Chapter 12

The more general statement of how an object loses heat, known as Newton's law of cooling, is that the rate of decrease of temperature of an object is continuous and proportional to the difference between the temperature of the object and that of the surrounding medium. Suppose that the surrounding medium is so vast that the heat it absorbs from the object does not appreciably change the medium's temperature so that it remains constant. Suppose that the object is initially at a temperature of \(100^{\circ}\) and loses temperature continuously at the rate of \(1 / 100\) per minute of the difference between its temperature at time \(t\) and the constant temperature \(T_{0}\) of the surrounding medium. Derive the formula that relates the temperature of the object and time. Ans. \(T-T_{0}=\left(100-T_{0}\right) e^{-0.01 t}\).

Estimate the integral part of the following: (a) \(\log _{e} 3\). Ans. 1 (b) \(\log _{e} 9\). (c) \(\log _{e} 100\). Ans. 4 . (d) \(\log _{e} 1\).

Given \(\log 7=0.8451\), find the following: (a) \(\log 70\). Ans. \(1.8451\). (b) \(\log 700\). (c) \(\log 0.7 .\) Ans. \(-1+0.8451\). (d) \(\log 0.07\) (e) \(\log 0.007 .\) Ans. \(-3+0.8451\).

Suppose that an object loses temperature at the rate of \(0.01\) of the existing temperature not continuously but at the end of each minute. If the temperature is initially \(100^{\circ}\), derive the formula that relates the temperature of the object and the time. Ans. \(T=100(0.99)^{t}\). To fit the physical situation \(t\) can take on only the values \(0,1,2, \ldots\), where the numbers refer to minutes.

Find the slope of \(y=e^{x}\) at \(x=0\).

Suppose that 2 is used as the base of a system of logarithms. What then are the following: (a) \(\log 8 .\) Ans. \(3 .\) (b) \(\log 16\). (c) \(\log 1 .\) (d) \(\log \frac{1}{2}\) Ans. \(-1\).

Having found the time required to fall \(219.5\) feet under the conditions of Exercise 2 use that result to find the velocity that the object acquires in falling that distance. This velocity should be compared with the velocity of \(1000 \mathrm{ft} / \mathrm{sec}\) with which the object was projected upward so as to reach the maximum height of \(219.5\) feet. Ans. \(75.2 \mathrm{ft} / \mathrm{sec}\).

Suppose that an object, initially at a temperature of \(100^{\circ}\), loses temperature continuously at the rate of \(0.01^{\circ}\) per minute. Derive the formula that relates the temperature and the time. Ans. \(T=100-\) \(0.01 t\).

Graph the function \(y=e^{-x^{2}}\).

Look up in the table of natural logarithms the following quantities: (a) \(\log 3 . \quad\) Ans. \(1.0986 .\) (b) \(\log 5 .\) (c) \(\log 10 . \quad\) Ans. \(2.3026\) (d) \(\log 0.1\). (e) \(\log 0.5 .\) Ans. \(-0.6931\).

Suppose that an object is thrown downward with an initial velocity \(V\) in a medium whose resistance is proportional to the velocity. (a) Find the expression for the velocity as a function of time and discuss the motion for the cases \(V>32 / k, V=32 / k\), and \(V<32 / k\). Ans. \(\quad v=\frac{32}{k}-\left(\frac{32}{k}-V\right) e^{-k t}\). (b) What is the terminal velocity in each case? Explain physically why the answers are correct. Ans. \(32 / k\).

Suppose that the population of a town increases at the net rate per year of \(0.03466\) of its population at the beginning of the year. Find the formula that relates the population and time. Ans. \(P=\) \(P_{0}(1.03466)^{t}\).

Find the integral when the derived function has the following values: (a) \(d y / d x=e^{-x} . \quad\) Ans. \(y=-e^{-x}+C\). (b) \(d y / d x=x e^{x^{2}}\) (c) \(d y / d x=e^{\sin x} \cos x\). Ans. \(y=e^{\sin x}+C\). (d) \(d y / d x=\frac{e^{-1 / x}}{x^{2}}\). (e) \(f^{\prime}(x)=e^{-x / 2}\).

