1 / 5

Diving problem Math 1475 Prof. Mingla

Diving problem Math 1475 Prof. Mingla. Group 3 Randy Medina Abdullah Zafar Julia Chang Henry Lam Ahmed Taher Joel Silva Jonathan Camacho. What can you say about the tangent line to the graph at the lowest point?.

benjamin
Download Presentation

Diving problem Math 1475 Prof. Mingla

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Diving problem Math 1475 Prof. Mingla Group 3 Randy Medina Abdullah Zafar Julia Chang Henry Lam Ahmed Taher Joel Silva Jonathan Camacho

  2. What can you say about the tangent line to the graph at the lowest point? The tangent line at the lowest point of the graph had a constant slope as the submarine slowed down transitioning into going upwards. Also, the tangent line to the lowest point will be horizontal making the slope equal to 0.

  3. B. After how many minutes will the submarine reach the lowest point on the dive? a(t)=500cos(t/2)+125t-564 => a(t’)=-250sin(t/2)+125 -250sin(t/2)+125=0 => -250sin(t/2)=-125 => sin(t/2)=(-125/-250) ½=sin(t/2) => sin^-1(½)=pi/6, 1/2sin=pi/6 & 10pi/6 5pi/6*2=10pi/6=5.24

  4. C. What is the greatest depth the submarine will reach on this dive? Round to nearest foot a(5.24)=500cos(5.24/2)+125(5.24)-564 a(5.24)=-433.51+655-564 a(5.24)=-342.51ft a(5.24)=-343ft

  5. D. What is the greatest rate at which the submarine’s depth will increase during the dive. Find the 2nd derivative, to find the acceleration. a(t)=125cos(t/2) a(0)=125cos(0/2) a(0)=125cos(0) a(0)=125

More Related