Mathematics OpenStudy (anonymous):

The exponential decay law dy dt = −λy, λ > 0 may be used to model decay of a radioactive isotope, or any other quantity that is decreasing at a rate proportional to its own magnitude. This is just the exponential growth model with a negative growth rate k = −λ. As is often done, I have put the sign into the differential equation so that the parameter λ is positive. In this question take the unit of time to be years. λ is called the decay rate. (a) By substitution into the differential equation verify that (as expected) y(t) = ce−λt is a solution of the exponential decay law for any constant c. OpenStudy (anonymous):

The problem says to show by substitution, so just plug $y(t)=Ce^{-\lambda t}$ and its derivative into the differential equation $\frac{dy}{dt}=-\lambda y$ and show that it is true. OpenStudy (anonymous):

the part of the question i needed help with got cut off .. idk if you'll still be able to help (b) At what time T (expressed in terms of λ) will y(t) be half of its initial value y(0) = c? T is called the half life of the substance. (c) For the solution obtained in (a), show that y(t0 + T) = 1 2y(t0), no matter what t0 is. In other words y(t) decreases by half its magnitude over any time interval of length T. (d) If T = 6000 years, what is λ (expressed as a decimal)? OpenStudy (anonymous):

For (b), since you know y(0)=C and y(t)=Ce^(-lambda * t), then y(t) will be half that value when y(t)=C/2. Thus, just solve $\frac{C}{2}=Ce^{-\lambda T}$ for T. For (c), you are just showing that $y(t_0+T)=\frac{1}{2}y(t_0)$ which you can solve by just using the value for T calculated in part (b). For (d), since you know that $\frac{1}{2}=e^{-\lambda \cdot 6000}$ (as T=6000), just solve for lambda. OpenStudy (anonymous):

in b when solving for T, you take ln of both sides right? but on the right hand side of the equation do you include C in the ln calculation of do you pull it out infront? OpenStudy (anonymous):

Just start off by dividing both sides by C, then there is no need to worry about it OpenStudy (anonymous):

oh alright, thanks! OpenStudy (anonymous):

so the answer would be -ln(1/2) over lambda? OpenStudy (anonymous): OpenStudy (anonymous):

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