Find the numbers whose natural logarithms are the following: (a) \(-0.5108\). (b) \(0.7885\). Ans. \(2.2\) (c) \(3.4012\). (d) \(4.3820\). Ans. 80 .

Suppose that an object is projected upward with an initial velocity of \(1000 \mathrm{ft} / \mathrm{sec}\) and air resistance is neglected. Find the time to reach maximum height and the maximum height. Ans. \(125 / 4 \mathrm{sec}\); \(15,625 \mathrm{ft}\).

Suppose that the population of a town was 5000 twenty years ago and that it increased continuously at a rate proportional to the existing population. Suppose that the population reached 15,000 at the end of the twenty years. What formula relates the population and the time? Ans. \(P=5000 e^{0.06 t}\).

Find \(y\) when \(y^{\prime}\) has the following values: (a) \(\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\). Ans. \(y=\log \left(e^{x}+e^{-x}\right)+C\). (b) \(\frac{e^{2 x}-1}{e^{2 x}+1}\) (c) \(\frac{e^{2 x}}{1+e^{2 x}} \quad\) Ans. \(y=\frac{1}{2} \log \left(1+e^{2 x}\right)+C\).

On the same set of axes sketch \(y=\log _{10} x\) and \(y=\log _{e} x\).

Evaluate: (a) \(\int e^{4 x} d x . \quad\) Ans. \(=\frac{1}{4} e^{4 x}+C .\) (b) \(\int \frac{e^{1 / x^{4}}}{x^{3}} d x\). (c) \(\int e^{-x^{2}+3} x d x\) (d) \(\int\left(e^{x}+2\right)^{2} d x\). (e) \(\int\left(e^{x}+1\right)^{4} e^{x} d x\) (f) \(\int \frac{e^{2 x}}{e^{2 x}+5} d x\).

On the same set of axes sketch the following pairs of curves. Base \(e\) is understood. (a) \(y=\log x\) and \(y=3 \log x\). (b) \(y=\log x\) and \(y=\log x+2\). (c) \(y=\log x\) and \(y=\log (x+3)\). (d) \(y=\log x\) and \(y=3 \log (x+2)\). (e) \(y=\log x\) and \(y=\log (5-x)\) for \(x<5\).

Sugar in water dissolves continuously at a rate proportional to the undissolved amount. If the amount of sugar is initially 200 grams and 100 grams are dissolved in 2 minutes, how long does it take to dissolve 150 grams? How long does it take to dissolve all 200 grams? Ans. \(4.1 \mathrm{~min} ; \infty\).

Find the area under the curves \(y=x e^{x^{2}}\) from \(x=0\) to \(x=1\).

Given the function \(y=\log a x\), where the notation indicates the logarithm to base \(a\), find \(y^{\prime}\). Suggestion: Rewrite \(y=\log _{a} x\) in terms of \(\log _{e} x\). Ans. \(y^{\prime}=\) \(1 / x \log _{e} a\).

Suppose that the population of a town increases at an annual rate of \(r\) people per hundred per year. Find the equivalent rate if the same population is assumed to increase continuously and at a rate proportional to the existing population. Ans. \(k=\log _{\mathrm{e}}(1+r)\).

Bacteria in a culture increase continuously at a rate proportional to the number already attained. If the proportionality constant is \(0.2\) per minute, find the number of bacteria in \(t\) minutes if there are 100 to start with. Ans. \(N=100 e^{0.2 t}\).

Graph the function \(y=e^{-x}\).

Graph the function \(y=e^{-x} \sin x\).

A radioactive substance such as uranium emits particles from the nucleus of each atom. After emitting this particle the atom is no longer uranium (it becomes thorium). We say then that the uranium atom has decayed or disintegrated. There are billions of atoms even in a small amount of uranium, and in any small interval of time some of the atoms disintegrate. Because some disintegration is continually taking place, it is reasonable to speak of the instantaneous rate of change with respect to time of the number of atoms of uranium. Moreover, even though two different atoms may emit particles at quite different instants, the number of atoms is so large that we can assume that the rate at which the atoms emit is proportional to the number of atoms present. Then if \(N\) is the number of uranium atoms present at time \(t\), the rate of decrease of \(N\) per unit of time (in years) is $$ \dot{N}=-k N $$ where \(k\) is the proportionality constant. If there are \(10^{12}\) uranium atoms present at \(t=0\), and if half of these are present after \(4,500,000,000\) years, find the value of \(k . \quad\) Ans. \(k=1.6 \cdot 10^{-10} .\)

