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antideritatives...

Originally posted by: CarlKillerMiller
The antiderivative is

(-1/2)e^(-t^2)

I think.

EDIT:
Whoops, forgot the negative.
Click to expand...

Pretty sure you're wrong, the exp() shouldnt change...

I believe its -e^(-t^2)/2

 
Originally posted by: RGUN
Originally posted by: CarlKillerMiller
The antiderivative is

(-1/2)e^(-t^2)

I think.

EDIT:
Whoops, forgot the negative.
Click to expand...

Pretty sure you're wrong, the exp() shouldnt change...

I believe its -e^(-t^2)/2
Click to expand...

I think that's the exact same thing that I wrote, you just moved the 2 to a denom and moved the negative sign to the numerator.
 
Originally posted by: CarlKillerMiller
Originally posted by: RGUN
Originally posted by: CarlKillerMiller
The antiderivative is

(-1/2)e^(-t^2)

I think.

EDIT:
Whoops, forgot the negative.
Click to expand...

Pretty sure you're wrong, the exp() shouldnt change...

I believe its -e^(-t^2)/2
Click to expand...

I think that's the exact same thing that I wrote, you just moved the 2 to a denom and moved the negative sign to the numerator.
Click to expand...


Yeah I went ahead and took the liberty to read your t^2 as t/2
 
Originally posted by: CarlKillerMiller
Originally posted by: RGUN
Originally posted by: CarlKillerMiller
The antiderivative is

(-1/2)e^(-t^2)

I think.

EDIT:
Whoops, forgot the negative.
Click to expand...

Pretty sure you're wrong, the exp() shouldnt change...

I believe its -e^(-t^2)/2
Click to expand...

I think that's the exact same thing that I wrote, you just moved the 2 to a denom and moved the negative sign to the numerator.
Click to expand...

I almost did the same thing. He probably read it when you had the mistake and didn't notice it was fixed in the edited quote.
 
I request help again 🙁 Thanks so far!

so if P'(t) = te^(-t^2) <--- this equals the rate of growth of profit in millions

t = time

3rd year total profit was $10,000

Total profit function would then be

-.5e^(t^2)

then?

graphing them both out doesn't really make anything that looks right.... 🙁


 
Originally posted by: djmihow
I request help again 🙁 Thanks so far!

so if P'(t) = te^(-t^2) <--- this equals the rate of growth of profit in millions

t = time

3rd year total profit was $10,000

Total profit function would then be

-.5e^(t^2)

then?

graphing them both out doesn't really make anything that looks right.... 🙁
Click to expand...

When you take the anti-derivative, you have a constant of integration.

P'(t) = te^(-t^2)
Integrating:
P(t) = -0.5e^(-t^2) + C

They tell you that in the 3rd year, the profit was 10,000.

So, P(3) = 10,000.

Solve for C and you have the profit function.
 
Originally posted by: djmihow
If the deritative is te^(-1t^2)


then the anti deritative should be e^(-1t^2) right?

t = time in this case.
Click to expand...

First, the word you're looking for is "derivative." Second, Integral is the proper term, I hate "anti deritative."

The best way to test your integral solution is to take its derivative.

(d/dt)(e^(-t^2)) = -2t*e^(-t^2)

You can just adjust your final solution by dividing by -1/2

So the actual answer is (-1/2)*e^((-1)*t^2)
 
Originally posted by: djmihow
I request help again 🙁 Thanks so far!

so if P'(t) = te^(-t^2) <--- this equals the rate of growth of profit in millions

t = time

3rd year total profit was $10,000

Total profit function would then be

-.5e^(t^2)

then?

graphing them both out doesn't really make anything that looks right.... 🙁
Click to expand...

Total profit as a function of time is P(t) where I assume t is in years

P'(t) = te^(-t^2)
You need to find P(t), so integrate
P(t) = -0.5*e^(-t^2) + C

Total profit in the 3rd year was 10,000. Plug in 10,000 for P(t) and 3 for t to find C

The 10,000 is only useful for finding C. Your final solution should not have 10,000 in it. Your final solution should look like

P(t) = -0.5*e^(-t^2) + (whatever number you found for C)
 
Originally posted by: Eeezee
Originally posted by: djmihow
I request help again 🙁 Thanks so far!

so if P'(t) = te^(-t^2) <--- this equals the rate of growth of profit in millions

t = time

3rd year total profit was $10,000

Total profit function would then be

-.5e^(t^2)

then?

graphing them both out doesn't really make anything that looks right.... 🙁
Click to expand...

Total profit as a function of time is P(t) where I assume t is in years

P'(t) = te^(-t^2)
You need to find P(t), so integrate
P(t) = -0.5*e^(-t^2) + C

Total profit in the 3rd year was 10,000. Plug in 10,000 for P(t) and 3 for t to find C

The 10,000 is only useful for finding C. Your final solution should not have 10,000 in it. Your final solution should look like

P(t) = -0.5*e^(-t^2) + (whatever number you found for C)
Click to expand...

ok, so whatever number I get for C should make a 10,000 dollar profit?

 
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