## anonymous 5 years ago 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.

1. 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.

2. 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)?

3. 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.

4. 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?

5. anonymous

Just start off by dividing both sides by C, then there is no need to worry about it

6. anonymous

oh alright, thanks!

7. anonymous

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

8. anonymous