Welcome to an introduction to differentials.
In this lesson, we'll define differentials,
determine the differentials of functions,
use differentials to make approximations,
and finally, estimate propagated and relative error.
So we know that dy/dx is equal to f prime of x,
where the given value of x, this represents
the slope of the tangent line, and therefore
differential y represents the change
of y along the tangent line,
and differential x represents the change
of x along the tangent line.
If we were to solve this equation for differential y,
or dy, we'd multiply both sides of the equation by dx,
and therefore we have differential y, or dy,
equals f prime of x times dx, or differential x.
More formally, let y equal f of x represent a function
that is differentiable in an open interval containing x.
The differential of x, or dx, is any nonzero real number.
The differential of y, or dy, is, as we stated above,
equal to f prime of x times dx.
And because the tangent line can be viewed
as a linear approximation, or tangent line
approximation to a function, we can use
differentials to approximate function values.
We can say delta y, the change of y of a function,
is approximately equal to differential y,
the change of y along a tangent line,
which again is equal to f prime of x times dx.
So the change in y of the function, delta y,
is approximately equal to differential y,
or f prime of x times dx, in most cases,
as long as differential x, the change of x, is small.
Let's look at this graphically.
Let's consider this blue function and this red tangent line.
Notice how here the point of tangency
looks to be at x equals one, this point here.
Notice how as long as we're close to x equals one,
the y values of the red tangent line
would be good approximations for the blue function values.
If we enlarge the graph around this point,
it would give us this graph here on the right.
And let's say our goal was to estimate
the function value, given here by this second point.
Notice how if we were to take
the first function value here and add differential y,
the change of y along the red tangent line,
that function value would be approximately
the same as f of x plus the true change of y, or delta y.
This small vertical distance here would be the error,
but notice how by using differential y,
we can make an approximation for
the function value given by the blue function.
So we can say that differential y is approximately delta y,
which means the change of y along the tangent line,
differential y, is approximately the change of y
of the original function, which is delta y.
And dx equals delta x because the change in x
of the tangent line, differential x, is the same
as the change of x of the function, given by delta x.
So this delta x here is the change of x for the function,
but it's also the change of x along the tangent line,
so delta x equals differential x.
So as we mentioned earlier, differentials
can be used to approximate function values.
To make a function approximation, we can use
this formula here, where f of the quantity x plus delta x
would be approximately equal to f of x plus differential y,
where x would be the point of tangency, so we'd evaluate
the function of the point of tangency, and then add
differential y, where again differential y is equal
to f prime of x times dx.
So let's use differentials
to estimate the square root of 16.4.
The first thing we would do
to find the function we're going to use,
and because we're trying to find a square root,
we'd let our function f of x
be equal to the square root of x,
which equals x to the one-half.
Next, we're looking for a convenient value of x
that's close to 16.4, and because 16's a perfect square,
we would let x equal 16, but because the true input
is 16.4, differential x, which equals delta x,
would be equal to positive 0.4.
And therefore, f of the quantity 16 plus 0.4,
which represents the square root of 16.4,
would be approximately equal to f of 16 plus
differential y, where differential y would be
equal to f prime of 16 times delta x.
So notice how in order to make this approximation,
we need to determine f of 16 as well as
differential y, which is f prime of 16 times dx.
Let's go ahead and find differential y.
Remember, differential y is equal
to f prime of x times dx, so the derivative
of x to the one-half would be one-half x
raised to the power of negative one-half,
and then we have times dx, which would be equal
to one over two square root x times dx.
So now we need to find f 16,
which would be equal to the square root of 16, or four,
and then we also need to determine differential y
when x equals 16, so we'd have f prime of 16 times dx,
which equals one over two square root 16 times dx,
where dx is 0.4, four-tenths.
Well, four-tenths simplifies to two-fifths,
so without a calculator, this simplifies again,
we have a common factor of two, so this
simplifies to one over, square root of 16
is four, so we have one-twentieth.
So we have f of 16.4
would be approximately equal to f of 16,
which we know is four, plus differential y,
which is one-twentieth, so we have
four and one-twentieth, which would be 4.05,
one-twentieth is equal to five-hundredths.
Now let's go ahead and compare our approximation
to the value of square root 16.4 on the calculator.
So the actual value of square root 16.4 from the calculator
is 4.04969, which, notice how it's very close
to our approximation of 4.05 using differentials.
Let's look at the tangent line that we used
in order to make this approximation.
This red tangent line is a tangent line to f of x
equals square root x, x equals 16, this point here,
and notice how this line is a very good approximation
for the function values near x equals 16,
and that's why our approximation was so good.
We can even compare differential y to delta y
at x equals 16.4 because the tangent line
is such a good approximation around x equals 16.
Now let's look at two examples where we're
given a function and asked to determine differential y.
