Chapter 6

Differentiate a. \(\quad P(t)=t^{4}+e^{2 t}\) b. \(\quad P(t)=t^{4} e^{2 t}\) c. \(\quad P(t)=\frac{t^{4}}{e^{2 t}}\) d. \(P(t)=\left(e^{2 t}\right)^{4}\) e. \(P(t)=e^{2 t^{4}}\) f. \(P(t)=\left(t^{2}+1\right)^{4}(5 t+1)^{7}\) g. \(\quad P(t)=(\ln t)^{3}\) h. \(\quad P(t)=\left(e^{3 t} \ln 2 t\right)^{4}\) i. \(\quad P(t)=\frac{\ln t}{t}\) j. \(\quad P(t)=t \ln t-t\) k. \(P(t)=\frac{1}{\ln t}\) l. \(P(t)=e^{(\ln t)}\) m. \(P(t)=\frac{5 t^{2}-2 t-7}{t^{2}+1}\) n. \(P(t)=\frac{(t+2)^{2}}{t^{2}+2}\)

Use the chain rule to differentiate \(P(t)\) for a. \(P(t)=e^{\left(-t^{2}\right)}\) b. \(P(t)=\left(e^{t}\right)^{2}\) c. \(P(t)=e^{2 \ln t}\) d. \(P(t)=\ln e^{2 t}\) e. \(P(t)=\ln (2 \sqrt{t})\) f. \(P(t)=\sqrt{2 \ln t}\) g. \(P(t)=\sqrt{e^{2 t}}\) h. \(P(t)=\sqrt{e^{\left(-t^{2}\right)}}\) i. \(\quad P(t)=\left(t+e^{-2 t}\right)^{4}\) j. \(P(t)=\left(1+e^{\left(-t^{2}\right)}\right)^{-1}\) k. \(P(t)=\frac{3}{4}\left(1-x^{2} / 16\right)^{1 / 2}\) 1\. \(P(t)=(t+\ln (1+2 t))^{2}\)

The word differentiate means 'find the derivative of'. Differentiate a. \(P(t)=\frac{e^{-3 t}}{t^{2}}\) d. \(P(t)=t^{2} e^{2 t}\) g. \(P(t)=\left(e^{t}\right)^{5}\) b. \(P(t)=e^{2 \ln t}\) e. \(P(t)=e^{t \ln 2}\) h. \(P(t)=\frac{t-1}{t+1}\) c. \(P(t)=e^{t} \ln t\) f. \(P(t)=e^{2}\) i. \(P(t)=\frac{3 t^{2}-2 t-1}{t^{-} 1}\)

Compute \(P^{\prime}\) for: a. \(\quad P(t)=t^{2} e^{t}\) b. \(P(t)=\sqrt{t} e^{\sqrt{t}}\) c. \(\quad P(t)=\frac{t}{1+t^{2}}\) d. \(P(t)=\frac{t+1}{t-1}\) e. \(\quad P(t)=e^{t} \sqrt{1+t}\) f. \(\quad P(t)=t \ln t-t\) g. \(\quad P(t)=t e^{t}-e^{t}\) h. \(\quad P(t)=t^{2} e^{t}-2 t e^{t}+2 e^{t}\) i. \(\quad P(t)=\frac{\sqrt{t}}{\ln t}\) j. \(\quad P(t)=e^{t} \ln t\) k. \(P(t)=\frac{1}{\ln t}\) l. \(\quad P(t)=e^{(t \ln t)}\) m. \(P(t)=10 \frac{e^{0.2 t}}{9+e^{0.2 t}}\) n. \(\quad P(t)=\frac{20}{1+19 e^{-0.1 t}}\)

In Chapter 7 we show that \([\cos (t)]^{\prime}=-\sin (t) .\) Use this formula and \([\sin t]^{\prime}=\cos t\) written earlier in this chapter to differentiate: $$ \begin{array}{ll} \text { a. } y(t)=\cos (2 t) & \text { b. } y(t)=\sin \left(\frac{\pi}{2} t\right) \\ \text { c. } y(t)=e^{\cos t} & \text { d. } y(t)=\cos \left(e^{t}\right) \\ \text { e. } y(t)=\sin (\cos t) & \text { f. } y(t)=\sin \left(\cos e^{t}\right) \\ \text { g. } y(t)=\ln (\sin t) & \text { h. } y(t)=\sec t=(\cos t)^{-1} \\ \text {i. } y(t)=\ln \left(\cos \left(e^{t}\right)\right) & \text { j. } y(t)=\ln \left(\cos \left(e^{\sin t}\right)\right) \end{array} $$

