Using the Increasing/Decreasing Test HELP

[AnnaSuxCalc]

Using the Increasing/Decreasing Test **HELP**

Homework Statement

I have to find the max number of bacteria between t=4 and t=8.5 using the Increasing/Decreasing Test:
(a) Find the largest open intervals (a,b) within the original time interval (4,8.5) during which N'(t)>0 for all t belonging to (a,b) --> during these, N(t) is increasing.
(b) Find the largest open intervals (c,d) within the original time interval (4,8.5) during which N'(t)<0 for all t belonging to (c,d) --> here N(t) is decreasing.
(c) Use (a) & (b) to find the time at which there will be the max number of bacteria.
(d) Use part (c) to find the max number of bacteria.

Homework Equations

N(t) = 2 + sin²t - sint, for 4<t<8.5

The Attempt at a Solution

OKAY soooo...
First of all I did the derivative of N(t):
N'(t) = sin2t - cos t

Now, I know that for the Increasing/Decreasing Test:
--> if N'(t) is positive on the interval N(t) is increasing there
--> if N'(t) is negative on the interval N(t) is decreasing

My problem seems to always be that I don't know what the question is really asking for, I do not understand (a) & (b)...do I have to find the Vertex here?!??!
I know that once I find (c), the time (t) I just plug that into find N(t), I DON'T plug it into N'(t) right?!
PLEASE HELP with (a) and (b)

 

[Mark44]

Homework Statement

I have to find the max number of bacteria between t=4 and t=8.5 using the Increasing/Decreasing Test:
(a) Find the largest open intervals (a,b) within the original time interval (4,8.5) during which N'(t)>0 for all t belonging to (a,b) --> during these, N(t) is increasing.
(b) Find the largest open intervals (c,d) within the original time interval (4,8.5) during which N'(t)<0 for all t belonging to (c,d) --> here N(t) is decreasing.
(c) Use (a) & (b) to find the time at which there will be the max number of bacteria.
(d) Use part (c) to find the max number of bacteria.

Homework Equations

N(t) = 2 + sin²t - sint, for 4<t<8.5

The Attempt at a Solution

OKAY soooo...
First of all I did the derivative of N(t):
N'(t) = sin2t - cos t

Now, I know that for the Increasing/Decreasing Test:
--> if N'(t) is positive on the interval N(t) is increasing there
--> if N'(t) is negative on the interval N(t) is decreasing

My problem seems to always be that I don't know what the question is really asking for, I do not understand (a) & (b)...do I have to find the Vertex here?!??!
I know that once I find (c), the time (t) I just plug that into find N(t), I DON'T plug it into N'(t) right?!
PLEASE HELP with (a) and (b)

You've already found the derivative, N'(t). For a), find any open intervals in (4.5, 8) where N'(t) > 0. IOW, solve the inequality sin(2t) - cos(t) > 0.
For b, it's almost the some, but you're solving the inequality sin(2t) - cos(t) < 0.
There's not really increasing and decreasing tests. All you're doing is finding the intervals where the function's derivative is positive (where the function is increasing) and where the derivative is negative (where the function is decreasing). If the derivative is positive, the slope of the tangent line is positive, so the graph is going up.

For c, if you know that N'(t) > 0 for (4.5, 6), and N'(t) < for (6, 8), then you know that N was increasing on (4.5, 6) and then decreasing on (6, 8), so must have reached a high point at t = 6. I'm just making these numbers up as examples.

For d, substitute the value of t you found into the function N (not the derivative). If you substituted that value of t into the derivative, you should get a value of 0.

 

[AnnaSuxCalc]

ok so for part (a) & (b):
sin2t - cost > 0
2sintcost - cost > 0
cost (2sint - 1) > 0
therefore:
cost > 0 OR 2sint > 1 --> sint > 1/2
soooo:
t > π/6, π/2, 5π/6, 3π/2
?
:uhh:

 

[Mark44]

Keep in mind you're working with inequalities, not equations.
cost(2sint - 1) > 0 ==> both factors are positive OR both factors are negative
If both factors are positive, then cos t > 0 AND sin t > 1/2
What's the solution set? Think intervals.

If both factors are negative, then cos t < 0 AND sin t < 1/2
What's the solution set? Again, think intervals.

For both of the preceding it would be helpful to look at the graphs of sin x and cos x.

 

[AnnaSuxCalc]

Keep in mind you're working with inequalities, not equations.
cost(2sint - 1) > 0 ==> both factors are positive OR both factors are negative
If both factors are positive, then cos t > 0 AND sin t > 1/2
What's the solution set? Think intervals.

If both factors are negative, then cos t < 0 AND sin t < 1/2
What's the solution set? Again, think intervals.

For both of the preceding it would be helpful to look at the graphs of sin x and cos x.

