MATH1120: Mathematics for Engineering, Science and Technology 2

1. Let \(Q(t)\) be the amount (in milligrams) of thorium \(234\) present at time \(t\) (in days). Then

   \begin{equation*} \dfrac{dQ}{dt} = -kQ\text{,} \end{equation*}
   where \(k\) is a positive constant and
   \begin{equation*} Q(0) = 100, Q(7) = 82.04\text{.} \end{equation*}
   The DE is separable but if we write it as
   \begin{equation*} \dfrac{dQ}{dt} + kQ = 0\text{,} \end{equation*}
   then we see that it is also linear. The integrating factor is
   \begin{equation*} I(t) = e^{\int k \, dt} = e^{kt}\text{.} \end{equation*}
   Multiplying both sides by this integrating factor gives
   \begin{equation*} \dfrac{d}{dt} \left( e^{kt} Q \right) = 0\text{,} \end{equation*}
   and hence the general solution is
   \begin{equation*} Q(t) = Ae^{-kt}\text{.} \end{equation*}
   Using \(Q(0) = 100\) gives \(A = 100\) and from \(Q(7) = 82.04\) we obtain
   \begin{equation*} 82.04 = 100 e^{7k} \end{equation*}
   from which
   \begin{equation*} k = \dfrac{\ln (0.8204)}{7} = 0.0283 \: (\text{ to 4 dp})\text{.} \end{equation*}
   Thus
   \begin{equation*} Q(t) = 100 e^{-0.0283t}\text{.} \end{equation*}

2. The time when the amount of isotope has decayed to \(50\) milligrams is the value of \(t\) that satifies

   \begin{equation*} 50 = 100 e^{-kt}\text{,} \end{equation*}
   which is
   \begin{equation*} t = \dfrac{\ln(0.5)}{-k} = \dfrac{\ln(2)}{k} \approx 24.5 \, \text{ days}\text{.} \end{equation*}

Let \(C(t)\) be the concentration (in g/L) of the salt in the water in the tank at time \(t\) minutes after the mixing began and let \(A(t)\) be the amount (in grams) of salt in the tank at time \(t\text{.}\) Then \(C(0) = 0\) and \(A(0) = 0\text{.}\)

Now, salt is coming into the tank at the rate

and it is leaving the tank at the rate

Thus \(A(t)\) will satisfy the DE

Using \(A(t) = 200 C(t)\) this becomes

which can be written as

This is a first order linear DE and hence can be solved by multiplying the equation by the integrating factor

In doing this we obtain

and hence

Thus the general solution is

Using \(C(0) = 0\) gives \(K = -25\) and so the required concentration function is

Figure 11.3.3 shows a graph of this function. It can be seen that as \(t \to \infty\text{,}\) \(C \to 25\text{.}\) Thus as time goes by the concentration of salt in the tank water approaches the concentration of the salt in the brackish water coming into the tank.

Let \(v(t)\) be the velocity of the ball at time \(t\) seconds after it was fired from the cannon and let the downward direction be the positive direction as shown in Figure 11.3.5.

The motion of the ball will be governed by Newton's second law of motion which says

where \(F\) is the net force acting on the ball, \(m\) is its mass and \(a\) is its acceleration. Once the ball leaves the cannon the forces acting on the ball are gravity, which we will denote by \(F_G\) and air resistance, denoted by \(F_R\text{.}\) Now with the positive direction as shown in Figure 11.3.5,

where \(g\) is the acceleration due to gravity. In the initial phase where the ball is going up the air resistance will be acting against the motion, (i.e. in the positive direction). Thus, since the velocity \(v\) is negative

In the second phase where the ball is falling the air resistance will again be acting against the motion (i.e. in the negative direction). In this phase the velocity is positive and so once again

Thus the net force acting on the ball is

Since

Newton's second law gives

or

with

This is a first order linear DE in \(v(t)\text{.}\) To solve, multiply both sides of the DE by the integrating factor

to get

Integrating both sides gives

and hence, using \(m=50\text{,}\) \(g=9.8\) and \(v(0) = -10\text{,}\)

Thus we have determined the velocity function for the ball. To use this to determine the speed of the ball when it hits the ground we must first determine the time at which the ball hits the ground. To do this we will need the displacement function which we can obtain by integrating the velocity function. So, letting \(s(t)\) be the displacement of the ball at time \(t\text{,}\) we have

Using \(s(0)=0\) gives \(K=-108\) and so the displacement function is

Now \(s(t) = 100\) when

Figure 11.3.6 shows a plot of the function

and as can be seen there is only one positive solution. Solving this equation numerically (using Newton’s method for example), shows this solution to be

Thus the velocity of the ball when it hits the ground will be