Surge Functions and finding Y co-ordinates

[kais]

Hey guys, was wondering if i could get some help on this calculus question

For the general surge function Ate^-bt verify by differentiation that

(a) the turning point occurs at t= 1/b

(b) the point of inflection occurs at t= 2/b

(c) find the Y co-ordinates of the turning point and point of inflections

So far i have the first derivative which is ate^-bt = ae^-bt - bate^-bt = ae^-bt(1-bt)
finding zeroes means that ae^-bt will never equal zero so no solution , but 1-bt=0 give t=1/b

then for second part ive found the second derivative which ive written ae^-bt(1-bt)= -abe^-bt - abe^-bt + ab^2te^-bt = abe^bt(2-bt)

setting that to zeroes you end up with 2-bt=0 t=2/b

Now im not sure how i plug this back into the equation to get the y coordinates? this is the part im stuck on. Any help would be appreciated
also im not sure how to plug that graph into wolfram alpha, if someone could send me a link that would be awesome. And last thing if anyone knows how to format the equations better on this forum that would be awesome!

Cheers in advance!

 

[stapel]

For the general surge function Ate^-bt...

As posted, this is not a function, since there is no "equals" nor a function name. Also, lacking grouping symbols, only the "-b" is in the exponent. However, I suspect that "t" is the variable. Also, in what follows, you use "a" rather than "A", so I suspect the function is actually as follows:

. . . . .\(\displaystyle f(t)\, =\, ate^{-bt}\)

Kindly please reply with corrections, if the above is wrong.

verify by differentiation that

(a) the turning point occurs at t= 1/b

I will take "turning point" to be as defined here.

(b) the point of inflection occurs at t= 2/b

(c) find the Y co-ordinates of the turning point and point of inflections

So far i have the first derivative which is ate^-bt = ae^-bt - bate^-bt = ae^-bt(1-bt)
finding zeroes means that ae^-bt will never equal zero so no solution , but 1-bt=0 give t=1/b

I think this means that you did the following:

. . . . .\(\displaystyle f(t)\, =\, ate^{-bt}\)

. . . . .\(\displaystyle f'(t)\, =\, (a)\left(e^{-bt}\right)\, +\, (at)\left(-be^{-bt}\right)\, =\, ae^{-bt}\, -\, abte^{-bt}\)

. . . . .\(\displaystyle ae^{-bt}\, -\, abte^{-bt}\, =\, 0\)

. . . . .\(\displaystyle ae^{-bt}\, =\, abte^{-bt}\)

. . . . .\(\displaystyle \mbox{We know that }\, e^{-bt}\, \mbox{ is never zero, so we can divide thru:}\)

. . . . .\(\displaystyle a\, =\, abt\)

. . . . .\(\displaystyle a\, -\, abt\, =\, 0\)

. . . . .\(\displaystyle a(1\, -\, bt)\, =\, 0\)

. . . . .\(\displaystyle \mbox{By definition, we have }\, a\, \neq\, 0,\, \mbox{ so:}\)

. . . . .\(\displaystyle 1\, -\, bt\, =\, 0\)

. . . . .\(\displaystyle \dfrac{1}{b}\, =\, t\)

then for second part ive found the second derivative which ive written ae^-bt(1-bt)= -abe^-bt - abe^-bt + ab^2te^-bt = abe^bt(2-bt)
setting that to zeroes you end up with 2-bt=0 t=2/b

I think your steps were as follows:

. . . . .\(\displaystyle f'(t)\, =\, ae^{-bt}\, -\, abte^{-bt}\)

. . . . .\(\displaystyle f''(t)\, =\, \left(-abe^{-bt}\right)\, -\, \left(abe^{-bt}\, -\, ab^2te^{-bt}\right)\, =\, ab^2te^{-bt}\, -\, 2abe^{-bt}\)

. . . . .\(\displaystyle ab^2te^{-bt}\, -\, 2abe^{-bt}\, =\, 0\)

. . . . .\(\displaystyle \mbox{We have }\, a\, \neq\, 0,\, b\, \neq\, 0,\, \mbox{ and }\, e^{-bt}\, \neq\, 0,\, \mbox{ so:}\)

. . . . .\(\displaystyle b^2t\, -\, 2b\, =\, 0\)

. . . . .\(\displaystyle b(bt\, -\, 2)\, =\, 0\)

. . . . .\(\displaystyle bt\, -\, 2\, =\, 0\)

. . . . .\(\displaystyle t\, =\, \dfrac{2}{b}\)

Now im not sure how i plug this back into the equation to get the y coordinates? this is the part im stuck on.

This is where having a proper functional statement is helpful. You have found the inputs, being the t-coordinates. Now plug those into the function in order to find the corresponding outputs, being the y- (or f-) coordinates.

. . . . .\(\displaystyle \mbox{turning point of }\, f(t)\, =\, ate^{-bt}\, \mbox{ at }\, t\, =\, \dfrac{1}{b}\, \mbox{ is given by:}\)

. . . . .\(\displaystyle f\left(\dfrac{1}{b}\right)\, =\, (a)\left(\dfrac{1}{b}\right)e^{-b\, \cdot\, \dfrac{1}{b}}\, =\, \dfrac{a}{b}\, e^{-\dfrac{b}{b}}\, =\, \dfrac{a}{be}\)

The same process is required for the remaining portion of the exercise.

 

[kais]

Thanks a lot stapel, right what i wanted to know

on a side note how do your write your formulas like that, so in the future if i need a question is dosent look so messy?