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8/10/2019 How to Solve Problems in Calculus Solution
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1. SolutionThe volume of a sphere of radius r is V(r ) =
V(r ) = 4
Ans. V(5) = 100
2. Solution(a) Let s represent the bicycles positionrelative to the end of the road.
4. Solutionince the position of the bicycle is !no"n atevery point in time# the instantaneousvelocity can be computed.
Let s(t) =
Then v(t) = s(t) = 2t
$t % ocloc!# t = % so v (%) = s(%) = 6 mi/h.
&.. Solution
'hen t = %# its positionx = 1. Theparticles instantaneous velocity isdetermined usin the derivative.
v = dx/dt = 4*t + %
'hen t = %, v = 117 in/sec.
,. Solution
v = dh/dt = -%2t 2&,.
'hen t = 1/, v = 64 ft/sec. The neativevelocity indicates that the pro0ectile ismovin in the neative direction (do"n"ard)at a speed of ,4 ftsec.
. . Solution'e let m(x) represent the mass of thesection of the rod of lenthx measured fromits leftmost end. The description of the
problem tells us that m(x) = k . ince the
total mass of the rod is 24 slus# m() = 24.
m() = 24
k = 24
k = *'e represent the density function by (x).(x) = m(x).
m(x) = * = *
(x) = m(x) = 4
(4) = 2 slugs/ft
*. Solution
I(t) = Q(t) = % - 4t &
I(2) = %(2)2- 4(2) & = 9 ampees
. Solutiondn/dt = ,t2 1/t 1
'hen t = %, dn/dt = !5
1/. Solution
'hen t = 2/
'ater is drainin at the rate of 4/ allonsper minute. The neative sin indicates thatthe volume of "ater is ettin smaller.
12.Solution3or comparison# the eact cost to producethe &1st item is
C(&1) - C(&/) = 5/.//1(&)% /./2&(&1)2%(&1) &6 - 5/.//1(&/)% /./2&(&/)2%(&/) &6
= %&&.,, - %42.&/
= "1#.176
1%. SolutionThe rate of depreciation is the rate at "hichvalue is lost.7f V(t) represents the value ofthe machine after t years# V(t)representsthe rate at "hich its value chanes.
V(t) = 2//t + %///V(&) = 2000
The machine depreciates at the rate of
82/// per year after & years.
14. Solution
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The surface area of the bois the sum of theareas S = 2x2 4xyince its volume is 1//in%#x2y = 1// and y =1///x2 . 7t follo"s that
S(x) = 4x - 4//
The rate of chane is determined by lettinx= &.
1&. Solutionince revenue = (number of items sold)(priceper item)#
R(x) = (&// - 2x - /.1x2)x = &//x + 2x2- /.1x3
R(x) = &// - 4x - /.%x2
R(1/) = &// - 4/ + %/ = 4#0
1,. Solution
9bservin a riht trianle# "e use thetheorem of :ythaoras to obtain arelationship bet"eenx andy;
x2y2 = 2/2uestion.?efore "e can compute dy/dt# ho"ever# "eneed to @ndy at this instant. This isaccomplished by usin the e>uationobtained in step %.
The neative derivative means thaty isdecreasin. The top of the ladder is fallin atthe rate of #/4 ft/sec.
1. olution
The distance bet"een the cars is increasinat the rate of 2 mi/h.
1*. olution
ince the plane travels 4*/ mih# it "ill have
Ao"n * miles in 1,/ hour (4*/ B 1,/ = *).incex = *# the value of! is easilydetermined by the :ythaorean theorem.
$ns is 465.67 mi/h
1. olution
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2/. olution
ince dAdtis neative# the area is
d"#r"asin at the rate of 16 in2/sec.
21. olution
The crossCsectional area of the "ater isA =(h2) (x %/). The volume of "ater V = l B
A. l is the lenth of the trouh soV = l (h/2)(x %/). ince the units in the problem mustbe consistent#"e ta!e l = 2// cm. Thus V =1//h(x %/). incex is not mentioned ineither the DEivenF or the D3indF in step 2# "eshould eliminatex from this e>uation. Toaccomplish this# "e observe similartrianles.
7n this fiure GA$C is similar to GA%&. incetheir correspondin sides are proportional#
'e substitute ddt = -2/// and h = 2/ intothe result
The minus sin indicates that the "ater levelis fallin at the rate of 0.24 cm/min.22. olution
?y observin that GA$C is similar to GA%& inthe diaram to the riht# "e see that
The "ater level is risin at the rate of 5.09ft/min.
2%. olution
?ecause drdtis neative# r is shrinkin at therate of0.01## in/min.
24. olution
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The volume is shrin!in at the rate of 1.5in#/min.2&. olution
T"o hours after the cars leave their initialpoint#x = */ miles(2 B 4/ = */) andy = ,/miles (2 B %/ = ,/). 7t follo"s that
The distance bet"een the cars is increasin atthe rate of 50 mi/h.
2,. olution
$ns"er is 62.61 $m/h.
2. olution
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incey = 1///x#y = 1/ "henx = 1/. Theminimum perimeter = 2x 2y = 40.
%4. olution
= 1/ is absurd.
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%he ma&imum 'olume of 4!6in#occurs"henx = %.
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