Math 20B Assignment 1 Due: April 13th, 2018 It is prefered that answers be expre
ID: 2886872 • Letter: M
Question
Math 20B Assignment 1 Due: April 13th, 2018 It is prefered that answers be expressed as exact values in the simplest terms. For example, 3 is preferred to elog3 , and both are preferred to either 2.718281.098612 or 2.999997; likewise, e2 is preferred to 7.389056. You should assume that all quantities given are exact values unless otherwise stated 0. I strongly recommend drilling integration by substitution until you are comfortable performing it. There are several online am Alpha Problem Generator. You do not need to submit anything for tools to help you practice, such as the problem 0. 1. Suppose that the mass of a sample of Uranium-238 was found to be changing at a rate of 2og 2)elin alion years A. By how much will the mass of the uranium change over the course of a billion years beginning at year 13.5 billion? B. Suppose that at time 13.5 billion years, the remaining mass of uranium was found to be 9/8 kg. How much uranium was present at time 0? 2. Compute the following indefinite integrals via substitution ???-20 3. Find the total area trapped between the curves sin(2x) and cos(x) from 0 to 2?. Note that the curves intersect at the x- coordinates ^, , and ^. Try sketching the region, but you need not submit your sketch , 4. Consider a sphere of radius 5, with a vertical cylinder of radius 4 drilled through its centre; the resulting solid is a ring A. Find the cross-sectional area of the solid as a function of height y above the centre of the sphere, and use this to find the volume of the solid. B. The same solid may be constructed as a volume of revolution. Find two curves so that when the area trapped between them is rotated about the y-axis, the result is the solid we are considering. Then verify that this approach yields the same volume. (Hint: one of the curves should be a vertical line.)Explanation / Answer
1.
We are given that the mass of Uranium is changing at rate of = -2(log(2))e^[(-t/45)*log(2)] kg/billion years
Here 't' is the time in billions of year
A. We need to find the change in the mass of Uranium over the course of a billion years starting from 13.5 billion
that is from t = 13.5 to t = 14.5
The the difference in time is = 1 billion year
=> In order to find the change in the mass of Uranium we plug , t = 1 in given rate
=> Change of mass = -2(log(2))e^[(-1/45)*log(2)]
= log(1/4)*e^[log[(2)^(-1/45)]
= log(1/4)*(2)^(-1/45)
= log(0.25)*(2)^(-1/45)
Hence the mass of Uranium changes by = log(0.25)*(2)^(-1/45) kg in 1 billion years time
B. We are given the rate of change of mass per billion year as :
dM/dt = -2(log(2))e^[(-t/45)*log(2)]
=> dM = -2(log(2))e^[(-t/45)*log(2)]dt
=> dM = -2(log(2))e^[log(2)^(-t/45)]dt
=> dM = [-2(log(2))*(2)^(-t/45)]dt
Integrating both the sides:
=> Integral dM = Integral [-2(log(2))*(2)^(-t/45)]dt
M = -2(log(2))*Integral [(2)^(-t/45)]dt
M = -2(log(2))*[(2)^(-t/45)]*[1/log(2)*(-1/45)] + C
or M = -2(log(2))*[(2)^(-t/45)]*[-45/log(2)] + C
M = -2[(2)^(-t/45)]*[-45] + C
M(t) = 45*(2)^(1-t/45) + C , C is the integration constant
Lets use the initial conditions which are given to us to find the value of 'C' : when t = 13.5 , M is = 9/8 kg
That is, M(13.5) = 9/8
=> M(t) = 9/8 = 45*(2)^(1-13.5/45) + C
=> C = - 71.9777
or C = - 72 (approx value)
Hence the mass function is :
M(t) = 45*(2)^(1-t/45) - 72 -------> (1)
Finally we need to find how much uranium was present at t = 0 years
Lets plug t = 0 in equation (1) and find M(0)
=> M(t) = 45*(2)^(1-0/45) - 72
M(t) = 18 kg
Hence at t = 0 the amount of uranium present is = 18 kg