I am not to certain about the topic but i think it is defferentiation

N

Nikita.M

New member
Joined
Mar 4, 2020
Messages
1
I have always struggled with maths and would love for some assistant. Grateful if you can let me know the method use. I tried using basic solving skills howver it did not work out so well. I am willing to learn so grateful for the asst.
 

Attachments

  • Question 4.PDF
    97 KB · Views: 7
It is formally a question about a limit. Informally, it asks what does p get close to when t gets very big.

[MATH]p = \dfrac{6t^2 + 5t}{(t + 1)^2} = \dfrac{6t^2 + 5t}{t^2 + 2t + 1}.[/MATH]
Suppose t = 10, Then p = (600 + 50)/11^2, about 5.4.

Suppose t = 100. Then p = (60000 + 500)/ 101^2, about 5.9

Suppose t = 1000. Then p = (6,005,000)/1001^2, about 5.99.

So it looks as though the answer is that in the long term, p will approximate 6.

Here is how to confirm that answer in a semi-rigorous way.

[MATH]p = \dfrac{6t^2 + 5t}{t^2 + 2t + 1} = \dfrac{6t^2 + 12t + 6 - 12t - 6 + 5t}{t^2 + 2t + 1} = [/MATH]
[MATH]\dfrac{6(t^2 + 2t + 1)}{t^2 + 2t + 1} + \dfrac{-\ 7t - 6}{t^2 + 2t + 1} = 6 -\dfrac{7t + 6}{(t + 1)^2}.[/MATH]
Now it should be intuitively obvious that the fraction is positive so subtracting it from 6 means p will never reach 6. But it should also be obvious that t squared grows a lot faster than 7t. If t is 1000, then t^2 is a million, which is a lot bigger than 7000. So the fraction is closing in on zero. Thus, the limit is 6 - 0 = 6.

This is not a formal proof, but it gives the idea of what a limit is and how to find it if it exists. (It does not have to exist.)

Hope this helps.
 
Another way of looking at this after you have multiplied out the denominator, is to divide both numerator and denominator by \(\displaystyle t^2\), the largest power: \(\displaystyle \frac{6+\frac{5}{t}}{1+ \frac{2}{t}+ \frac{1}{t^2}}\). Now the point is that while "infinity" is difficult to work with, "0" is not! A t goes to infinity, \(\displaystyle \frac{1}{t}\) and \(\displaystyle \frac{1}{t^2}\) go to 0.
 
HallsofIvy said:
Another way of looking at this after you have multiplied out the denominator, is to divide both numerator and denominator by \(\displaystyle t^2\), the largest power: \(\displaystyle \frac{6+\frac{5}{t}}{1+ \frac{2}{t}+ \frac{1}{t^2}}\). Now the point is that while "infinity" is difficult to work with, "0" is not! A t goes to infinity, \(\displaystyle \frac{1}{t}\) and \(\displaystyle \frac{1}{t^2}\) go to 0.
Click to expand...
Working with infinity is fun!, not hard!
But for the sake of the OP, it is best to do as Sir Halls said in his post.
 
topsquark said:
I can't open the pdf. Is it just me?

-Dan
Click to expand...

The content is as follows:

Q4

(i) A business manager determines that \(t\) months after production begins on a new product the number of units produced will be \(P\) thousand where:

[MATH]P(t)=\frac{6t^2+5t}{(t+1)^2}[/MATH]
What happens to production in the long run?
 
You must log in or register to reply here.
Top