Exercise 6.2 .4 The Doppler effect. You are standing 100 meters south of a straight train track on which a train is traveling from west to east at the speed 30 meters per second. See Figure 6.2 .4 . Let \(y(t)\) be the distance from the train to you and \(|x(t)|\) be the distance from the train to the point on the track nearest you; \(x(t)\) is negative when the train is west of the point on the track nearest you. a. Write \(y(t)\) in terms of \(x(t)\). b. Find \(y^{\prime}(t)\) for a time \(t\) at which \(x(t)=-200\) c. Time is measured so that \(x(0)=0 .\) Write an equation for \(x(t)\). d. Write an equation for \(y^{\prime}(t)\) in terms of \(t\). e. The whistle from the train projects sound waves at frequency \(f\) cycles per second. The frequency, \(f_{L},\) of the sound reaching your ear is $$ f_{L}(t)=\frac{331.4}{331.4+y^{\prime}(t)} f \frac{\text { cycles }}{\text { second }} $$ \(331.4 \mathrm{~m} / \mathrm{s}\) is the speed of sound in air. Draw a graph of \(f_{L}(t)\) assuming \(f=500\). Derivation of Equation 6.18 for the Doppler effect. A sound of frequency \(f\) traveling in still air has wave length \((331.4 \mathrm{~m} / \mathrm{s}) /(f\) cycles \(/ \mathrm{s})=(331.4 / \mathrm{f}) \mathrm{m} /\) cycle. If the source of the sound is moving at a velocity \(v\) with respect to a listener, the wave length of the sound reaching the listener is \(((331.4+v) / f) \mathrm{m} /\) cycle. These waves travel at \(331.4 \mathrm{~m} / \mathrm{sec},\) and the frequency \(f_{L}\) of these waves reaching the listener is $$ f_{L}=\frac{331.4 \mathrm{~m} / \mathrm{s}}{((331.4+v) / f) \mathrm{m} / \text { cycle }}=\frac{331.4}{331.4+v} f \frac{\text { cycles }}{\text { second }} $$ High frequency sound waves may be used to measure the rate of blood flow in an artery. A high frequency sound is introduced on the skin surface above the artery, and the frequency of the waves reflected from the arterial flow is measured. The difference in frequencies emitted and received is used to measure blood velocity. A train and track with listener location. As drawn, \(x(t)\) is negative

Air is being pumped into a spherical balloon at the rate of \(1000 \mathrm{~cm}^{3} / \mathrm{min}\). At what rate is the radius of the balloon increasing when the volume is \(3000 \mathrm{~cm}^{3} ?\) Note: \(V(t)=\frac{4}{3} \pi r^{3}(t)\).

