Updated 2013 2.3-10: ======= Prior to chute open, the velocity model is v'=-32-0.15v, v(0)=0. The distance model is y'=v(t), y(0)=10000. The answers are v(20) approx -202 and y(20) approx 7084. After chute open, the velocity model is w'=-32-1.5w, w(0)=v(20). The distance model is x'=w(t), x(0)=y(20). Some maple solving is required, or a graphing calculator with zoom, to find the answer t approx 326. The book's answer is 300 min (5 min) plus 47 seconds. This is obtained by adding 20 sec to the 326 sec found after chute open. =====maple example===== eq:= -120*exp(-1.5*t)+21*t; solve(eq=7000,t); # solve an equation for t # This code must be modified to solve problem 2.3-10 2.3-20: ======= The solution uses book formulas. There is no solving of a DE, because that has been done already in the textbook, section 2.3. The model is v'=-g-r v^2, v(0)=160 where g=32 and r=1/800. This model is solved on page 101. Do not solve the model; use formulae (13), (14) page 103 with v0=160, g=32, rho=1/800. The symbol y(t) is the height of the bolt measured from the ground, so y(0)=0. The maximum height is obtained for a value of t found by setting the velocity zero, using equation (13). Because tan(theta)=0 at theta=0, then the tangent argument in (13) must be zero: C_1 - t sqrt(rho g) = 0. Solve this equation for t, then make t into a decimal number using C_1 = arctan(v_0 sqrt(rho/g)). Use equation (14) with the computed value of t to find the maximum height. Answer 277.26 ft published in the BOB is correct. 2.3-22: ======= The formulas in the textbook are used. No solving of the DE is required. The model is v'=-g+r v^2 where g=32 and r=0.075 (Downward Motion, page 104). Because the parachutist bails, v(0)=0 and y(0)=10000=y0, where y(t) is the distance from the ground. Equations (16) and (17) require v(t)<0, which means y'(t)<0 or y(t) decreasing. Opposite coordinates in which y(t) increases and measures the distance from the airplane would require a different model, not found in the textbook. Use formulae (17), (18) to answer the questions. Do not solve the DE; that is done in the book. Formula (17) gives 0=y0-(1/r)ln(u(t)) where y0=10000, r=0.075, u(t)=cosh(c2-t sqrt(rg))/cosh(c2). Solve for t to find the flight time. The terminal velocity is given by (18), the logic from (16), in which tanh(theta) approaches 1 at theta= -infinity. DEFINITIONS. cosh(u) is the hyperbolic cosine, defined by cosh(u)=(exp(u)+exp(-u))/2 sinh(u) is the hyperbolic sine, defined by sinh(u)=(exp(u)-exp(-u))/2 tanh(u)=sinh(u)/cosh(u) arctanh(u) is defined by arctanh(tanh(x))=x Similarly for arcsinh, arccosh. maple details ======= u:=t->cosh(c2-t*sqrt(rho*g))/cosh(c2); eq:=10000-(1/rho)*ln(u(t)); solve(eq=0,t); # solve an equation for t maple end ======