## 1.9 Problem Set

**General**

- An application of both the definite integral and exponential growth is modelling demand for natural resources. Assume \(P(t)\) represents the rate of use of a resource at time \(t\geq0\), and \(P(t)\) represents the rate of consumption at time \(t = 0\). The exponential model then gives \(P(t) = P_0 e^{kt}\). The constant \(k\) can be defined as the rate of increase in the use of this resource. Let \(A(T)\) represent the amount of resource used during the interval \([0,T]\), \(T\geq0\).

- Write the definite integral representing \(A(T)\) in terms of a general function \(P(t)\).
- Evaluate the integral for general \(T\geq0\) by substituting the above exponential function for \(P(t)\).
- In 1973 (say \(t = 0\)), the world use of copper was estimated to be 6.99 x 10
^{9}kg, and the demand was increasing at an exponential rate of 8% per year. How many tons of copper will then be used from 1973 to 1983? (Hint: First find \(P_0\), \(k\), \(T\) and then use your answer to b.) - If the rate of growth in demand for copper remains at 8% and no new reserves are discovered, when will the world supply of copper be exhausted? Assume the reserves in 1973 are 336 x 10
^{9}kg, and no recycling occurs.

- Five thousand trout, each weighing 80 grams, are planted in a lake. The population size, \(N(t)\), declines exponentially according to the equation \(N(t) = 5000e^{-0.5t}\) where \(t\) is measured in months. Each fish grows according to the formula \(W(t) = 10000 (1 - 0.8e^{-0.05t})^3\) where \(W\) is the weight in grams.

- Check that both \(N(t)\) and \(W(t)\) satisfy the given initial conditions.
- Write the formula for the biomass (total weight of all the planted trout) in the lake as a function of \(t\), and calculate the biomass at 12 months.
- What is the average biomass for the first 12 months? What was the initial biomass? Sketch how you think the graph of biomass versus time would appear. Mark on the vertical axis the initial, final and average biomass values.

- Assume that a radioactive element disintegrates so that the number (\(N\)) of atoms present at a given time (\(t\)) decreases at a rate proportional to \(N\) itself. Let \(k\) be the constant of proportionality and \(N_0\) be the number of atoms present at \(t = 0\) (thus \(N(0) = N_0\)).

- With the derivative representing the rate of change of \(N\), use the above defined parameters and variables to write an equation for the rate of decrease in the number of radioactive atoms with passing time.
- Solve this differential equation for \(N(t)\) so that the solution satisfies the initial condition, \(N(0) = N_0\). (Hint: first separate variables. Do not forget the integration constant!)
- Determine the “half-life” : the time (\(t_1\)) such that \(N(t_1) = 0.5 N_0\). At \(t =t_1\), half of the original number of radioactive atoms have disintegrated (decayed).
- We now make certain assumptions which will allow us to estimate the age of a piece of wood by radiocarbon dating:

- living organic material contains the carbon isotopes \(C^{12}\) and in a fixed proportion independent of time,
- the \(C^{12}\) atoms do not disintegrate,
- the \(C^{14}\) atoms disintegrate radioactively with a half-life of 5568 years, and
- the \(C^{12}\) and \(C^{14}\) atoms are not replaced once the organism dies, even though the \(C^{14}\) atoms are being lost through disintegration. A sample of wood in an American Indian cliff dwelling is measured to have 87% of the \(C^{14}\) isotope expected in living wood. Estimate the age of the sample.

- The survival and health of intertidal organisms often depend on the amount of time they are exposed, i.e. not covered by sea water. A given position on the beach can be identified by its tidal height: the height of the tide when the water’s edge just touches that spot. A model of tidal height as a function of time can then be used to approximate the length of time between successive high and low tides (or vice versa) during which a position on the beach is exposed.

The following data were taken from tide tables for Port Townsend, Washington for the day of maximum tide difference for daylight low tides.

Time | Height (m) |
---|---|

4:35 | 2.47 |

11:49 | -0.82 |

19:40 | 2.77 |

For each of the two times intervals 4:35-11:49, 11:49-19:40, the model uses the following cosine function fitted to the tide data: \[H = a \cos (b(T-T_h)) + H_a,\: H_a=\frac{high\:tide + low\:tide}{2}\]

Sketch by hand a graph of the cosine function with the maximum at the high tide level and a minimum at the low tide level for the first time interval to aid in interpreting the equation. Consider the general form of a cosine function: \(y=a \cos(bx+c)+d\), where the amplitude is \(|a|\); the period is \(2\pi/b\); the phase shift in \(-c/b\); and the vertical shift in \(d\). Use this information to determine what the constants a, b, \(T_h\) represent in the provides tidal cosine function? Calculate these constants and \(H_a\) for each of the two time intervals.

The constant \(H_a\) seems to be the average tide height. Verify this for the first time interval by evaluating the appropriate definite integral. Prove that the average height \(\overline H\) for the two intervals combined can be written as the weighted average of the average heights \(\overline H_1\), \(\overline H_2\) for both subintervals. That is, show that (using decimal hours)

\[\overline H= \frac{1}{15.09} \int^{19.67}_{4.58} H(t)dt= \frac{7.24}{15.09} \overline {H_1} + \frac{7.85}{15.09} \overline H_2\]

- For a point on the beach at the 0.0 meter tide level, how long is it exposed between these successive high tides? (Hint: find the inverse function T=g(H), but be careful with the second interval.)

**Least Squares**

Write equation (1.13) using \(y(x_i) = Ax_i + B\). Evaluate the two first partial derivatives and, by setting both equal to zero, obtain two linear equations in A and B.

- For the following data, derive the “least squares” line and prove that for your choices for A and B, the sum of squares is at a true minimum: \[f(x,y) \text{ is at a minimum at } x_o, y_o \text{ if, for } x=x_0, y=y_0 \]

- \(\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0\),
- \(\frac{\partial^2 f}{\partial x^2}-\frac{\partial^2 f}{\partial y^2}>0\), and
- \(\frac{\partial^2 f}{\partial^2 x}>0\).

Data from figure 1.8 in text

- Now determine a “best” exponential fit to the above data. First transform the desired function \(y = Axe^{Bx}\) into a linear function (with respect to \(x\)): \(h(x,y) = f(A,B) + g(A,B)x\). A neater format is \(h= f + gx\). Make a table of values for \(x\) and \(h\). Determine the least squares line through this new data set and, in the process, calculate \(f\), \(g\), and thus \(A\), \(B\).