Uuhm...okay...I already drew out the graphs of the two curves..and it's not occurring to me:grumpy:!
for cos t > π/2 & 3π/2 --> cost is decreasing from π/2 to π and increasing from π to 3π/2
for sin t >π/6 & 5π/6 --> sint is increasing from π/6 to π and decreasing from π to 5π/6

omg I'm sooo missing something here

 

[jgens]

Put your answers in terms of intervals like Mark said. Then you need to consider the domain of time given, namely (4, 8.5), deducing on what intervals the functions are either greater or less than zero within the domain (4, 8.5)

 

[Mark44]

Not to further complicate your life, but you're only interested in values of t between 4 and 8.5.

for cos t > π/2 & 3π/2 --> cost is decreasing from π/2 to π and increasing from π to 3π/2
for sin t >π/6 & 5π/6 --> sint is increasing from π/6 to π and decreasing from π to 5π/6

No, cos t is decreasing on (0, pi) and is increasing on (pi, 2*pi), right? Decreasing means the graph is heading down, increasing means it's heading up. Decreasing doesn't mean that the graph is below the horizontal axis, and increasing doesn't mean that the graph is above the hor. axis. I think you might be confusing those.
(I don't know what "cos t > pi/2 & 3pi/2" means!)
Also, cos t is decreasing on (2pi, 3pi) and on (4pi, 5pi), and on (6pi, 7pi), etc.
sin t > 1/2 on (pi/6, 5pi/6) and on (2pi + pi/6, 2pi + 5pi/6) and on (4pi + pi/6, 4pi + 5pi/6), etc.
sin t < 1/2 on (0, pi/6) and on (5pi/6, pi) and on (2pi + 0, 2pi + pi/6) and on (2pi + 5pi/6, 3pi) etc.

 

[AnnaSuxCalc]

oh wow, I seem to be too stupid for this one today, I think I'll try some more tomorrow
I think what might be throwing me off a bit might be that on the x-axis for the curves the values are $$\pi$$ /2 etc and not just real numbers like (4, 8.5) lol ya...its been a couple of years since high school calculus I know this stuff is probably elementary :shy:

 

[AnnaSuxCalc]

Keep in mind you're working with inequalities, not equations.
cost(2sint - 1) > 0 ==> both factors are positive OR both factors are negative
If both factors are positive, then cos t > 0 AND sin t > 1/2
What's the solution set? Think intervals.

If both factors are negative, then cos t < 0 AND sin t < 1/2
What's the solution set? Again, think intervals.

For both of the preceding it would be helpful to look at the graphs of sin x and cos x.

Okay so I'm still trying to solve/understand this question...
Just to summarize, this is ALL I have (sadly):
sin2t - cost > 0
2sintcost - cost > 0
cost(2sint - 1) > 0
therefore --> cost >0 AND sint>1/2

ALSO:
sin2t - cost < 0
2sintcost - cost < 0
cost(2sint - 1) < 0
therefore --> cost < 0 AND sint < 1/2

I know that:
cos^-1(0) = $$\pi$$/2 and 3$$\pi$$/2
AND
sin^-1(1/2) = $$\pi/6$$ and 5$$\pi$$/6

Okay so...Mark, when you say "If both factors are positive, then cos t > 0 AND sin t > 1/2" what do you mean by 'factors'?
ALSO, when you said that (for example) cos is increasing on 2$$\pi$$ to 3$$\pi$$ but when I look at my graph starting from 2$$\pi$$ the graph goes down until 5$$\pi$$/3 where it goes up until 3$$\pi$$ ... I really NEED to understand this very soon

(I don't know why the $$\pi$$ symbol looks like a 'power' but in no case is it meant to the power/exponent of any number)

 

[Mark44]

Okay so I'm still trying to solve/understand this question...
Just to summarize, this is ALL I have (sadly):
sin2t - cost > 0
2sintcost - cost > 0
cost(2sint - 1) > 0
therefore --> cost >0 AND sint>1/2

That's only part of the story. If two factors--expressions that are multiplied together -- produce a positive product, then both of them must be positive OR both of them must be negative. E.g., (-2)(-3) = +6.
So if cost(2sint - 1) > 0, then
cost > 0 and 2sint - 1 > 0
OR
cost < 0 and 2sint - 1 < 0

ALSO:
sin2t - cost < 0
2sintcost - cost < 0
cost(2sint - 1) < 0
therefore --> cost < 0 AND sint < 1/2

I know that:
cos^-1(0) = $$\pi$$/2 and 3$$\pi$$/2
AND
sin^-1(1/2) = $$\pi/6$$ and 5$$\pi$$/6

Okay so...Mark, when you say "If both factors are positive, then cos t > 0 AND sin t > 1/2" what do you mean by 'factors'?

See above.

ALSO, when you said that (for example) cos is increasing on 2$$\pi$$ to 3$$\pi$$

You're misquoting me. I said that cos t is decreasing on (2pi, 3pi), not increasing. Please reread what I wrote in post 7.

but when I look at my graph starting from 2$$\pi$$ the graph goes down until 5$$\pi$$/3 where it goes up until 3$$\pi$$ ... I really NEED to understand this very soon

(I don't know why the $$\pi$$ symbol looks like a 'power' but in no case is it meant to the power/exponent of any number)

 

[Mark44]

I'm starting to think that the reason you're having such a hard time in calculus is that you had a hard time in the classes before it. Calculus builds on the classes that come before it, so if you didn't do well in those foundational classes, you will be floundering in Calculus.

 

[AnnaSuxCalc]

I'm starting to think that the reason you're having such a hard time in calculus is that you had a hard time in the classes before it. Calculus builds on the classes that come before it, so if you didn't do well in those foundational classes, you will be floundering in Calculus.

I did well in my high school calculus and the calculus I am taking now is supposed to be an introduction, it is the only calculus I have to take so I just need to get through it now!
I understand that maybe you are getting frustrated helping me but what can I do, I do try these myself first, its not like I'm just trying to rely on you guys' help. :(