In this video I'm going to do one more example
of using a Linear Approximation
to approximate the value of
the square root of 15.9
the idea is kind of the same as in
the other example
again, I'm just going to draw a little graph
just to ....
hopefully make a little extra sense
out of what's going on
the graph I'm going to draw ...
we're trying to take the square root of a number ...
so I'm going to draw
the square root function
and 'Y equals the square root of X'
just looks something like that ...
so again
what we're trying to do is
we're trying to take 15.9
and we want to plug it into our function
and get the true Y value
here's 15.9
comma
the square root of 15.9
whatever that value is
I would like to know that exactly but
since I can't figure it out exactly ...
in this case we're going to
use a Tangent Line Approximation
so ... the idea is
I'm taking a number close to 15.9
that I can take the square root of
like 16
so this would just be the point
16 comma 4
well if I can find the equation
of that Tangent line
so the Tangent Line ...
here's my ...
rough looking little Tangent Line
the idea is
the Y value on the Tangent Line
is close to the true Y value
so again, what we're going to do is
we're just going to find the equation
for that Tangent Line
and then once I have the equation
of that Tangent Line
we'll just plug in the value
15.9
and that will be our Approximation
so the equation of the Tangent Line
it's going to be Y minus the Y co-ordinate
we'll have to figure out the slope
X minus the X co-ordinate
to get the slope again
we just have to take the derivative ...
in this case we'll plug in the X co-ordinate
of 16
well we've got our function up here
f of X is just X to the one half
or square root of X
when we take the derivative of that
we'll just get
1 over 2
X to the ...
well we have to subtract 1 so
one half minus 1
will just give us to the
negative one half
and then we can rewrite this as
1 over 2
we can put the X to the one half
in the denominator
and that's the same thing as
1 over 2 times
the square root of X
well OK easy enough
that means if I plug in 16
into my derivative formula
we'll have
1 over 2 times
the square root of 16
well the square root of 16
is just 4
so we've got
2 times 4 or
one 8th
as our slope ...
all I'm going to do is just
rewrite the equation
of my Tangent Line here
fill it in completely
so we have Y minus 4 equals ...
so the slope is going to be
one over 8
and then X minus 16
and I'm going to clean this up a little bit
I'm just going to put it in
Slope Intercept form
so if we distribute
the 'one 8th times X'
that will just give us
one 8th X
we'll take one 8th times
negative 16
well that will give us
negative 16 over 8 or
negative 2
and then we'll add to that
positive 4
so we'll add the 4 over
and it says hey!
Y will be equal to
one 8th X
negative 2 plus 4
will just give us positive 2
so now we've got the nice little equation
of our Tangent Line
so the equation of our Tangent Line here
is Y equals one 8th X ...
plus 2
and now we're going to do
just like we said
I would like to know
the true Y value
which is
when we plug 15.9 into
the square root function
well in this case
since we're approximating
I'm just going to plug that
same X value
15.9
into our Line Equation
and that'll be our Approximation
so it says we'll get
1 over 8
times 15.9
plus 2
it was probably actually easier
to evaluate it
before we multiplied things out
so I'm going to go back
and actually plug it in there
this line
and this line
are the same
but notice if we plug it in here we'll get
Y minus 4
we'll just get 1 over 8
and then we'll get 15.9
minus 16
when we plug it in
well i think that's a little bit easier to do
so we have one 8th
let's see ..
15.9 minus 16
that'll be negative point 1
well let's see if we can't clean this up
without a calculator
so negative point 1 ...
well that's just negative 1 over 10
if we rewrite it as a fraction
so if we simplify
on the right side
I guess we would get
negative 1 over 80
and then we can add
positive 4 to that
we can keep simplifying here
so I can guess we can get
common denominators
multiply top and bottom
by 80
so if we multiple top and bottom
by 80
4 times 80 is going to give us 320
so it looks like at the end ...
so 320 plus a negative 1
that'll give us 319
over 80 ...
I've got a calculator right here beside me
I'm going to go ahead
and turn that into a decimal
so 319 divided by 80 ...
I'm getting 3 point 9 8 7 5
and then a bunch of zeroes
so the idea is, again
we've approximated
so we've found the Y value on our line
the Y value on the line again is
3 point 9 8 7 5
so we can say
3 point 9 8 7 5
is roughly equal to
the square root of 15 point 9
and just to see how close we are
again since I have my little calculator here
so if we take the square root of
15.9
I'm getting this to be
3 point 9 8 7 4 8 0 4 1
da da da da da ...
so, you know
pretty close
it agrees with the
first three decimal places exactly
9 8 7
and then a ...
we've got a 5
and here we've got a 4 8
it looks like our approximation is in fact
pretty darn good
again though
you can tell from the geometry ...
we could have easily ...
another number I could take the square root of
would be 25
but the idea is
you don't want to use that Tangent Line
because
by the time you get to
sort of the Y value that you're interested in ...
that line is going to be
certainly much further away
than the Tangent line at this point
using 16 comma 4