Module 19 - Further Calculus 1 (self Study)

SCHOOL OF MATHEMATICS MATHEMATICS FOR PART I ENGINEERING Self Study Course

MODULE 19

FURTHER CALCULUS I

Module Topics 1. Chain rule for partial derivatives 2. Higher partial derivatives 3. Total differentials and small errors

A:

Work Scheme based on JAMES (THIRD EDITION)

1. The basic idea of partial differentiation was introduced at the end of Module 3, and a few simple examples were considered there. This module explores further features of partial derivatives, since the latter arise in many engineering situations. Study the paragraph in the lower half of p.630 which defines the partial derivative of a function of n variables, and then work through Example 9.28. In this Example f depends on the variables x, y and z, and a partial derivative with respect to one of the variables is found by differentiating with respect to that variable keeping all the other variables fixed. You should remember that fx , fy , . . . are alternative notations ∂f ∂f , , . . . respectively. for the partial derivatives ∂x ∂y 2. Read section 9.7.2 on directional derivatives, starting on p.631. The relationship z = f (x, y) gives a surface in three dimensions. You can easily draw a tangent plane at any point on the surface, but any vector lying in this plane is a tangent to the surface so there is no unique slope at a point on the surface unless a direction is specified. J. considers the general situation on p.631-633 and derives the general result (9.26). An illustration of this result is given in Example 9.29, which you should carefully look at. 3. Read the theory below the horizontal line on p.633. Figure 9.28 illustrates the special cases of formula ∂f (9.26) which arise when α = 0 and α = π/2. Note that is the gradient of the line produced when the ∂x ∂f is the gradient of the line produced when surface z = f (x, y) intersects a plane y = constant , whereas ∂y the surface z = f (x, y) intersects a plane x = constant . It should be noted that the angle α shown on Figure 9.28, and in the caption below the figure, is different from the α used elsewhere in this section. Therefore, you should replace α in both Figure 9.28(a) and the caption by a different symbol, γ say. ***Do Exercises 57, 58 on p.634*** 4. Study the first 8 lines of section 9.7.4 on p.635. If z = f (x, y) and both x and y depend on the two variables s and t, then the partial derivatives of z with respect to s and t are found using the chain rule (9.27). You may wish to look at the proof of the formula in the middle of that page. –1–

Work through Example 9.30. 5. In some circumstances z = f (x, y) but x and y both depend only on the single variable t. In these situations the chain rule (9.27) reduces to the form df ∂f dx ∂f dy dz = = + . dt dt ∂x dt ∂y dt Observe that (9.27) contains only partial derivatives but that ordinary derivatives appear in the above expression. The type of derivative used clearly depends on whether the variable being differentiated depends on one, or more than one, variable. Work through Example 9.30 which illustrates the special form of the chain rule displayed above. ***Do Exercises 60(a), 61 on p.637*** ∂2f ∂y∂x means differentiation of the function f with respect to x and then differentiating the answer with respect ∂f ∂f and . to y . It does NOT mean the product of the two first derivatives ∂x ∂y 6. Study section 9.7.6 on calculating higher order partial derivatives. It should be emphasised that

For most functions occurring in engineering problems the mixed second derivatives are equal, i.e. the order in which the two partial derivatives are performed does not matter. This is a very useful property. Work through Examples 9.33 and 9.34, which are both straightforward. 7. Next work through Example 9.35, which is a much more difficult Example. The first derivatives are easily calculated, and the answers appear at the top of p.640. The determination of second derivatives is not so simple. Expanding the analysis near the top of p.640, we obtain         ∂ ∂f ∂f ∂ ∂f ∂ ∂f ∂ ∂f fxx = = 2x +y = 2x + y . (1) ∂x ∂x ∂x ∂s ∂t ∂x ∂s ∂x ∂t Both terms on the right-hand side of (1) involve differentiation of products of functions, and will be treated in turn. For the first term          ∂f ∂f ∂ ∂f ∂ ∂ ∂f ∂ ∂f 2x = (2x) + 2x =2 + 2x , ∂x ∂s ∂x ∂s ∂x ∂s ∂s ∂x ∂s and the second term on the right-hand side of the last equation can be calculated using the chain rule ∂z ∂s ∂z ∂t ∂z = + ∂x ∂s ∂x ∂t ∂x

(2)

∂f , as shown in J. The calculation of the second partial derivative ∂s on the right-hand side of equation (1) is easier, since y is kept fixed when differentiating partially with respect to x, so     ∂f ∂ ∂f ∂ y =y . ∂x ∂t ∂x ∂t

with z replaced by the partial derivative

The latter partial derivative can then be calculated using the chain rule (2) with z replaced by bining the answers and rearranging then leads to the result (9.28).

