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picture1_Calculus Pdf 168707 | Sgw Limits Solns


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File: Calculus Pdf 168707 | Sgw Limits Solns
math19 calculus summer2010 practice problems on limits and continuity 1 atankcontains10litersofpurewater saltwatercontaining20gramsofsaltperliter is pumpedintothetankat2litersperminute 1 express the salt concentration c t after t minutes in g l 2 whatisthelong termconcentrationofsalt ...

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                Math19: Calculus                                                                        Summer2010
                              Practice Problems on Limits and Continuity
                 1 Atankcontains10litersofpurewater. Saltwatercontaining20gramsofsaltperliter
                 is pumpedintothetankat2litersperminute.
                    1. Express the salt concentration C(t) after t minutes (in g/L).
                    2. Whatisthelong-termconcentrationofsalt, i.e., limt→∞C(t)?
                Solution:
                    1. The concentration is, in units of g/L,
                                               C(t) =     total salt    = 20·2·t = 20t
                                                        total volume       10+2·t         5+t
                    2. The long-term concentration is, in units of g/L,
                                         lim    20t = lim 20t · 1/t = lim                20     = 20
                                        t→+∞5+t          t→+∞5+t 1/t            t→+∞5/t+1
                                                                    1
             2 Findthevaluesof a and b that make f(x) continuous for all real x.
                                          x
                                         be +a+1,         x ≤ 0
                                   f (x) = ax2 +b(x+3),   0 < x ≤ 1
                                         
                                         acos(πx)+7bx, x >1
            Solution: Wenotethatthefunctionsarecontinuousontheirdomains,sowecheckthatthe
            left- and right-hand limits agree at the boundary x-values. At x = 0,
                                  lim f(x) = lim bex +a+1 = b+a+1,
                                 x→0−       x→0−
                                  lim f(x) = lim ax2+b(x+3) = 3b,
                                 x→0+       x→0+
            so b + a +1 = 3b, and a = 2b−1. Next, at x = 1,
                                lim f(x) = lim ax2+b(x+3) = a+4b,
                               x→1−       x→1−
                                lim f(x) = lim acos(πx)+7bx = −a+7b,
                               x→1+       x→1+
            soa+4b=−a+7b,and2a=3b.Solvingthislinearsysteminaandbyields b = 2anda = 3
            as the only solution.
                                                   2
        3 Sketchthegraphofafunction f withthefollowingproperties:
          • limx→1 f(x) = 2, but f(1) = 1
          • limx→3 f(x) = +∞
          • limx→2+ f(x) = −1, limx→2− f(x) = 3
          • limx→+∞ f(x) = −2
          • limx→−∞ f(x) = −∞
        Solution: Answers may vary, but here is a representative solution:
                   y
                  3
                  2
                  1
                                                   x
                         1     2    3
                 −1
                 −2
                               3
             4 Showthattheequation√x−5=        1  hasatleast one real solution.
                                             x+3
            Solution: Let f(x) = √x−5−   1 ,sothat f(x) = 0 if and only if x is a solution to the
                                       x+3
            equation. Then f is defined and continuous for all x ≥ 5. Evaluating f at 5 and at 6, we
            see that
                 f (5) = √5−5−   1   =−1<0       and     f (6) = √6−5−   1   =8>0.
                                5+3      8                              6+3    9
            BytheIntermediateValueTheorem,thereissomecintheinterval (5,6) sothat f(c) = 0,
            so f has at least one root.
               (In fact, it is possible to reduce this equation to the cubic polynomial equation (x −
            5)(x +3)2 −1 = 0, and it is unpleasant but not impossible to find its roots exactly; the
            only valid root of the original equation is
                                s         √        s         √
                     c = −1 + 1 3 1051−15 249 + 1 3 1051+15 249 ≈ 5.01556... .)
                           3  3         2         3        2
                                                  4
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...Math calculus summer practice problems on limits and continuity atankcontainslitersofpurewater saltwatercontaininggramsofsaltperliter is pumpedintothetankatlitersperminute express the salt concentration c t after minutes in g l whatisthelong termconcentrationofsalt i e limt solution units of total volume long term lim findthevaluesof a b that make f x continuous for all real be ax acos bx wenotethatthefunctionsarecontinuousontheirdomains sowecheckthatthe left right hand agree at boundary values bex so next soa anda solvingthislinearsysteminaandbyields as only sketchthegraphofafunction withthefollowingproperties limx but answers may vary here representative y showthattheequation hasatleast one let sothat if to equation then dened evaluating we see bytheintermediatevaluetheorem thereissomecintheinterval has least root fact it possible reduce this cubic polynomial unpleasant not impossible nd its roots exactly valid original s...

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