Could someone please explain how to do these two problems? They are extremely li

Question

Could someone please explain how to do these two problems? They are extremely likely to be on my midterm tomorrow and I would greatly appreciate it if someone could explain how to do them!

4. Radium goes through radioactive decay, with a half-life of 1599 years. What percent of a present amount of radioactive radium will remain after 900 years? [Hint: You do not need to know what the initial amount is.]

5. Find the equation of the function f whose derivative is f'(x) = 6sqrt(x)-10 and whose graph passes through the point (4,2).

Explanation / Answer

4.

If N be the amount at time t and N0 be the initial amount, we have,

dN/dt = -kN where k is some positive constant called decay consatnt.

This gives, dN/N = -k dt

Integrating both the sides we get, ln N = -kt + C

At t = t_half, we have N = N0/2

Thus, ln N0/2 = -k*t_half + C

C = ln N0/2 + k*t_half

Thus, ln N = -kt + ln N0/2 + k*t_half...................1

At t= 0 we have N = N0

Thus, ln N0 = 0 + ln N0/2 + k*t_half

ln (N0 / (N0/2)) = k*t_half

ln 2 = k*t_half

k = (ln 2) / t_half

Thus, k = (ln 2) / 1599

Using eqn 1, ln N = ln (N0/2) + k(t_half - t)

ln (N/(N0/2)) = ((ln 2) / 1599) *(1599 - 900) = 0.303

2N/N0 = exp 0.303 = 1.354

N/N0 = 0.677 or 67.7 %.

Hence, after 900 years, 67.7 % will remain.

5.

f'(x) = 6 sqrt x - 10

Let y = f(x)

So, dy/dx = f'(x)

So, dy/dx = 6 sqrt x - 10

Integrating both the sides, y = 6*(1/2 + 1)*x^(1/2 + 1) + 10x + C

y = 6*(x^1.5)/1.5 + 10x + C

y = 4*x^1.5 + 10x + C.............1

As curve passes through (4,2) we get

2 = 4*4^1.5 + 10*4 + C

C = -70

Thus, from eqn 1 we get that

y = 4*x^1.5 + 10x - 70

Thus, f(x) = 4*x^1.5 + 10x - 70

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