Chapter 6: Problem 3
Explain the meaning of doubling time.
/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}
Learning Materials
Features
Discover
Chapter 6: Problem 3
Explain the meaning of doubling time.
All the tools & learning materials you need for study success - in one app.
Get started for free
A power line is attached at the same height to two utility poles that are separated by a distance of \(100 \mathrm{ft}\); the power line follows the curve \(f(x)=a \cosh (x / a) .\) Use the following steps to find the value of \(a\) that produces a sag of \(10 \mathrm{ft}\) midway between the poles. Use a coordinate system that places the poles at \(x=\pm 50.\) a. Show that \(a\) satisfies the equation \(\cosh (50 / a)-1=10 / a.\) b. Let \(t=10 / a,\) confirm that the equation in part (a) reduces to \(\cosh 5 t-1=t,\) and solve for \(t\) using a graphing utility. Report your answer accurate to two decimal places. c. Use your answer in part (b) to find \(a\) and then compute the length of the power line.
At Earth's surface, the acceleration due to gravity is approximately \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) (with local variations). However, the acceleration decreases with distance from the surface according to Newton's law of gravitation. At a distance of \(y\) meters from Earth's surface, the acceleration is given by where \(R=6.4 \times 10^{6} \mathrm{m}\) is the radius of Earth. a. Suppose a projectile is launched upward with an initial velocity of \(v_{0} \mathrm{m} / \mathrm{s} .\) Let \(v(t)\) be its velocity and \(y(t)\) its height (in meters) above the surface \(t\) seconds after the launch. Neglecting forces such as air resistance, explain why \(\frac{d v}{d t}=a(y)\) and \(\frac{d y}{d t}=v(t)\) b. Use the Chain Rule to show that \(\frac{d v}{d t}=\frac{1}{2} \frac{d}{d y}\left(v^{2}\right)\) c. Show that the equation of motion for the projectile is \(\frac{1}{2} \frac{d}{d y}\left(v^{2}\right)=a(y),\) where \(a(y)\) is given previously. d. Integrate both sides of the equation in part (c) with respect to \(y\) using the fact that when \(y=0, v=v_{0} .\) Show that $$ \frac{1}{2}\left(v^{2}-v_{0}^{2}\right)=g R\left(\frac{1}{1+y / R}-1\right) $$ e. When the projectile reaches its maximum height, \(v=0\) Use this fact to determine that the maximum height is \(y_{\max }=\frac{R v_{0}^{2}}{2 g R-v_{0}^{2}}\) f. Graph \(y_{\max }\) as a function of \(v_{0} .\) What is the maximum height when \(v_{0}=500 \mathrm{m} / \mathrm{s}, 1500 \mathrm{m} / \mathrm{s},\) and \(5 \mathrm{km} / \mathrm{s} ?\) g. Show that the value of \(v_{0}\) needed to put the projectile into orbit (called the escape velocity) is \(\sqrt{2 g R}\)
Archimedes' principle says that the buoyant force exerted on an object that is
(partially or totally) submerged in water is equal to the weight of the water
displaced by the object (see figure). Let \(\rho_{w}=1 \mathrm{g} /
\mathrm{cm}^{3}=1000 \mathrm{kg} / \mathrm{m}^{3}\) be the density of water and
let \(\rho\) be the density of an object in water. Let \(f=\rho / \rho_{w}\). If
\(0
Suppose the cells of a tumor are idealized as spheres each with a radius of \(5 \mu \mathrm{m}\) (micrometers). The number of cells has a doubling time of 35 days. Approximately how long will it take a single cell to grow into a multi- celled spherical tumor with a volume of \(0.5 \mathrm{cm}^{3}(1 \mathrm{cm}=10,000 \mu \mathrm{m}) ?\) Assume that the tumor spheres are tightly packed.
Use a calculator to evaluate each expression or state that the value does not exist. Report answers accurate to four decimal places. a. \(\coth 4\) b. \(\tanh ^{-1} 2\) c. \(\operatorname{csch}^{-1} 5\) d. \(\left.\operatorname{csch} x\right|_{1 / 2} ^{2}\) e. \(\left.\ln \left|\tanh \left(\frac{x}{2}\right)\right|\right|_{1} ^{10}\) f. \(\left.\tan ^{-1}(\sinh x)\right|_{-3} ^{3}\) g. \(\left.\frac{1}{4} \operatorname{coth}^{-1} \frac{x}{4}\right|_{20} ^{36}\)
What do you think about this solution?
We value your feedback to improve our textbook solutions.