Show that if a phenomenon obeys the law \(y=e^{k x}\), then for successive values of \(x\) that are in arithmetic progression the corresponding values of \(y\) are in geometric progression. Suggestion: Suppose that the successive values of \(x\) are \(0, h\), \(2 h, \ldots\), and calculate the corresponding values of \(y\).

Evaluate: (a) \(\int \frac{3 x}{x^{2}+2} d x\). Ans. \(\frac{3}{2} \log \left(x^{2}+2\right)+C\). (b) \(\int \frac{x^{2}}{1-x^{3}} d x\). (c) \(\int \frac{x+1}{x^{2}+2 x+5} d x\). (d) \(\int\left(\frac{1}{2 x-5}-\frac{1}{2 x+3}\right) d x\). (e) \(\int \tan \frac{x}{2} d x\).

In a certain chain of nuclear reactions that take place in a nuclear reactor plutonium decays to uranium 235 and the uranium decays to thorium. The amount of uranium derived from plutonium at any time \(t\) is given by \(u=P\left(1-e^{-\lambda t}\right)\) where \(P\) is the original amount of plutonium. The amount of uranium that has decayed at any time \(t\) is given by \(U=u\left(1-e^{-\lambda t}\right)\). Let us assume that \(P=3400\) grams and that \(\lambda=\frac{1}{10} \log 2\). If the amount of uranium present at any time exceeds what is called the critical mass, namely 800 grams, the reactor will explode. Is the reactor safe? Ans. Disaster.

Evaluate \(\lim _{h \rightarrow 0} \frac{e^{h}-1}{h}\). Ans. 1 .

Find the area between the curve \(y=1 / x\), the \(x\)-axis and the ordinates at \(x=1\) and \(x=10\). Ans. \(2.3026\).

A tank contains 100 gallons of salt water in which 150 pounds of salt is dissolved. Salt water containing one pound of salt per gallon enters continuously at the rate of \(2 \mathrm{gal} / \mathrm{min}\). If the mixture in the tank is \(\mathrm{kept}\) uniform by stirring and the mixed salt water flows out continuously at the rate of \(2 \mathrm{gal} / \mathrm{min}\), what is the amount of salt in the tank at the end of 1 hour?

Show that \(y=D e^{k x}\) satisfies the equation \(d y / d x=k y\).

Show by mathematical induction that for positive integral \(n\) $$ \frac{d^{n}}{d x^{n}} \log x=(-1)^{n-1} \frac{(n-1) !}{x^{n}} . $$ The left-hand side means the \(n\)th derivative of \(\log x\) and \((n-1) !\) means 1. \(2.3 \ldots(n-1)\).

The half-life of a radioactive mass of atoms \(N(t)\), which disintegrate according to the law \(N(t)=N(0) e^{-k t}\) where \(t\) is time in years, is the time required for \(N(t)\) to equal \(N(0) / 2\). Find the half-life in terms of \(k\). Ans. \(0.693 / k\).

Evaluate \(\lim _{x \rightarrow 1} \frac{\log x}{x-1}\).

The population of the earth was about 3 billion in 1970 . It is estimated that the world's population is increasing continuously at a rate of \(2 \%\) per year of the existing population. When will the population of 20 billion be reached?

A drug is injected into the blood and is gradually absorbed and then eliminated. The concentration \(y\) of the drug at any time \(t\) is given by the formula \(y=\left[A /\left(C_{2}-C_{1}\right)\right]\left(e^{-C_{1} t}-e^{-C_{2} t}\right)\), where in \(A, C_{1}\) and \(C_{2}\) are positive constants. Find the time \(t\) at which the concentration is a maximum. Ans. \(t=\left[1 /\left(C_{1}-C_{2}\right)\right] \log \left(C_{1} / C_{2}\right)\).