y = 2xe x + x 2 e x at (0, 3). solution: Since y is implicitly related to x we have to use implicit differentiation: 3 6y = 0 y = 1 2 x ln(b) ln(b)

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Download "y = 2xe x + x 2 e x at (0, 3). solution: Since y is implicitly related to x we have to use implicit differentiation: 3 6y = 0 y = 1 2 x ln(b) ln(b)"

Bethany Houston
5 years ago
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1 4. y = y = + 5. Find th quation of th tangnt lin for th function y = ( + ) 3 whn = 0. solution: First not that whn = 0, y = (1 + 1) 3 = 8, so th lin gos through (0, 8) and thrfor its y-intrcpt is 8. y = 3( + ) ( + ) y (0) = 3() ( 1 + ) = 1 Th slop of th lin is 1, and abov w found th y-intrcpt to b 8. Thus th quation w nd is y = Givn y + + y = 8, find dy d at (0, 3). solution: Sinc y is implicitly rlatd to w hav to us implicit diffrntiation: Plugging in th point (0, 3) givs us y (y + y) + yy = 0 1(0 y + 3) + (0) 6y = 0 3 6y = 0 y = 1 W will mostly b intrstd in y =, but w can considr othr bass. Thorm 1.1 Lt b > 0, b 1. If y = b, thn y = b ln(b). proof: W can rwrit b : b = [ ln(b)] = ln(b) So by th chain rul y = ln(b) ln(b) = [ ln(b)] ln(b) = b ln(b) This complts th proof. Eampls: 1. y = 3 5 y = ln(3)3 5 (5 4 )

2 . y = 54 y = ln(5)(54 )4 5 4 ( 1 1/) = 54 (8 ln(5) 1) 3/ 4.5 Drivativs of Logarithms Sinc w know th drivativ of ponntial functions, it s natural to want to know th drivativ of logarithms. W can find ths drivativs asily using implicit diffrntiation. Suppos y = ln(). Thn by th ruls of logs = y. Using implicit diffrntiation on this givs us 1 = y y But y =, so y = 1. This dsrvs a bo: y = 1 y If f() = ln(), thn f () = 1 Eampls: 1. Lt y = ln(). Thn y = ln() + 1 = + ln(). Lt y = ln(3 + 7). Thn y = 1 (6 + 7) Lt y = ln( 3 + 7). Thn w can tak th drivativ dirctly or w can notic first that w can simplify this with th laws of logs: y = ln( 3 + 7) = ln(( 3 + 7) 1/ ) = 1 ln(3 + 7) So that y = (3 + 7) = Suppos y = ln(). Find th quation of th tangnt lin at =.

3 solution: Not that whn =, y = 1. W tak th drivativ so that w can find th slop: ) ln()(1) y = ( 1 = 1 ln() Plugging in = givs us 1 ln() = 0. So th slop is 0, and our quation so far is y = 0 + b. Plugging in th point (, ) 1 givs us b = 1, so th quation of th tangnt lin is y = Eponntial Growth and Dcay Many naturally occuring systms involv a quantity whos rat of chang is proportional to how much of that quantity is prsnt. Bactrial growth is a prfct ampl - th mor bactria thr ar in a colony, th fastr th colony grows du to cll division. Cass lik this ar modlld by th quation dy dt = ky That is, th rat of chang of y is proportional to y (hr k is a fid constant). In this cas, it can b shown that y has th form y = C kt Whr C is th initial valu and k is th constant of proportionality. If k > 0, w say that y grows ponntially and if k < 0 w say that y dcays ponntially. Not that th rason w say C is th initial valu is that whn t = 0, y = C 0 = C. Also not that in othr placs (spcifically som of th old midtrms) you will s y = Cb t for som b instad. Th two ar quivalnt if you tak b = k. Eampls: 1. Our formula for continuous compounding A(t) = P rt is a typical ampl of ponntial growth - at all tims th rat of chang of A(t) is proportional to th intrst rat: da dt = ra. Th numbr of bactria in a crtain cultur is 160,000 at 1pm. At 3pm, th count is 30,000. Assuming th population is growing ponntially, find th population at 7pm.

4 solution: Sinc w ar assuming ponntial growth, w us th quation P (t) = P 0 kt whr P 0 is th initial population. Taking 1pm to b t = 0, P 0 = 160,000. To b abl to us this formula w nd to find k, and to do this w us th fact that at t =, P (t) = 30,000: P (t) = kt P () = k = k = k = ln() k = ln() So P (t) = ln() t. At 7pm, t = 6, so P (6) = ln() = 1,80, Radioactiv substancs dcay ponntially, maning thir mass dcrass in an ponntial fashion. Th half-lif of a radioactiv substanc is th tim it taks for its siz to b half of th initial amount. Th half-lif of a crtain radioactiv substanc is 5 yars. If 0 grams wr prsnt in 1985, how much will b lft in 05? solution: W again assum that P (t) = P 0 kt. A half-lif of 5 yars mans that whn t = 5, P (t) will b half of th initial amount, P 0 : P 0 = P (5) = P 0 5k 1 = 5k ln( 1 ) = 5k k = ln( 1 ) 5 = ln() 5 If w tak 1985 as t = 0, thn P 0 = 0. Thus W want P (40) = 0 ln() grams. P (t) = 0 ln() 5 t

5 4. Th numbr of rabbits on an island grows ponntially and taks 1 yar to doubl. How long dos it tak to tripl? solution: W again assum that P (t) = P 0 kt. W ar givn that at t = 1, P (t) will b doubl th initial population, i P 0. Thus P 0 = P 0 k = k k = ln() Thus P (t) = P 0 ln()t. W want t such that P (t) = 3P 0. Thus w solv 3P 0 = P 0 ln()t 3 = ln()t ln(3) = ln()t t = ln(3) ln() 1.58y

Where k is either given or determined from the data and c is an arbitrary constant.

Where k is either given or determined from the data and c is an arbitrary constant. Exponntial growth and dcay applications W wish to solv an quation that has a drivativ. dy ky k > dx This quation says that th rat of chang of th function is proportional to th function. Th solution is