Section 13.4

1. Problem Let f(x, y) = sin (x⁴ + 3 y) where x = 5 t and y = t² + 1 and denote f(x(t), y(t)) ≡ F(t). Evaluate df/dt using the chain rule:

∂f

∂x

∂f

∂y

Note : Actually, (1.1) is not the end of the "chain differentiation story," for in computing [(∂f)/(∂x)], we set x⁴ + 3 y = u, so that again applying chain differentiation,

∂f

∂x

(sin u)

∂u

∂x

= etc.,

and similarly for [(∂f)/(∂y)].

(Solution)

Straightforward, no tricks, i.e.,

= (cos (x⁴ + 3 y) 4 x³) 5 + (cos (x⁴ + 3 y) 3)(2 t).

2. Problem Let f(x, y)=exp x y, and denote f(x(t), y(t)) = F(t). Evaluate dF/dt in each, using the chain rule.

(a) x = t² − 1, y = sin 3 t

(Solution) (a)

∂F

∂x

∂F

∂y

(y e^xy)

∂x

∂t

+(x e^xy)

∂y

∂t

(y e^xy) (2t)+(x e^xy) (3 cos 3 t).

(1)

8. Problem Show that u = f(x + c t) + g(x − c t) satisfies the partial differential equation c² u_xx − u_tt = 0 (the wave equation), where c is a constant and f and g are arbitrary twice-differentiable equation.

(Solution) u = f(a) + g(b) where a ≡ x + c t and b ≡ x − c t. u_x = f_a a_x + g_b b_x = f_a + g_b, u_xx = f_aa+g_bb and u_t = f_a c− g_b c. Therefore, u_tt = f_aa c² + g_bb c² = c² (f_aa + g_bb) = c² u_xx.

Section 13.5

Please note that when a function is to be expanded by te Taylor series about (a,b), the final expression must be in the form of ∑_i ∑_ja_ij (x − a)ⁱ (y − b)^j. Note also that you can bypass using the Taylor series formula for most of the cases by using (or combining) well-known series such as geometric series.

1. Problem Expand the given function about the indicated point a, through third-order terms. Note: (x − a)ⁿ is the n th order.

(b) e^{−2 x}, a=5.

(d) ln x, a = 2

(e) 1/(1 + x²), a = 1.

(f) 1/(1 + x²), a = −1.

(g) sin x, a = 2.

(i) x (x − 1)², a = 1.

(j) x³ (x⁴ − 1) + 5, a = 0.

(Solution) (b)

e^{−2 x}

e^{−2( (x − 5) + 5 )} = e⁻¹⁰ e^{−2 (x − 5)}

e⁻¹⁰ (1 −2(x − 5) +

2² (x − 5)²

−

2³ (x − 5)³

+ …).

(2)

(d)

ln x

ln ((x − 2) + 2) = ln (2 (1 +

x − 2

)) = ln 2 + ln (1 +

x − 2

)

ln2 + (

x − 2

) −

(

x − 2

)² +

(

x − 2

)³−

(

x − 2

)⁴ + …

(e)

1+x²

1+(x−1+1)²

1 + (x−1)²+ 2(x−1)+1

(define t ≡ (x−1))

⎛
⎜
⎝

1+ ( t +

t²

)

⎞
⎟
⎠

⎛
⎝

1 −

⎛
⎝

t +

t²

⎞
⎠

⎛
⎝

t +

t²

⎞
⎠

−

⎛
⎝

t +

t²

⎞
⎠

+ …

⎞
⎠

⎛
⎝

1 −

⎛
⎝

(x−1)+

(x−1)²

⎞
⎠

⎛
⎝

(x−1)+

(x−1)²

⎞
⎠

−

⎛
⎝

(x−1)+

(x−1)²

⎞
⎠

+…

⎞
⎠

−

x − 1

(x − 1)²

−

(x − 1)⁴

+ …

(f) Note: In this problem and (e), using directly the Taylor series formula is faster.

1 + x²

1 + x

(1 + x)²

−

(1 + x)⁴

+ …

(g)

sin x

sin ((x − 2) + a)

sin (x − 2) cos 2 + cos (x − 2) sin 2

cos 2

⎛
⎝

(x − 2) −

(x − 2)³

(x − 2)⁵

− …

⎞
⎠

+ sin2

⎛
⎝

1 −

(x − 2)²

(x − 2)⁴

− …

⎞
⎠

(i)

x (x−1)² = ((x − 1) + 1)(x − 1)² = (x − 1)³ + (x − 1)²

(j)

x³ (x⁴ − 1) + 5 = 5 − x³ + x⁷.

