any problems.
 
  • #11
Missy said:
Ok sorry, it was a misunderstanding of context. I have found out that my powers have calculated incorrectly because I was using a poor and unbranded calculator. This is why I was having trouble solving the formula. I will re attempt it in the next 48 hours and let you know if I have any problems.

There are several free on-line calculators available. Alternatively, if your computer is loaded with a spreadsheet, you can use that.
 
  • #12
Ray Vickson said:
Just TELL us what values you finally obtained. How can we possibly be of help if you won't give us the information?

Missy said:
Ok, firstly, please could you atleast be civil. I don't appreciate the tone in your message, I would have happily given you the answers had you asked me nicely. I'm here for help with my homework because I don't know how to solve it to get the right answers, I would appreciate it if you reconsidered your approach to me from here forward.
IMO, Ray WAS being civil, if a bit curt, which is understandable given the number of posts in this thread asking you for more information.
 

Related to Drawing a Graph using Integral Equation

1. How do I know which integral equation to use for a specific graph?

The specific integral equation to use for a graph depends on the type of graph you are trying to draw. Generally, if the graph is a continuous function, you can use the fundamental theorem of calculus to find the integral equation. If the graph has discontinuities, you may need to use specific integral equations for those types of functions. It is important to have a good understanding of different types of functions and their corresponding integral equations to determine which one to use.

2. What is the purpose of using an integral equation to draw a graph?

The purpose of using an integral equation is to find the relationship between the function and its derivative. This allows us to represent the graph of a function as the area under the curve, making it easier to visualize and analyze the function. It also helps in solving problems related to optimization and finding areas and volumes.

3. Can an integral equation be used to draw any type of graph?

Yes, an integral equation can be used to draw any type of graph as long as the function is continuous. However, if the function has discontinuities, special techniques may need to be used to accurately draw the graph.

4. Is it necessary to know the derivative of a function to draw its graph using an integral equation?

Yes, it is necessary to know the derivative of a function to draw its graph using an integral equation. The integral equation represents the relationship between the function and its derivative, so it is important to have this information in order to accurately draw the graph.

5. Are there any limitations to using an integral equation to draw a graph?

One limitation of using an integral equation to draw a graph is that it may not work for functions with infinite discontinuities or functions that are not continuous. In these cases, other methods may need to be used to accurately represent the function's graph. Additionally, using an integral equation may be more time-consuming and complex for certain types of functions, so it may not always be the most efficient method of graphing.

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