In example 1, we're given y equals the quantity
three x to the fourth plus two raised to the three-halves,
and asked to determine differential y.
Let's first find the derivative function
and then write it in differential form.
So dy/dx is going to be equal to
the derivative of the given function.
Notice how we have to apply the chain rule
where the inner function would be
three x to the fourth plus two, so if u
is equal to three x to the fourth plus two,
notice how u prime, or du/dx would be 12 x to the third,
so dy/dx would be equal to,
if we multiplied by three-halves,
we'd have three-halves u to the three-halves minus one,
that's one-half times u prime, which would be equal
to three-halves times the quantity three x to the fourth
plus two to the one-half, times u prime,
which is 12 x to the third.
Notice here, we can simplify the two and the 12,
there's one two in two and six twos in 12,
so this is dy/dx, let's write this in differential form,
we can say differential y is equal to, simplifying here,
we'd have 18 x to the third times the quantity
three x to the fourth plus two to the one-half times dx.
So this would be differential y.
Number two's a little bit different.
We're given a function and asked to determine
dy when x equals four and dx equals 0.03.
So again, let's first find our derivative.
So dy/dx is equal to, the derivative
of four x minus five raised to the third,
again, notice how we have to apply the chain rule,
where u would be equal to four x minus five
and u prime is equal to four, so we can think of this
as u to the third, and therefore the derivative
would be three times u to the second times u prime,
which would be three times the quantity four x
minus five to the second, times u prime, which is four,
which means differential y is equal to, simplifying,
we'd have 12 times the quantity four x
minus five to the second times dx.
But now we're asked to evaluate this when x equals four,
and dx equals 0.03, so we'd have dy equals
12 times four times four minus
five squared times dx, which is 0.03.
Let's go ahead and go to the calculator to get this value.
So 12 times a quantity, this would be 16 minus five,
that's 11 squared times .03,
which is 43.56.
Now remember, this value tells us
the change of y along the tangent line
from x equals four to x equals 4.03.
Now let's talk about propagated error.
If a measure value x is used
to compute another value f of x,
the difference of f of the quantity x plus delta x
and f of x, which is delta y, is the propagated error.
So the difference of these two function values
is equal to delta y, which is the propagated error, but now
we know that delta y is approximately differential y,
and therefore we can use differential y
to estimate the propagated error.
Let's look at an example.
The radius of a ball bearing is measured to be 0.8 inches.
If the measurement is correct within 0.02 inches,
estimate the propagated error.
So again, to estimate the propagated error,
we'll use, in this case, differential V.
So the volume of a sphere is equal
to four-thirds pi r cubed,
and therefore to find differential V,
we first find the derivative with respect to r,
then multiply by differential r.
Notice for the derivative here,
we'd multiply by three, and then subtract one
from the exponent, giving us four pi r squared.
So differential V is equal to four pi r squared
times differential r, and from the given information,
we know r, the radius, is equal to 0.8 inches, and now,
because the measurement is correct within 0.02 inches,
this means delta r or differential r
would be equal to plus or minus 0.02.
Notice how because it says within,
it could be over or under by this amount.
So our estimate for the propagated error,
which is differential V, is going to be equal
to four times pi times r squared, or 0.8 squared,
times differential r, which is plus or minus 0.02.
And now I'll go to the calculator.
So we have four pi times .8 squared
times, of course we won't enter the plus or minus,
but we do include it in the answer.
So .02, close parenthesis, Wnter.
So the error is going to be approximately,
so the estimate for our propagated error
is going to be plus or minus, let's say, 0.1608.
And because this represents the error
in volume, this would be cubic inches.
Now I'll finish by talking about relative error.
The propagated error is given in relative terms
if we compare differential V with V.
So the ratio differential V divided by V
is called the relative error in this situation.
Now let's go ahead and find this.
This will give us a percent of error.
So we have differential V divided by V,
which, in this case, differential V
is four pi r squared, times dr, or differential r,
and the volume is four-thirds pi r cubed.
We want to evaluate this ratio
when r is equal to 0.8 inches
and differential r is equal
to plus or minus 0.02 inches.
Before we perform this substitution,
though, let's simplify this ratio.
These fours would simplify to one,
pi over pi simplifies to one,
looking at the rs, notice how we have
one more factor of r in the denominator,
so r squared would simplify to one,
r cubed would simplify to r to the first.
So we have differential r divided by,
our denominator is one-third r, remember,
dividing by one-third is equivalent
to multiplying by the reciprocal,
so multiplying by three over one would give us
three differential r divided by r.
So the relative error in this situation would be
three times plus or minus 0.02
divided by r, which is 0.8.
And now we'll go back to the calculator one more time.
Our numerator would be three times .02,
and then divided by .8,
which gives us 0.075,
so plus or minus 0.075,
converting to a percent
would be plus or minus 7.5 percent.
I hope you found this helpful.