Download LESSON 21 - Math @ Purdue

LESSON 21: REVIEW OF DIFFERENTIALS
MARCH 8, 2017
Suppose a company produces a smoothie which contains crushed pineapple and
that the amount of profit the company makes on the sale of these smoothies depends
on the amount of pineapple used. Suppose that the price of pineapple increases and
the company wants to determine how it affects the overall profit if they reduce the
amount of pineapple in the smoothie. Depending on how complicated the formula
that describes the company’s profit is, this could be a daunting task. It is in this
type of situation that differentials can be useful. Differentials approximate how much
changing an input will change the output of a function. In this case, the input would
be the amount of pineapple in the smoothie and the output is the company’s profit.
√
Let’s consider a less complicated situation. Consider the function f (x) = x.
√
√
We know that f (9) = 9 = 3, but what is f (9.1) = 9.1? Obviously, if you have
a calculator this is easy. But there are some functions that even calculators have
trouble handling, which is where differentials come in.
Let x = 9 and x + ∆x = 9.1, that is, ∆x = .1. ∆x is the actual change in the
input x. Our goal is to approximate how this change in the input affects the output
function, that is, f (9.1) = f (x + ∆x). For this, we use calculus. Write
∆y = f (x + ∆x) − f (x) = f (9.1) − f (9) =
√
√
9.1 − 9.
∆y is the actual change the function f (x), which is our goal. In an ideal world, we
could compute this directly for any potential ∆x. But, in general, this is difficult to
compute even with a calculator so we settle for an approximation of ∆y instead.
Observe that
∆y
f (x + ∆x) − f (x)
=
∆x
∆x
(1)
looks a lot like a derivative. In fact, the only difference between equation (1) and
an actual derivative is that we need to take the limit as ∆x → 0. Because limits
deal with things getting really close together, if our ∆x is small we can make an
approximation of
∆y
∆x
using this derivative. We can write this like
f (x + ∆x) − f (x)
dy
∆y
=
≈ f ′ (x) =
.
∆x
∆x
dx
More helpfully, we have
(2)
∆y ≈ f ′ (x)∆x.
This just means that we can approximate the change in the function by taking the
change in the input and multiplying it by√the derivative of the function. Let’s apply
this to the example above. Since f (x) = x, we have
1
f ′ (x) = √ .
2 x
1
2
MATH 16020
Hence, by equation (2),
√
√
1
.1
1
9.1 − 9 = ∆y ≈ f ′ (9)∆x = √ (.1) =
= .
2(3)
60
2 9
This tells us that
√
9.1 ≈ 3 +
Using a calculator, we find
1
≈ 3.01666667
60
√
9.1 ≈ 3.0166207.
So our approximation is pretty good.
Note 1. We call dx and dy differentials. By the nature of derivatives (in particular, because we would assume that ∆x → 0), the smaller ∆x is, the better the
approximation of ∆y.
Think of ∆ as the actual change and d as the infinitesimal change. This is why
we use dx in an integral but not ∆x because ∆x is “too” big.
We can apply much of this thinking to functions of more than 1 variable as well.
This time, however, we consider how changes in x, y change z = f (x, y). Our notation
will be essentially the same with the goal of approximating
∆z = f (x + ∆x, y + ∆y) − f (x, y).
The total differential is given by
∂z =
∂z
∂z
dx +
dy = fx (x, y)dx + fy (x, y)dy.
∂x
∂y
We can use this formula to approximate ∆z. As with before, we think of ∆x ≈ dx
and ∆y ≈ dy. Hence,
∆z ≈
(3)
∂z
∂z
∆x +
∆y.
∂x
∂y
Let’s do an example that’s in the same vein as above. Suppose we have z =
√
f (x, y) = x2 + y 2 . Then if x = 3, y = 4,
f (3, 4) =
√
(3)2 + (4)2 =
√
9 + 16 =
√
25 = 5.
What if we wanted to find f (3.1, 3.8)? Then, ∆x = 3.1 − 3 = .1 and ∆y = 3.8 − 4 =
−.2. We also need to find the derivatives of f with respect to x and y:
)
∂ (√ 2
2
fx (x, y) =
x +y
∂x
2x
= √
2 x2 + y 2
x
=√
2
x + y2
MATH 16020
fy (x, y) =
3
)
∂ (√ 2
x + y2
∂y
2y
= √
2 x2 + y 2
y
=√
.
2
x + y2
Therefore, by equation (3) above,
y
x
∆x + √
∆y
∆z ≈ √
x2 + y 2
x2 + y 2
=√
3
(3)2
+
(4)2
(.1) + √
4
(3)2
+ (4)2
(−.2)
4
3
= (.1) + (−.2)
5
5
3
8
=
−
50 50
5
=−
50
1
= − = −.1
10
This implies
∆z = f (3.1, 3.8) − f (3, 4),
and so
f (3.1, 3.8) = f (3, 4) + ∆z ≈ 5 + (−.1) = 4.9.
√
Plugging it into a calculator, (3.1)2 + (3.8)2 ≈ 4.9041. So our approximation wasn’t
too far off.