6. Problem In the sentence following (9), we speak of assuming the continuity of the derivative f⁽ⁿ⁾(x). That comment begs a question as to how a derivative can fail to be continuous. For instance, suppose f(x) is the "ramp function" x H(x), where H(x) is the Heaviside step function. Then f′(x) = H(x) has a jump discontinuity at x = 0. However, at that point the original function f(x) = H(x) has a "kink" and is not differentiable. Thus, it is hard to imagine how a derivative f′(x) can exist at a point x, yet fail to be continuous there. Nonetheless, the function

g(x) =

⎧
⎪
⎨
⎪
⎩

x² sin

x ≠ 0

x=0

shows that such behavior is possible because g′(x) exists at x = 0 but is not continuous there. The problem that we pose is for you to verify that g′(x) exists at x=0 but is not continuous there. Hint: Evaluate g′(0) and g′(x) for x ≠ 0, and show that lim_x→0 g′(x) doesn't equal g′(0).

Solution

g′(0)

lim
h→ 0

g(h) − g(0)

lim
h→ 0

h² sin (1/h)

lim
h→ 0

h sin

Note that 0 ≤ | hsin (1/h)| ≤ 1 so when h→ 0, | hsin (1/h)| → 0 and g′(0)=0. For x ≠ 0, g′(x)=2 x sin(1/x)−cos(1/x) and although x sin(1/x) goes to 0 as x→ 0, the cos(1/x) term is indefinite so g′(x) is not continuous at x=0.

Section 13.6

1. Problem Given f(x, y) = 0 and a point (x₀, y₀) such that f(x₀, y₀) = 0, see if the conditions of Theorem 13.6.1 are met. If so, develop the implicit function y(x) in a Taylor series about x₀, through second-order terms, as we did in Example 4.

(b) x² + 4y² − 4 = 0; (0, 1)

Solution f(x, y) = x² + 4 y²−4. [(∂f)/(∂y)] = 8 y. Thus, [(∂f)/(∂y)]_{(x,y) = (0,1)} = 8. Yes, there does exist a function y = y(x) in the neighborhood of (0, 1). By differentiating f(x, y) with respect to x, one obtains 2 x + 8 y y′ = 0 or y′ = −[(x)/(4y)]. Thus, y′(0) = 0. Differentiating the above with respect to x again gives 2 + 8 y′y′+ 8 y y" = 0 which leads to y" = − [(1 + 4 (y′)²)/(4 y)]. Thus, y"(0) = −1/4. Finally, using the Taylor series formula,

y(x) = y(0) + y′(0) x + y"(0)/2! x² + … = 1 − 1/8 x² + ….

2 Problem In each case, find y′(x) and y"(x).

(a) x y − y³ = 1 (d) x y e^y = 1

Solution (a) Differentiating the both sides of x y − y³ = 1 with respect to x yields y + x y′− 3 y² y′ = 0 which is solved for y′ as y′ = −y/(x − 3 y²). Differentiating with respect to x again gives y′+ y′+ x y" − 6 y (y′)² − 3 y² y" = 0 which can be solved for y" as y" = (6 y (y′)² − 2y′)/(x − 3 y²).

(d) Differentiating the both sides of x y e^y=1 with respect to x yields y e^y+ x y′e^y + x y e^y y′ = 0. Therefore,

y′(x) = −

x (1 + y)

Differentiating the above with respect to x again gives

⎛
⎝

−

x (1 + y)

⎞
⎠

′

−

y′x (1 + y) − y (1 + y + x y′)

x² (1 + y)²

−

y′x − y − y²

x² (1 + y)²

(3)

3. Problem Solve for y_x|_z (i.e., holding z fixed) and z_x|_y.

(a) x y + sin (x + z) − z² = 5

Solution

For y_x, it is important to interpret y(x, z) as the dependent variable and x and z as independent variables.

Differentiating the both sides with respect to x yields y +x y_x + cos (x + z) = 0 which can be solved for y_x as

y_x = − (y + cos (x + z))/x.

Similarly, for z_x, differentiating with respect to x yields y + cos (x + z)(1 + z_x) − 2z z_x = 0 which can be solved for z_x as

z_x = (y + cos (x+z)) /(2 z − cos (x + z)).

4. Problem Apply Theorem 13.6.2 to see if implicit functions u(x, y) and v(x, y) exist in some neighborhood of (x₀, y₀) given the values (x₀, y₀, u₀, v₀) satisfying the two relations.

(a) x − u cos v = 0, y − u sin v = 0 ;(0, 0, 0, 0)

(b) x − u cos v = 0, y − u sin v = 0; (0, 2, 2, π/2).

