Tuesday, December 9, 2014

Single Variable Calculus, Chapter 7, Review Exercises, Section Review Exercises, Problem 62

Suppose that the function c(t)=k(eatebt)

a.) Show that limtc(t)=0


limtc(t)=limtk(eatebt)=limtk(1eat1ebt)=k(1ea()1eb())=k(11)=k(00)=0



b.) Find c(t)


If c(t)=k(eatebt) thenc(t)=k(eat(a)ebt(b))c(t)=k(bebtaeat)


c.) When is the rate equal to 0?

When c(t)=0


0=k(bebtaeat)0=bebtaeataeat=bebtab=ebteatab=et(b(a))ab=et(ab)


If we take the natural logarithm, we have


ln(ab)=t(ab)(lne)ln(ab)=t(ab)(1)t=ln(ab)ab

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