Download presentation

Presentation is loading. Please wait.

Published byParker Surrey Modified over 3 years ago

1
Exponential Decay Exponential Radioactive Decay

2
8/14/2013 Exponential Applications 2 Exponential Decay Radioactive Decay Radioactive isotopes of some elements such as 14 C, 16 N, 238 U, etc., decay spontaneously into more stable forms ( 12 C, 14 N, 236 U, 232 U, etc.) Decay times range from a few microseconds to thousands of years Decay rate proportional to amount present Same proportion decays in equal time

3
8/14/2013 Exponential Applications 3 Exponential Decay Radioactive Decay Decay measurement Might not measure whole decay time Can measure limited decay, then calculate half-life Half-life = time for decay to half original measured amount Model with exponential function

4
8/14/2013 Exponential Applications 4 Exponential Decay Radioactive Decay (continued) C A A A(3k) = (½)A(2k), …, A(nk) = A 0 (½) n, … = ( A 0 (½) ) (½) = A 0 (½) 2 = (A 0 (½) 2 )(½) = A 0 (½) 3 A(4k) = A 0 (½) 4 A(2k) = (½)A(k) Initial amount of radioactive substance Q is A 0 and decays to A(x) after x years After half-life of k years, A(k) = ½A 0 … or just A 0 (½) After 2k years amount is

5
8/14/2013 Exponential Applications 5 Exponential Decay Radioactive Decay (continued) A, …, A(nk) = A 0 ( ½ ) n, … A(4k) = A 0 ( ½ ) 4 After n half-lives, x = nk so the amount A(x) = A(nk) = A 0 ( ½ ) n Since x = nk, then n = x/k and A(x) = A 0 ( ½ ) x/k left is

6
8/14/2013 Exponential Applications 6 Exponential Decay Radioactive Decay (continued) A life can be modeled by an exponential function: Initial amount is f(0) = C and, for half-life k, = C((½) 1/k ) x = C(½) x/k a = (½) 1/k f(x) = C a x f(k) = (½)C = C a k Dividing out constant C gives: a k = ½ Solving for a, we get and thus f(x) = C a x Question: What does this look like graphically ? Alternate View: just recognize that half-

7
8/14/2013 Exponential Applications 7 Exponential Decay Radioactive Decay Graph A0A0 A0A0 1 2 A0A0 1 4 A(t) t k2k3k 4k5k6k 0 A(t) = amount at A 0 = initial amount k = half-life A(t) = A 0 (½) t/k Since n = t/k A(t) = A 0 (½) n = A(0) A(t) reduced by half in each half-life After n half-lives t = nk A(t) = A 0 (½) t/k A0A0 1 8 A0A0 1 16 A0A0 1 32 time t

8
8/14/2013 Exponential Applications 8 Think about it !

9
8/14/2013 Exponential Applications 9 Solving Equations: Examples 1. World population 195019601970198019902000 20102020203020402050 15 14 13 12 11 10 9 8 7 6 5 4 3 2 t P(t) 1950 2.5098 1960 3.0000 1970 3.5859 1980 4.2862 1990 5.1233 2000 6.1239 2010 7.3199 2020 8.7495 2030 10.458 2040 12.500 2050 14.942 t P( t) P1P1 P2P2 P3P3 ( x 10 9 ) P(t) = 3(1.018) t–1960

10
8/14/2013 Exponential Applications 10 Spare Parts Slide 2 2 | | | ± ½ ½ ½ ¼ ¼ ¼

Similar presentations

OK

Objectives: I will be able to… Graph exponential growth/decay functions. Determine an exponential function based on 2 points Solve real life problems.

Objectives: I will be able to… Graph exponential growth/decay functions. Determine an exponential function based on 2 points Solve real life problems.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on fans and blowers Ppt on cash reserve ratio Ppt on say no to drugs Ppt on telephone exchange in indian railways Ppt on spices industry Ppt on any topic of maths for class 9 Ppt on computer software and languages Download ppt on classification of animals Ppt on electricity for class 10th board Ppt on current account deficit greece