- This topic has 5 replies, 4 voices, and was last updated 2 years, 10 months ago by .
You must be logged in to reply to this topic.
Viewing 6 posts - 1 through 6 (of 6 total)
You must be logged in to reply to this topic.
You must be logged in to reply to this topic.
Leader: Randy Medina
Abdullah Zafar
Julia Chang
Henry Lam
Abdullah Saleh
Ahmed Taher
Joel Silva
We’ll be discussing the problems of problem 2 to agree upon a response
Alright so what I’ve understood from these problems was that for
a) The tangent line to the graph at the lowest point of the dive is that it’s located between the interval of 4 and 6 and that the lowest point stays constant for a bit before the submarine goes upward.
b) To find how many minutes till the submarine reaches its lowest point just substitute the t with the numbers between 4 and 6, as it shows it looks like the lowest point is near 5, so I divided each minute into 4 segments (5.25- for 5 minutes and 15 seconds, 5.5 for 5 min and 30 sec etc.) and I got 5.25 as the lowest point: a(5.25)=500cos(5.25/2)+125(5.25)-564, a(5.25)=-434.75+656.25-564, a(5.25)=-434.75+92.25=-342.5, while a(5)=-339.57 and a(5.5)=-338.65 which both are less than the value of a(5.25)
c) To find the greatest depth of which there submarine traveled we need to know at which time it traveled the deepest which was solved at the previous question (b), a(5.25)= -342.5 rounded to the nearest foot is -343ft.
d) I don’t understand this question, from what I can assume from reading the question I’m thinking it has to do with instantaneous rate.
If ya disagree let me know where you think I’ve messed up
I agree with your answers. I got similar answers
It’s Jonathan Camacho
My name is not posted, but i would still like to join the group. I also didn’t understand part d you included, but I also got around the same answers as you did.
Sure thing jonathan you can join the group, and if anyone wants to contact me, I’m going to try and be more frequent here, but in case I’m not my personal email is randym0203@gmail.com, also I still haven’t looked up the answer for number 4, I’ll look into it tonight and post anything I was able to understand. I’ll review my answers from the notes and research and update anything that needs clarification,
C) What i did was get the value from problem B and plugged it in for t.
a(t)=500cos(t/2)+125t-564
a(5.24)=500COS(t/2)+125t-564
a(5.24)=500COS(5.24/2)+125(5.24)-564
a(5.24)= -433.513+655-564
a(5.24)=-342.513
D.I think i got the answer for the last one.
Basically what i did was 0-(342.513)/7.634-5.24 = 50.1067 Rate of change
How i got 0, i used the original formula a(t)=500cos(t/2)+125t-564 and finding the t value.
You must be logged in to reply to this topic.
New York City College of Technology, C.U.N.Y
300 Jay Street, Library Building - 4th Floor
Our goal is to make the OpenLab accessible for all users.