Solution (a) f₁ ≡ x − u cos v and f₂ ≡ y − u sin v.

Thus,

∂(f₁, f₂)

∂(u,v)

⎢
⎢
⎢
⎢

− cos v

u sin v

− sin v

−u cos v

⎢
⎢
⎢
⎢

= u (cos² v + sin² v) = u.

Substituting (u, v) = (0, 0) yields 0 from which no conclusion can be drawn.

(b) This time, substituting (u, v) = (2, π/2) yields 2. Thus, implicit functions, u(x, y) and v(x, y), exist.

5. Problem Evaluate u_y.

(a) x − y + u² + v² = 1, x + y + u³ e^v = 2

Solution

(a) By differentiating the both formulas with respect to y, one gets

−1 +2 u u_y + 2v v_y

1 + 3 u² u_y e^v + u³ e^v v_y

which can be solved for u_y as

u_y =

u³ + 2 e^−v v

2 u⁴ − 6 u² v

6. Problem Evaluate the indicated Jacobian(s).

(a)

f(u, v) = 3 u v², g(u, v) = u² − v² ;

∂(f, g)

∂(u, v)

Solution (a)

J(u,v) =

∂(f, g)

∂(u, v)

⎢
⎢
⎢
⎢

3 v²

6 u v

2 u

− 2 v

⎢
⎢
⎢
⎢

= − 6 v³ − 12 u² v.

10. Problem (Chain rule) Recall from the calculus that if u = u(x(s)), then

which result is an example of the chain rule; equation (44) is a special case of (10.1), where s is u.

If u and v are functions of x and y, and x and y, in turn, are functions of r and s, then (10.1) generalizes to

∂(u, v)

∂(x, y)

∂ (x, y)

∂(r, s)

∂(u, v)

∂(r, s)

Similarly,

∂(u, v, w)

∂(x, y, z)

∂ (x, y, z)

∂(r, s, t)

∂(u, v, w)

∂(r, s, t)

and so on. Prove (10.2). Hint: Proceed essentially as we did in (50)-(51).

Solution

(a) As suggested by the hint, if you expand the both sides and compare term by term using the chain differentiation rule, the identity can be proven (albeit tedious algebra). However, if you remember that ∂(x, y)/∂(u, v) is the scaling factor from the (u, v) coordinate system to the (x, y) coordinate system as discussed in class, the left hand side can be interpreted as the scaling factor from (u, v) to (x, y) multiplied by another scaling factor from (r, s) to (u, v), which should be equal to the scaling factor from (r, s) to (x, y) which is the right hand side.

If you want to pursue long algebra, here is a brief procedure: Note that x and y are functions of r and s through u and v, i.e.,

x( u(r, s), v(r, s))

(4)

y( u(r, s), v(r, s))

(5)

Thus,

∂x

∂r

∂x

∂u

∂r

∂x

∂v

∂r

∂y

∂s

∂y

∂u

∂s

∂y

∂v

∂s

∂x

∂s

∂x

∂u

∂s

∂x

∂v

∂s

∂y

∂r

∂y

∂u

∂r

∂y

∂v

∂r

(chain differentiation). Therefore, the right hand side is (x_u u_r + x_v v_r)(y_u u_s + y_vv_s) −(x_u u_s + x_v v_s)(y_u u_r + y_v v_r) which is identical to the right hand side upon expansion.

11. Problem Verify (10.2), above, for these cases.

(a) x = ucos v, y = usin v, and u = r + s, v = r²+ s².

Solution

(a) Easy.

Section 13.7

13. Problem (a) Find the point on the line x + 2y = 2 that is closest to the origin, by the method of elimination and also by Lagrange's method. Prove that your result is indeed a minimum.

Solution

(a) Employ the Lagrange multiplier method. F ≡ x² + y² + λ(x + 2 y − 2). ∂F/∂x = 2 x + λ = 0,∂F/∂y = 2 y + 2 λ = 0,∂F/∂λ = x + 2 y −2 = 0. Solving for (x, y, λ) yields λ = −4/5, x = −λ/2, y = − λ.