∂f . Com∂t

In an analogous way the second derivative fyy can be found, expression (9.29), leading to the required result at the bottom of the page. –2–

***Do Exercises 71, 76 on p.641*** 8. Continue on p.641 by studying section 9.7.8 on the total differential and small errors, up to the small paragraph at the top of p.643. [Note that there is a misprint in the equation at the top of p.642. In the second term in the first numerator on the RHS of this equation replace f (x, y + ∆x) by f (x, y + ∆y).] You may find the theory in this section a little confusing so a summary of the main argument is provided below. Consider a function u = f (x, y) and let ∆x and ∆y be the increments in x and y. When both x and y are changed to x + ∆x and y + ∆y the corresponding change in u is represented by ∆u. J. defines the total differential by ∂f ∂f du = dx + dy . (3) ∂x ∂y The infinitesimals dx, dy and du and the corresponding increments ∆x, ∆y and ∆u are approximately the same, so ∂f ∂f ∆u ' ∆x + ∆y , (4) ∂x ∂y which is a very useful result when estimating small errors. Expressions (3) and (4) can be generalised in an obvious way to functions of more than two variables (see the bottom of p.642). 9. Work through Example 9.36. 10. Read carefully the paragraph below the horizontal line in the middle of p.643 and then work through Example 9.37. After using the appropriate version of result (3) above, you should note that in this Example the maximum value of ∆V is obtained when both ∆r and ∆h are positive. In an analogous way the minimum of ∆V occurs when both ∆r and ∆h are negative. In the final part of this Example the maximum value of V (= πr2 h) clearly occurs when both r and h take their maximum values, and conversely for the minimum of V . 11. Finally work through Example 9.38. Here the value of x is assumed to have some prescribed constant value whereas a and b are known only approximately. Note in the solution to 9.38 the meaning of relative error bounds for a and b. Using the standard procedure an expression for dy is obtained. In order to work with relative errors the equation is divided on the left-hand side by y and on the right-hand side by dy the equivalent expression ae−bx , leading to a formula for which has the form y dy = g − h = g + (−h). y A standard inequality, often called the triangle inequality, then implies dy ≤ |g| + |h| , y as used in the solution in J. Note that in applications you are often told the percentage change in a quantity. For instance, a 3% decrease ∆R × 100 = −3, or ∆R = −0.03R. in R means that R ***Do Exercises 83, 85 on p.645*** –3–

B:

Work Scheme based on STROUD (FIFTH EDITION)

Much of the material in this module is in S. programmes 10 and 11, although the topics are covered in a different order. Turn first to S. p.674 and work through frames 15–37. Next move to programme 11 on p.691 and work through frames 1–15. Remaining in programme 11, study frames 24–29 on extended use of the chain rule.

Specimen Test 19 1. (i) If f is a function of x and y, but x and y depend only on t, use the chain rule to write down an df in terms of derivatives of f, x and y. expression for dt (ii) Hence, if f (x, y) = xy − y 2 cos x, x = t2 and y = t, determine an expression for

df in terms of t only. dt

2. (i) If ω is a function of x and y and if each of the variables x and y depends on both r and θ, write down ∂ω ∂ω the general equations for and in terms of derivatives of ω, x and y. ∂r ∂θ ∂ω ∂ω and in the case where ω = x2 y +xy 2 , x = r cos θ (ii) Hence find (in terms of r and θ) the quantities ∂r ∂θ and y = r sin θ. For the function f (x, y) = y sin x + x2 ey obtain

3. (i)

4.

∂f , ∂x

(ii)

∂f , ∂y

(iii)

∂2f , ∂x∂y

(iv)

∂2f . ∂y∂x

If f is a function of x and y and the substitutions u = x + t and v = x − t are introduced (i) show that

∂f ∂f ∂f = + , ∂x ∂u ∂v

(ii) deduce an expression for

∂2f in terms of derivatives of f with respect to u and v. ∂x2

5. (i) Given that u = u(x, y), write down an approximate expression for the total increment ∆u in u due to increments ∆x and ∆y in x and y respectively. 1 2 πr h where h is the height and r is the radius of the 3 base. If h is decreased by 2% and r is increased by 2% use partial fractions to calculate the approximate percentage change in V .

(ii) The volume V of a right circular cone is V =

–4–