Examples : Approximations Video Lecture | Mathematics (Maths) Class 12 - JEE

hello friends this video an application
of derivatives part 17 has brought you
back exam Viacom normal fear from exam
before watching this video please make
sure that you have watched part 1 - part
16 let's take one more example we have
to find the approximate change in the
volume of the cube of site xmitter
caused by increasing the site by 1% this
is the Q by half and what we are saying
is that the side is X less was inside is
X I am increasing the site by one person
all sides are increasing with one person
so this became this this length also
increased the high tossing trees so new
cube which you get will be something
like this something like this right this
guy is also increasing air your degrees
actually so this is what you need a
chain like this so everything is
increased by one person so in this case
my Delta X is one percent of X that is 1
by hundred into X so my Delta X is X by
hundred correct I have to find change in
volume that is I have to find Delta B I
know B is equal to X you volume of you I
have to find in W if I just
differentiate this guy in respect to X
what I will get DV by DX is nothing but
3 X square or I can say that DV is
nothing but 3 X square DX now since I
have to find Delta V and assume of the
changes where it is I can say that Delta
V is nothing but 3 X square Delta X
correct
Delta X I know is this life let's put
the rather 3 X square into Delta X that
is 1 by hand right thanks and that is
nothing but 3 X cube by all right and
this is in meter so LC meter and that is
my answer
very simple see this is nothing but
simple and DV by DX and I am saying that
since DV is very less
I say DV is nothing but Delta be the
same concept which we use to derive the
formula correct please understand what
we have done here we knew V is equal to
X cube you wanted to find Delta V you
differentiate it this guy we got DV by
DX DX I know so I got D let's say for
example we have to find the approximate
change surface area of you this time is
surface area of the cube by xmitter
where the length is decreasing by X so
Delta X is actually minus minus 1 by 100
X Y 1% of X is the Delta X that is
nothing but minus of my hundred x
correct surface area has nothing but 6 X
square I have to find D s so let's
differentiate this guy by XV s by DX is
nothing but 2 of X right so my d s is
nothing but 2 well X DX
let's assume the change in surface area
is less there's nothing but Delta s is
nothing but 12 X Delta X and this is
nothing but 12 into X into Delta X is
only 1 minus 1 my hundred into X and
that is nothing but minus 12 by hundred
X square meter cube so I can see that
change in area is minus 12 100 X cube it
is a negative chain so I can say that it
is decrease by 12%
not 12 person you can say decrease by 12
I hundred into x1 and this if you
observe carefully this is nothing but
two by a hundred into 6x square or you
can say that 2 by 100 into s so you can
say that this is nothing but 2 percent
of surface area you can write 12 100 X
square also or you can say 2 percent of
surface area let's take for example the
radius of a sphere is measured 7 meter
within the error of point 0 to find the
approximate error in calculating volume
so in this we have to find error in
calculating volume please note that is
we have to find 10 w see if my b is
equal to FX sum function B plus Delta V
that we could do f of X plus Delta X so
if this is this is the error if this is
the error mind this will be de this is
the error in measurement the Delta will
be the error in volume correct so my V
is nothing but 4 by 3 PI R cube Delta R
is known because Delta R is this error
in measurement out radius Delta R has
point 0 to R is 7 all meter I have to
find Delta V is equal to 4 by 3 PI R DV
by D R is nothing but 4 by 3 PI into
you become three hours per cancel so DV
is nothing bad 4 PI R square dia or
Delta V is nothing but 4 PI R square
delta y-tan tile because i am assuming
that DV and D are very small so this is
nothing bad for pi is equal to seven
seven square Delta R point zero two so
you solve this 49 into 4 into point zero
2 this becomes 3 point 9 2 pi and this
is meter why because this guy is point
zero 2 meter and this guy is in meter
square symmetry this matter so error in
volume is the sky three point nine two
five meter so in this case also if you
see understanding the question was a
difficult part you to understand by
reading the question that you are asked
to find error in calculating volume that
is here to find Delta V where Delta R is
equal to point zero two because the
error in calculating radius is the
solution
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