# ch07_03 - 7-23 SECTION 7.3 . Direction Fields and Eulers...

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7-23 SECTION 7.3 . . Direction Fields and Euler’s Method 587 78. The differential equation y ± =− ay ln( y / b ) (for positive con- stants a and b ) arises in the study of the growth of some animal tumors. Solve the differential equation and sketch several mem- bers of the family of solutions. What adjective (e.g., rapid, moderate, slow) would you use to characterize this type of growth? 79. Use the technique of exercise 77 to show that solutions of y ± ln( y / b ) for positive constants a and b have at most one inﬂection point, which occurs at y = b / a . EXPLORATORY EXERCISES 1. Look up the census ﬁgures for the U.S. population starting in 1790. (You can ﬁnd this information in any library, in virtually any almanac or encyclopedia.) Plot the data on a graph of population versus time. Does this look like the so- lution of a logistic equation? Brieﬂy explain. If you wanted to model these data with a logistic function, you would need to estimate values for k and M .As shown in exercise 77, M equals twice the height of the inﬂection point. Explain why (for the logistic curve) the inﬂection point represents the point of maximum slope. Estimate this for the population data. To estimate r , note that for small populations, the logistic equation y ± = ry (1 y / k ) . Then r equals the rate of exponential increase. Show that for the ﬁrst 50 years, the U.S. population growth was approximately exponential and ﬁnd the percentage increase as an estimate of r .With these values of r and M and the initial population in 1790, ﬁnd a function describing the population. Test this model by comparing actual populations to predicted populations for several years. 2. An object traveling through the air is acted on by grav- ity (acting vertically), air resistance (acting in the direc- tion opposite velocity) and other forces (such as a motor). An equation for the horizontal motion of a jet plane is v ± = c f ( v ) / m , where c is the thrust of the motor and f ( v )i s the air resistance force. For some ranges of ve- locity, the air resistance actually drops substantially for higher velocities as the air around the object becomes tur- bulent. For example, suppose that v ± = 32,000 f ( v ), where f ( v ) = ± 0 . 8 v 2 if 0 v 100 0 . 2 v 2 if 100 <v .To solve the initial value problem v ± = 32,000 f ( v ) ,v (0) = 0, start with the initial value problem v ± = 32,000 0 . 8 v 2 (0) = 0. Solve this IVP ² Hint: 1 40,000 v 2 = 1 400 ³ 1 200 + v + 1 200 v ´ µ and de- termine the time t such that v ( t ) = 100. From this time forward, the equation becomes v ± = 32,000 0 . 2 v 2 . Solve the IVP v ± = 32,000 0 . 2 v 2 (0) = 100. Put this solution together with the previous solution to piece together a solution valid for all time. 3. Solve the initial value problems dy dt = 2(1 y )(2 y )(3 y ) with (a) y (0) = 0, (b) y (0) = 1 . 5, (c) y (0) = 2 . 5 and (d) y (0) = 4. State as completely as possible how the limit lim t →∞ y ( t ) depends on y (0).

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## ch07_03 - 7-23 SECTION 7.3 . Direction Fields and Eulers...

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