PROFESSOR: Hi.
Welcome back to recitation.
You've been talking about differentials.
And one of the things that you've shown is that some of
the stuff that we've been doing with derivatives can
also be phrased in terms of differentials.
That for many purposes they form a just a different
language to compute the same things.
So in particular one example of that is linear
approximation.
So I have a problem here that's a linear approximation
problem, but I'd like you to do it using the method of
differentials.
So in particular I'd like you to approximate the
square root of 21.
And the way I'd like you to do that is by using a linear
approximation to the function f of x equals the square root
of 10x minus x squared based at the point 2, x0 equals 2.
Note that, you know, why are we using this function to
approximate square root of 21?
Well because at 3, the function value at 3, f of 3,
is exactly equal to the square root of 21.
So to approximate the square root 21 means approximate the
function value at 3.
And so, you know, this function has a nice value at x
equals 2, so we can use linear approximation at 2 to try and
compute an approximation to the function at x equals 3.
So why don't you take a couple of minutes, work this out,
come back and we can work it out together.
All right, welcome back.
So we're doing linear approximation here.
And we're doing it with differentials.
So, in general, when we use the method of differentials to
compute a linear approximation, what we have is
that if y equals f of x, so if y is given as a function of x,
then we have that so if we want to change x a little bit
and figure out what the change in y is, we have this formula,
which is that f of x plus dx--
so the function value at a nearby point--
is approximately equal to y plus dy.
So this is our formula for linear approximation in terms
of differentials.
So in our case we have that our, that the point we want is
x equals 2 and y equals 4.
And so we have, we know was dx is also, because we want the
approximation at the point 3.
So we want dx to be equal to 1 for this approximation.
And so the question is, what is dy going to be?
So that's the value that we're after.
That tells us how much the function value changes.
So to compute dy, well dy is the
differential of the function.
So this is d of the square root of 10x
minus x squared, quantity.
OK, so to compute this differential now, we just use
our straight-forward rules for doing this.
So we take a derivative and then we
have dx's where necessary.
So this is d of square root of 10x minus x squared.
So the outermost function is the square root function.
So this is going to be 1/2 times 10x minus x squared to
the minus 1/2.
So that's 1 over the square root of 10x minus x squared.
And now I need to multiply by the derivative of the inside
function, which is 10 minus 2x dx.
So really it's 10dx minus 2xdx but I just, you know, pulled
that dx out in front.
So we have a differential equal to a differential.
OK, and so, but we want this value at this particular point
in question.
So at this particular point in question we have x equals 2
and dx is equal to 1.
So the value of dy that we want is equal to 1/2 times--
well this is again, OK, 10 times 2 minus 2 squared.
That's 16 to the minus 1/2 times--
so here x equals 2.
10 minus 4 is 6.
Times dx is just this 1 that we're interested in.
OK so 16 to the minus 1/2, 16 to the 1/2 is 4, so 16 to the
minus 1/2 is 1/4.
So this is 1/2 times 1/4 times 6.
So that's 6/8, which is 3/4.
So our dy is 3/4.
So that means our linear approximation that we're after
is just f of 3, f of x plus dx is approximately equal to our
value y that we, at our original point, which was 4
plus the dy.
So plus 3/4.
So the linear approximation that we get from this method
of differentials is exactly the same approximation that we
would get if we did this the other way.
But I think it's kind of nice to do it in this
differential form.
Somehow it seems a little more straightforward sometimes.
And OK, andso the answer that we get out of
it is 4 plus 3/4.
All right, so that's that.