We want to use differentials
to approximate the cube root of 25.5.
When using differentials we are actually using
the change in y of a tangent line to approximate
the change in y of an actual function.
So because we are trying to approximate
the cube root of 25.5, our function
is going to be the cube root function,
which we know we can write as, x to the power of one third.
But before we do this lets go ahead
and take a look at what's happening graphically.
In red, we have the graph of the
cube root function in the first quadrant
and we're trying to approximate
the function value at x equals 25.5, as we see right here.
But 25.5 is not a convenient value
to sub into the cube root function.
But notice how 27 is and it's very close to 25.5.
So we can use an x value of 27 to determine
the tangent line at x=27, and then use
the change in y of this blue tangent line
to make an approximation about the function value at x=25.5.
But since we're using x=27 we're going to have to let delta x,
or differential x representing
the change in x be negative one point five.
Now the reason that we're using
negative one point five is because
27 plus negative one point five
will give us an x value of 25.5.
So lets go back to the previous screen and set this up.
So again the idea here is we want to
make an approximation for f of 25.5.
To do this we're going to let x equal 27,
and let delta x equal negative one point five.
And to make this approximation,
we're going to evaluate the function
at the convenient value of 27.
But then we're going to add the
change in y along the tangent line,
which is differential y which is
equal to f prime of x times dx.
So we're going to add f prime of
27 times dx which is, negative one point five.
So notice to do this, we're going to
have to determine the derivative of the given function.
So lets go ahead and do that now.
F prime of x is going to be equal to
one third times x to the negative two thirds,
which we can write as one over
three times x to the two thirds power.
Now we have all the information
we need to determine this approximation.
F of 25.5 is approximately equal to
the function evaluated at 27 which is
the cube root of 27, plus the derivative
evaluated at 27, that will be
one over three times 27 to the two thirds power,
times differential x which is negative one point five.
Which I'll write as negative three halves.
Since this is in fraction form.
Now lets go ahead and simplify this,
it's going to be three plus 27
to the two thirds, that's the
cubit of 27 that's three squared, so that would be nine.
And then times three, that's one 27th.
Times negative three halves, notice how this simplifies.
There's one factor of three here,
and there's nine factors of three here.
So we have three plus, this is
going to be negative one eighteenth,
so we're going to have two and 17/18,
as our approximation for the cube root of 25.5.
So notice how we were able to make
a good approximation of the cubit
of 25.5 without the use of a calculator.
Lets go a head and convert this into a decimal, and then
we will compare it to what the
calculator would give for this function value.
So two plus 17/18, so it's approximately
two point nine four.
Lets see what the calculator gives us for the cubit of 25.5.
Press math and then option four.
So notice how our approximation is very
good using differentials without the use of a calculator.
So I'm sure you can use this to
impress your friends, this is pretty amazing stuff,
that you probably haven't been
able to do before, without a calculator.
I hope you found this helpful.