Recent content by Coltjb7

Sorry, I just totally avoided the sec somehow... du/2cos^2u =1/2 * 1/cosu * du/cosu =1/2 * sec * sec *du (1/2) sec^2u du 1/2 tanu du 1/2 tan(3x^2+2x)(3x+1)

So I did this two different ways: A) du/2cos^2u =1/2 * 1/tanu *du/tanu =du/2tan^2u =(3x+1)/2tan^2(3x^2+2x) OR B) 1/2 * 1/tanu *du/tanu =1/2 *cotu *du/tanu =cot(du)/2tan(u) =cot(3x+1)/2tan(3x^2+2x) Which one if wither is correct? Why?

Homework Statement Find the antiderivative: ∫[(3x+1)/cos^2(3x^2+2x)]dx Homework EquationsThe Attempt at a Solution I attempted to use "u substitution" but got stuck towards the end: u=3x^2+2x du=6x+2dx =2(3x+1)dx du/2=(3x+1)dx After my substitutions it looks like this: ∫(du)/2cos^2(u) =...

I think I have got it. Thank you guys very much!

Ohhh. Okay. Thank you. I was going about it wrong and set t=pi/2 and 3pi/2 then plugged that value into H'(t). Now I realize that I made a stupid mistake. I worked out those values and came up with: When pi/2=0.017214t-1.369247 Then following the same premise for 3pi/2...

So now i solved H'(t) for zero when t=pi/2 and 3pi/2. These are my new answers: when t=pi/2 there is a critical value at 108.7306 When t=3pi/2 there is a critical value at 36.24. So, is day number 108.7306 (do i round to day 109 or keep it at 108?) the day where the most daylight is being...

The original function. H(t)=3.3sin(0.017214t-1.369247)+12

See that's what I am not understanding. I did nothing with pi/2. I ignored that information and solved it using cos^-1

I understand I need to look at the derivative but how do I find a specific date where it is increasing/decreasing the most. She said we could use our calculators to do it, but I cannot remember how to do this on a calculator. I have the TI-84 Plus.

I honestly am not sure how to find the other solution. Mind helping me out? I would be using the intervals for my first derivative test. our teacher has us create a table like this: Interval ( we gather this from the critical values) Test Value (any number you want within each interval) Sign of...

So would my intervals go from (-infinity,170.79) and (170.79, infinity)?

Haha, yes I know about them but have not used them in a while. Would I do this: 0=cos(0.017214t-1.369247) cos^-1(0)=0.017214t-1.369247 [(cos^-1(0))+1.369247]/0.017214=t t=170.79

so what I have been trying to do is this: H'(t)=0.0568062cos(0.017214t-1.369247) Since anything multiplied by 0 is equal to zero I ignored the constant leaving me with this: 0=cos(0.017214t-1.369247). I know that cos=0 at pi/2 and 3pi/2 but am having trouble figuring out how to work this out.

It would tell me on what intervals the equation is increasing/decreasing. Honestly I am having trouble finding the Critical Values. Thank you for the welcome

Homework Statement My teacher created this following problem: The equation H(t)=3.3sin(0.017214t-1.369247)+12 estimates the number of daylight hours in Portland, OR for any day, t, during 2016. 1. Give the date in 2016 when the number of daylight hours is increasing the fastest. How many...