APPROXIMATIONS USING DIFFERENTIALS

4. APPROXIMATIONS USING DIFFERENTIALS

Let’s consider some examples.

Example 1. Find approximately the value of

√ If f (x) = 1/ x, the desired difference is ∆f = f (0.25 − 10 −20 √ ) − f(0.25). But ∆f is approximately df = d(1/ x) with x = 0.25 and dx = −10 −20 .

√ d(1/ x) = (−1/2)x −3/2 dx = (−1/2)(0.25) −3/2 (−10 −20 ) = 4 × 10 −20 .

Now why not just use a computer or calculator for a problem like this? First note that we are subtracting two numbers which are almost equal to each other. If your calculator or computer isn’t carrying enough digits, you may lose all accuracy in the subtraction (see Chapter 1, Section 15, Example 1). So it may take you more time to check on this and to type the problem into the computer than to find df which you can probably do in your head! However, there is another important point here which is shown in the next example. For theoretical purposes, we may want a formula rather than a numerical result.

Section 4 Approximations using Differentials 197

Example 2. Show that when n is very large

2 n − (n + 1) 2 ∼ = n 3

(∼ = means “approximately equal to”). If f (x) = 1/x 2 , the desired difference is ∆f = f (n) − f(n + 1). But ∆f is approximately df = d(1/x 2 ) with x = n and dx = −1.

. (This result is used in obtaining the “correspondence principle” in quantum me-

chanics; see texts on quantum physics.) Also see Problem 17.

Example 3. The reduced mass µ of a system of two masses m 1 and m 2 is defined by µ −1 =m −1 1 +m −1 2 . If m 1 is increased by 1%, what fractional change in m 2 leaves µ unchanged? Taking differentials of the equation and substituting dm 1 = 0.01m 1 , we find

0 = −m −2 1 dm 1 −m −2 2 dm 2 , dm 2 dm 1 0.01m 1 dm 2

2 m =− m 2 =− m 2 or

= −0.01m 2 /m 1 .

For example, if m 1 =m 2 ,m 2 should be decreased by 1%; if m 2 = 3m 1 ,m 2 should

be decreased by 3%; and so on.

Example 4. The electrical resistance R of a wire is proportional to its length and inversely proportional to the square of its radius, that is, R = kl/r 2 . If the relative error in length measurement is 5% and the relative error in radius measurement is 10%, find the relative error in R in the worst possible case.

The relative error in l means the actual error in measuring l divided by the length measured. Since we might measure l either too large or too small, the relative error dl/l might be either +0.05 or −0.05 in the worst cases. Similarly |dr/r| might be as large as 0.10. We want the largest value which |dR/R| could have; we can find

dR/R by differentiating ln R. From R = kl/r 2 we find ln R = ln k + ln l − 2 ln r.

In the worst case (that is, largest value of |dR/R|), dl/l and dr/r might have opposite signs so the two terms would add. Then we would have:

or 25%. R

198 Partial Differentiation Chapter 4

Example 5. Estimate the change in

when x changes from π/2 to (1 + ǫ)π/2 where ǫ << 1/10. Recall from calculus that

df /dx = (sin x)/x. Then we want df = (df /dx) dx with x = π/2 and dx = ǫπ/2. Thus

Note that the approximations we have been making correspond to using a Taylor series through the f ′ term. We can write Chapter 1 equation (12.8) with the

replacements x → x + ∆x, a → x, x − a → ∆x, to get

f (x + ∆x) = f (x) + f ′ (x)∆x + f ′′ (x)(∆x) 2 /2! + · · · . Dropping the (∆x) 2 and higher terms we have the approximation we have been

using: = ∆f = f (x + ∆x) df ∼ − f(x) ∼ =f ′ (x)∆x = f ′ (x)dx.

PROBLEMS, SECTION 4

1. Use differentials to show that, for very large n, 1 1 3 (n + 1) 3 − 3 n ∼ = − n 4 . √

√ a 2. Use differentials to show that, for large n and small a,

n+a− n∼ = √ .

Find the approximate value of 10 +5− 10 26 .

3. The thin lens formula is

where f is the focal length of the lens and o and i are the distances from the lens to the object and image. If i = 15 when o = 10, use differentials to find i when o = 10.1.

4. Do Problem 3 if i = 12 when o = 18, to find i if o = 17.5. 5. Let R be the resistance of R 1 = 25 ohms and R 2 = 15 ohms in parallel. (See Chapter

2, Problem 16.6.) If R 1 is changed to 25.1 ohms, find R 2 so that R is not changed. 6. The acceleration of gravity can be found from the length l and period T of a pen-

dulum; the formula is g = 4π 2 l/T 2 . Find the relative error in g in the worst case if the relative error in l is 5%, and the relative error in T is 2%.

7. Coulomb’s law for the force between two charges q 1 and q 2 at distance r apart is F = kq 1 q 2 /r 2 . Find the relative error in q 2 in the worst case if the relative error in q 1 is 3%; in r, 5%; and in F , 2%.

8. About how much (in percent) does an error of 1% in a and b affect a 2 b 3 ? 9. Show that the approximate relative error (df )/f of a product f = gh is the sum of

the approximate relative errors of the factors. 10. A force of 500 nt is measured with a possible error of 1 nt. Its component in a

direction 60 ◦ away from its line of action is required, where the angle is subject to an error of 0.5 ◦ . What is (approximately) the largest possible error in the component?

Section 5 Chain Rule or Differentiating a Function of a Function 199

11. Show how to make a quick estimate (to two decimal places) of p(4.98) 2 − (3.03) 2 without using a computer or a calculator. Hint: Consider f (x, y) = px 2 −y 2 .

12. As in Problem 11, estimate p(2.05) 3 2 + (1.98) 2 .

13. Without using a computer or a calculator, estimate the change in length of a space diagonal of a box whose dimensions are changed from 200×200×100 to 201×202×99.

14. Estimate the change in

if x changes from 0.7 to 0.71. 15. For an ideal gas of N molecules, the number of molecules with speeds ≤ v is given

by the formula

4a 3 N Z v

2 n(v) = 2 √ x 2 e −a x dx,

where a is a constant and N is the total number of molecules. If N = 10 26 , estimate the number of molecules with speeds between v = 1/a and 1.01/a.

16. The operating equation for a synchrotron in the relativistic range is

qB = ωm[1 − (ωR) 2 /c 2 ] −1/2 ,

where q and m are the charge and rest mass of the particle being accelerated, B is the magnetic field strength, R is the orbit radius, ω is the angular frequency, and

c is the speed of light. If ω and B are varied (all other quantities constant), show that the relation between dω and dB can be written as

ω 3 B ω [1 − (ωR/c) 17. Here are some other ways of obtaining the formula in Example 2.

(a) Combine the two fractions to get (2n + 1)/[n 2 (n + 1) 2 ]. Then note that for large n, 2n + 1 ∼ = 2n and n + 1 ∼ = n.

„ „1 « 1 1 −2 (b) Factor the expression as

n 2 1−

, expand 1+ by bino-

n mial series to two terms, and then simplify.