Exercise 6.2 .6 Consider a spherical ice ball that is melting. A reasonable model is: Mathematical Model. 1\. The rate at which heat is transferred to the ice ball is proportional to the surface area of the ice ball. 2\. The rate at which the ball melts is proportional to the rate at which heat is transferred to the ball. The volume, \(V,\) of a sphere of radius \(r\) is \(\frac{4}{3} \pi r^{3}\) and its surface area, \(S,\) is \(4 \pi r^{2}\). From 1 and 2 we conclude that the rate of change of volume of the ice ball is proportional to the surface area of the ice ball. a. Write an equation representative of the previous italicized statement. b. As the ball melts, \(V, r,\) and \(S\) change with time. Differentiate \(V(t)=\frac{4}{3} \pi r^{3}(t)\) to obtain $$ V^{\prime}(t)=4 \pi r^{2}(t) r^{\prime}(t) $$ c. Use your equation from (a) and the equation from (b) to show that $$ r^{\prime}(t)=K $$ where \(K\) is a constant d. Why should \(K\) be negative? e. Because \(K\) should be negative, we write $$ r^{\prime}(t)=-K $$ A good candidate for \(r(t)\) is \(r(t)=-K t+C \quad\) where \(C\) is a constant Why? f. Only discussion included in this part. With \(r(t)=-K T+C\) we find that \(W(t)=A(1-t / B)^{3} \quad\) where \(W(t)\) is the weight of the ball and \(A\) and \(B\) are constants. In order to test this conclusion, we filled a plastic ball about the size of a volley ball with water and froze it to \(-14^{\circ} \mathrm{C}\) (Figure 6.2 .6 ). One end of a chord (knotted) was frozen into the center of the ball and the other end extended outside the ball as a handle. We removed the plastic and placed the ball in a \(10^{\circ} \mathrm{C}\) water bath, held below the surface by a weight attached to the ball. At four minute intervals we removed the ball and weighed it and returned it to the bath. The data from one of these experiments is shown in Table 6.2 .6 and a plot of the data and of a cubic, \(y=3200(1-t / 120)^{3}\), is shown in the figure of Table 6.2 .6 . The cubic looks like a pretty good fit to the data, and we might argue that the data is consistent with our model. There are some flaws with the fit of the cubic, however. The cubic departs from the data at both ends. \(y(0)=3200\), but the ball only weighed \(3020 \mathrm{~g}\); the cubic is also above the data at the right end. g. We found that we could fit the data more closely with an equation of the form $$ W(t)=A(1-t / 100)^{\alpha} $$ where \(\alpha\) is closer to 2 than to 3 . Find values for \(A\) and \(\alpha\). Note: \(\ln W(t)=\ln A+\alpha \ln (1-t / 100)\). Then reasonable estimates of \(\ln A\) and \(\alpha\) are the coefficients of a line fit to the graph of \(\ln w(t)\) versus \(\ln (1-t / 100)\). h. If the data is not consistent with the model, in what way might the model be deficient?

Is there an example of two functions, \(u(x)\) and \(v(x),\) for which \([u(x) \times v(x)]^{\prime}=u^{\prime}(x) \times v^{\prime}(x) ?\)

Is there an example of two functions, \(u(x)\) and \(v(x),\) for which \(\left[\frac{u(x)}{v(x)}\right]^{\prime}=\frac{u^{\prime}(x)}{v^{\prime}(x)} ?\)

An examination of 1000 people showed that 41 were carriers (heterozygotic) of the gene for cystic fibrosis. Let \(p\) be the proportion of all people who are carriers of cystic fibrosis. We can not say with certainty that \(p=41 / 1000\). For any number \(p\) in [0,1] , let \(L(p)\) be the likelihood of the event that 41 of 1000 people are carriers of cystic fibrosis given that the probability of being a carrier is \(p\). Then $$ L(p)=\left(\begin{array}{c} 1000 \\ 41 \end{array}\right) p^{41} \times(1-p)^{959} $$ where \(\left(\begin{array}{c}1000 \\ 41\end{array}\right)\) is a constant \(^{1}\) approximately equal to \(1.3 \times 10^{73}\). a. Compute \(L^{\prime}(p)\). b. Find the value \(\hat{p}\) of \(p\) for which \(L^{\prime}(p)=0\) and compute \(L(\hat{p})\). The value \(L(\hat{p})\) is the maximum value of \(L(p)\) and \(\hat{p}\) is called the maximum likelihood estimator of \(p\).

An examination of 1000 people showed that 41 were carriers (heterozygotic) of the gene for cystic fibrosis. In a second, independent examination of 2000 people, 79 were found to be carriers of cystic fibrosis. Let \(p\) be the proportion of all people who are carriers of cystic fibrosis. For any number \(p\) in \([0,1],\) let \(L(p)\) be the likelihood of finding that 41 of 1000 people in one study and 79 out of 2000 people in a second independent study are carriers of cystic fibrosis given that the probability of being a carrier is \(p\). Then $$ L(p)=\left(\begin{array}{c} 1000 \\ 41 \end{array}\right) p^{41} \times(1-p)^{959} \times\left(\begin{array}{c} 2000 \\ 79 \end{array}\right) p^{79} \times(1-p)^{1921} $$ where \(\left(\begin{array}{c}1000 \\ 41\end{array}\right) \doteq 1.3 \times 10^{73}\) and \(\left(\begin{array}{c}2000 \\ 79\end{array}\right) \doteq 1.4 \times 10^{143}\) are constants. a. Simplify \(L(p)\). b. Compute \(L^{\prime}(p)\). c. Find the value \(\hat{p}\) of \(p\) for which \(L^{\prime}(p)=0\). The value \(L(\hat{p})\) is the maximum value of \(L(p)\) and \(\hat{p}\) is called the maximum likelihood estimator of \(p\).

