wtf...how is this right? math...

[ManBearPig]

rate at which something is produced is 25t/(7t^2 + 15)

t is in mins, 0<=t<=110

110
sumation symbol 25t/(7t^2 + 15)dt
0

=15.4268

WTF? how in the hell do you get that answer, it just makes no sense!

 

[Tiamat]

have you run it in MS Excel?

In any case, why dont you solve it analytically?

 

[ManBearPig]

i dunno how to run that.

i tried just putting the rate in my calc, graphing that, then hitting 2nd, trace, 7, 0 lower limit, 110 upper limit but it says error invalid. omggggggggg

 

[tranceport]

42

 

[Leros]

By doing a summation, you are only hitting the integers. You need to calculate the sum of EVERY SINGLE point (there are an infinite number of them) in between 0 and 100. So you need to take the integral of that function from 0 to 110.

Edit: Checked it out. Doing the integral got me 15.426781

Edit2: A summation got me a close approximation at 15.29536

 

[Sc4freak]

That's correct. I assume you're wanting to find the number of items produced in 110 minutes - in which case you simply need to integrate it between 0 and 110. Summation won't give you the correct answer.

 

[Tiamat]

Summation gives only an approximation. You want to integrate it, should be basic calculus here. Hint: You will get a ln term.

 

[Sassy Rabbit]

Got the same answer. Integrate the expression 25t/(7t^2 + 15) from 0 to 110.

 

[ManBearPig]

i thought to integrate you just did this though: "i tried just putting the rate in my calc, graphing that, then hitting 2nd, trace, 7, 0 lower limit, 110 upper limit but it says error invalid."

isn't that how you do it with the calc?

 

[Leros]

Use the integrate function?

 

[Tiamat]

Originally posted by: Kazaam
i thought to integrate you just did this though: "i tried just putting the rate in my calc, graphing that, then hitting 2nd, trace, 7, 0 lower limit, 110 upper limit but it says error invalid."

isn't that how you do it with the calc?

Dont use the calculator. Use your brain. It takes 20 seconds to find the antiderivative of the function you listed. Its taken you over 40 minutes with the calculator and failing.

Hint 2: Use u-substitution on the denominator...

 

[dighn]

Originally posted by: Kazaam
i thought to integrate you just did this though: "i tried just putting the rate in my calc, graphing that, then hitting 2nd, trace, 7, 0 lower limit, 110 upper limit but it says error invalid."

isn't that how you do it with the calc?

no you enter the equation into a ti-89 and it spits out 25/14(ln(7t^2 + 15))

or you could actually use your brain

 

[blinky8225]

Yeah, but your window has to be big enough. I'm assuming you're using a Ti 83. Make sure your window actually contains x from 0 to 110.

 

[ManBearPig]

ugh...its like she just gave us the fucking answer in the notes but didnt show us HOW to do it, everything else i get.

anyone wanna do 11t/(8t^2 + 16) limits t=0 and t=20 for me? i tried it in the calc and got 9.347 but that doesnt work!

 

[Tiamat]

Originally posted by: Kazaam
ugh...its like she just gave us the fucking answer in the notes but didnt show us HOW to do it, everything else i get.

anyone wanna do 11t/(8t^2 + 16) limits t=0 and t=20 for me? i tried it in the calc and got 9.347 but that doesnt work!

do you understand how to take the integral of that equation? I dunno if you want help with the actual calculus...

Pull 11/8 outside of the integral such that you get {11/8}Integral [t/t^2+2]dt

Let u = t^2 + 2

such that: du = 2tdt

so: dt = du/2t

Therefore: {11/8}Integral {[t/u]*[du/2t]}

Which is equivalent to: [11/16] Integral (1/u)du

Taking the integral of 1/u gives ln(u) where u = t^2 + 2

Substituting: the answer is [11/16] ln (t^2 + 2) which needs to be evaluated from t = 0 to t = 20.

Evaluating: 11/16 {ln (20^2 + 2) - ln (0^2 + 2)}

solution: 11/16{ ln (402) - ln(2)}

simplification: 11/16 {ln(201)} in the final answer.

 

[blinky8225]

Do you have a textbook? Read the chapter on u-substitution. You'll learn that you can't rely on the teacher for everything.

 

[ManBearPig]

i dont understand how to take the integral...we learned how to do reimann sums or whatever, left and right hand, over and under estimates, and how to find answers thru the calculator, but it wont work for this. and this problem is due tomorrow morning lol...damn her!

 

[WombRaider]

I'm glad I'm done with all the Calculus I'll ever need.

 

[Tiamat]

Originally posted by: Kazaam
i dont understand how to take the integral...we learned how to do reimann sums or whatever, left and right hand, over and under estimates, and how to find answers thru the calculator, but it wont work for this. and this problem is due tomorrow morning lol...damn her!

Check my above post, I just did out the entire problem so that you can hopefully learn it by tomorrow.

 

[ManBearPig]

Tiamat, thank you for explaining it! and thanks to the others for trying to.

 

[Tiamat]

Originally posted by: Kazaam
Tiamat, thank you for explaining it! and thanks to the others for trying to.

no prob. its hard in the beginning, but once you keep practicing, it becomes very easy. Riemann sums is the approximation attempt to solve the integral. The way i showed you yields the exact answer for the types of integrals that are typically solvable. For wierd functions, riemann sums is faster. For normal functions, analytical can be much faster.


I sent you PM just in case you had trouble figuring out the first problem in your OP.