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10. Since we are assuming continuity, either convert the monthly
payment into an annual payment or convert the yearly
interest rate into a monthly interest rate for 240 months. Then proceed as in Prob. 9.
12a. Using Eq. (15) we have - S = 800(1) or
S (800t), S(0) 100,000. Using an integrating factor and
integration by parts (or using a D.E. solver) we get S(t) t c. Using the I.C. yields c . Substituting this value into S, setting S(t) 0, and solving numerically for t yields t 135.363 months.
16a. This problem can be done numerically using appropriate D.E. solver software. Analytically we have
(.1.2sint)dt by separating variables and thus y(t) cexp(.1t.2cost). y(0) 1 gives c , so
y(t) exp(.2.1t.2cost). Setting y 2 yields ln2 .2 .1 .2cos, which can be solved numerically to give
2.9632. If y(0) , then as above,
y(t) exp(.2.1t.2cost). Thus if we set y 2 we get the
same numerical equation for and hence the doubling time has not changed.
18. From Eq.(26) we have 19 = 7 + (20-7) or
k = -ln = ln(13/12). Hence if (T) = 15 we get: 15 = 7 +
13. Solving for T yields
T = ln(8/13)/-ln(13/12) = ln(13/8)/ln(13/12) min.
19. Hint: let Q(t) be the quantity of carbon monoxide in the
room at any time t. Then the concentration is given by x(t) = Q(t)/1200.
20a. The required I.V.P. is dQ/dt = kr + P - r,
Q(0) = V. Since c = Q(t)/V, the I.V.P. may be rewritten Vc(t) = kr + P - rc, c(0) = , which has the solution c(t) = k + + ( - k - ).
20b. Set k = 0, P = 0, t = T and c(T) = .5 in the solution found
21a.If we measure x positively upward from the ground, then
Eq.(4) of Section 1.1 becomes m = -mg, since there is no air resistance. Thus the I.V.P. for v(t) is dv/dt = -g, v(0) = 20. Hence = v(t) = 20 - gt and x(t) = 20t - (g/2) + c. Since x(0) =
30, c = 30 and x(t) = 20t - (g/2)t + 30. At the maximum
height v(tm) = 0 and thus
tm = 20/9.8 = 2.04 sec., which when substituted in the equation for x(t) yields the maximum height.
21b. At the ground x(tg) = 0 and thus 20tg - 4.9tg + 30 = 0.
22. The I.V.P. in this case is m = v - mg, v(0) = 20,
where the positive direction is measured upward. dv
2 4a. The I.V.P. is m = mg - .75v, v(0) = 0 and v is
measured positively downward. Since m = 180/32, the D.E.
dv 2 _2t/i5
becomes = 32 - v and thus v(t) = 240(1-e / ) so
that v(10) = 176.7 ft/sec.
24b. Integration of v(t) as found in (a) yields
x(t) = 240t + 1800(e 2t/15-1) where x is measured positively down from the altitude of 5000 feet. Set
2 . 3
t = 10 to find the distance traveled when the parachute opens.
24c. After the parachute opens the I.V.P. is m = mg-12v,
v(0) = 176.7, which has the solution
v(t) = 161.7e + 15 and where t = 0 now represents
the time the parachute opens. Letting t^<^> yields the limiting velocity.
24d. Integrate v(t) as found in (c) to find
x(t) = 15t - 75.8e-32t/15 + C2. C2 = 75.8 since x(0) = 0,
x now being measured from the point where the parachute opens. Setting x = 3925.5 will then yield the length of time the skydiver is in the air after the parachute opens.
26a. Again, if x is measured positively upward, then Eq.(4) of
Sect.1.1 becomes m = -mg - kv.
mg mg -kt/m
26b.From part (a) v(t) = - + [v0 + ]e / . As k ^ 0
k 0 k
this has the indeterminant form of - + . Thus rewrite
v(t) as v(t) = [-mg + (v0k + mg)e-kt/m ]/k which has the
indeterminant form of 0/0, as k ^ 0 and hence L'Hopital's Rule may be applied with k as the variable.
27a. The equation of motion is m(dv/dt) = w-R-B which, in this problem, is
4 3 4 3 4 3
pa p(dv/dt) = pa pg - 6ppav - pa pg. The limiting
velocity occurs when dv/dt = 0.
27b. Since the droplet is motionless, v = dv/dt = 0, we have
4 3 4 3
the equation of motion 0 = ()pa pg - Ee - ()pa pg,
where p is the density of the oil and p' is the density
of air. Solving for e yields the answer.
28. All three parts can be answered from one solution if k
represents the resistance and if the method of solution
of Example 4 is used. Thus we have dv dv
m = mv = mg - kv, v(0) = 0, where we have assumed
the velocity is a function of x. The solution of this
I.V.P. involves a logarithmic term, and thus the answers to parts (a) and (c) must be found using a numerical procedure.
22 29b. Note that 32 ft/sec = 78,545 m/hr .
30. This problem is the same as Example 4 through Eq.(29).
In this case the I.C. is v(XR) = vo, so c =
2 1+X '
The escape velocity,ve, is found by noting that vo >
in order for v to always be positive. From Example 4, the escape velocity for a surface launch is ve(0) = V"2gR . We want the escape velocity of xo = XR to have the relation ve(^R) = .85ve(0), which yields
X = (0.85)-2 - 1 @ 0.384. If R = 4000 miles then xo = XR = 1536 miles.
31b. From part a) = v = ucosA and hence dt
x(t) = (ucosA)t + d1. Since x(0) = 0, we have d1 = 0 and
x(t) = (ucosA)t. Likewise = gt + usinA and