A bird searches bushes in a field for insects. The total weight of insects found after \(t\) minutes of searching a single bush is given by \(w(t)=\frac{2 t}{t+4}\) grams. Draw a graph of \(w .\) From your graph, does it appear that a bird should search a single bush for more than 10 minutes? It takes the bird one minute to move from one bush to another. How long should the bird search each bush in order to harvest the most insects in an hour of feeding?

Let $$ P(t)=\frac{u(t)}{v(t)}=u(t) \times(v(t))^{-1} $$ Use the product rule and power chain rule to show that $$ P^{\prime}(t)=\frac{u^{\prime}(t) v(t)-u(t) v^{\prime}(t)}{v^{2}(t)} $$

Let \(P(t)=u(t) \times v(t) .\) Then $$ \ln P(t)=\ln (u(t) \times v(t))=\ln u(t)+\ln v(t) $$ Compute the derivative of the two sides of Equation 6.11 using the logarithm chain rule and show that $$ P^{\prime}(t)=u(t) v^{\prime}(t)+u^{\prime}(t) v(t) $$

Let \(P(t)=u(t) / v(t)\). Then $$ \ln P(t)=\ln \left(\frac{u(t)}{v(t)}\right)=\ln u(t)-\ln v(t) $$ Compute the derivative of the two sides of Equation 6.12 using the logarithm chain rule and show that $$ P^{\prime}(t)=\frac{u(t) v^{\prime}(t)-u^{\prime}(t) v(t)}{v^{2}(t)} $$

A useful special case of the quotient formula is the reciprocal formula: If \(u(t)\) has a derivative and \(u(t) \neq 0\) and $$ P(t)=\frac{1}{u(t)} $$ then $$ P^{\prime}(t)=\frac{-1}{u^{2}(t)} u^{\prime}(t) $$ Prove the formula using logarithmic differentiation. That is, write $$ \ln P(t)=\ln \left(\frac{1}{u(t)}\right)=-\ln u(t) $$ and compute the derivatives of both sides using the logarithm chain rule. We write the formula as $$ \left[\frac{1}{u(t)}\right]^{\prime}=\frac{-1}{u^{2}(t)} u^{\prime}(t) \quad \text { Reciprocal Rule } $$

Exercise 6.1 .18 Sketch the graphs of the logistic curve $$ P(t)=\frac{P_{0} M e^{r t}}{M-P_{0}+P_{0} e^{r t}} $$ for a. \(r=0.5 \quad M=20 \quad P_{0}=1, \quad 15, \quad 20, \quad\) and \(\quad 30\) \(0 \leq t \leq 20\) b. \(\quad r=0.1 \quad M=20 \quad P_{0}=1, \quad 15, \quad 20, \quad\) and \(\quad 30 \quad M \leq t \leq 70\) c. \(\quad P_{0}=1 \quad M=20 \quad r=0.1, \quad 0.3, \quad 0.5, \quad\) and \(\quad 0.7 \quad 0 \leq t \leq 50\) d. \(P_{0}=1 \quad r=0.2 \quad M=10, \quad 15, \quad 20,\) and \(\quad 25 \quad r \leq t \leq 50\)

For what population size is the growth rate \(P^{\prime}\) of the logistic population function the greatest? The equation $$ P^{\prime}(t)=r P(t)\left(1-\frac{P(t)}{M}\right) $$ provides an answer. Observe that \(y=r p(1-p / M)=r p-(r / M) p^{2}\) is a quadratic whose graph is a parabola. The answer to this question is important, for the population size for which \(P^{\prime}\) is greatest is that population that wildlife managers may wish to maintain to provide maximum growth.