(x, y) = (2/5, 4/5)

16. Problem Find the point (x, y), on the given plane , that is closest to the point (1, −2).

(a) 2 x + y = 4

Solution (a) (x − 1)² + (y + 2)²→ min. subject to 2 x + y −4 = 0. F ≡ (x − 1)² + (y + 2)² + λ(2 x + y −4). From ∂F/∂x = 0, ∂F/∂y = 0,∂F/∂λ = 0, one obtains

x = 13/5, y = −6/5, λ = − 8/5

17. Find the point (x, y, z), on the given plane, that is closest to the point (2, 0, −1).

(a) x + y + z = 4

Solution (a) (x−2)² + y² + (z + 1)² → min. subject to x + y + z −4 = 0. The Lagrange multiplier method yields

x = 3, y = 1, z = 0, λ = −2

21. Problem (More than one constraint) (a) Extend the Lagrange method to cover cases that include more than one constraint. Specifically, explain how to solve the problem

f(x, y, z)

extremum,

g₁(x, y, z)

c₁,

g₂(x, y, z)

c₂

by that method, and explain the logic behind each step. HINT: You should end up extremizing f^* = f − λ₁g₁ −λ₂ g₂, where λ₁ and λ₂ are Lagrange multipliers.

(b) Apply that method to the problem

x² + y² + z² = extremum,

g₁

x² + 4 y² + 4 z² = 4,

g₂

x + y + z = 0.

Solution (b)

F ≡ x² + y² + z² + λ₁ (x² + 4 y² + 4 z² −4) +λ₂ (x + y + z).

The Lagrange multiplier method (∂F/∂x = 0, ∂F/∂y = 0,∂F/∂z = 0, ∂F/∂λ₁ = 0,∂F/∂λ₂ = 0) yields

(x, y, z, λ₁, λ₂)

(0, −

√2

, −

, 0)

(0,

√2

, −

√2

, −

, 0)

(

√3

, −

√3

, −

√3

, −

√3

)

(−

√3

, −

√3

)

(6)

The first two choice renders x² + y² + z² to 1 while the last two choices render x² + y² + z² to be 2. So choose the last two.

Section 13.8

1. Problem Apply the Leibniz rule:

(a)

⌠
⌡

t²

0

sin (t x²) dx

(d)

dα

⌠
⌡

−α

−2α²

e^αx³dx

Solution (a)

⌠
⌡

t²

0

sin ( t x²) dx

⌠
⌡

t²

0

cos (t x²) (x²) dx + (2 t) sin (t (t²)²)

⌠
⌡

t²

0

x² cos (t x²) d x + 2 t sin (t⁵)

(7)

(d)

dα

⌠
⌡

−α

−2 α²

e^αx³ d x

⌠
⌡

−α

−2 α²

x³ e^αx³ dx + (−1) e^α(−α)³ − (− 4α) e^{α(−2 α)³}

⌠
⌡

−α

−2 α²

x³ e^αx³ dx − e^−α⁴+ 4 αe^−8α⁷.

(8)

2. Problem Derive the Taylor series of the given function f(x) about x = 0, up to and including terms of second order, using the Leibniz rule to obtain f′(x) and f"(x).

(a)

⌠
⌡

x

0

e^−t² dt

Solution (a) Try changing the variable, x, with t, i.e.

f(t) ≡

⌠
⌡

t

0

e^−x² dx.

Using the Leibniz rule, one gets

f′(t)

⌠
⌡

t

0

∂

∂t

( e^−x² ) dx +

⎛
⎝

⎞
⎠

{ e^−x²_{x→ t}} − 0

0+ e^−t²−0

e^−t².

(9)

So f"(x) = −2 x e^−x² and f"′(x) = −2 e^−x² + 4 x² e^−x². f(0)=0, f′(0)=1, f"(0)=0, f"(0)=−2. Therefore,

f(x)

f(0)+ f′(0)x +

f"(0)

x² +

f"′(0)

x³ + …

x−

x³

+ …

3. Problem Show, by repeated differentiation of the formula

⌠
⌡

∞

0

e^−ax dx =

that ∫₀^∞ xⁿ exp (−a x) dx = n! for n = 0, 1, 2, 3, 4, ….

Solution

Differentiating the both sides with respect to a yields ∫₀^∞ (−x) e^{−a x} dx = (−1) a⁻² . Differentiating again yields ∫₀^∞ (−x)² e^{−a x} dx = (−1)(−2) a⁻³ and again to get ∫₀^∞ (−x)³ e^{−a x} dx = (−1)(−2)(−3) a⁻⁴ . Therefore, ∫₀^∞ (−x)ⁿ e^{−a x} dx = (−1)ⁿ n! a⁻⁽ⁿ⁺¹⁾. Setting a = 1 yields

∫₀^∞ xⁿ e^−x dx = n!

5 Problem (a) To evaluate

I =

⌠
⌡

∞

0

(ln x)²

1 + x³

dx,

differentiate the formula

⌠
⌡

∞

0

x^α

1 + x³

dx =

3 sin (

α+ 1

π)

a suitable number of times.

Thus, show that

I =

10 π³

(81 √3)

Solution

⌠
⌡

∞

0

x^a

1 + x³

dx =

3 sin

⎛
⎝

a + 1

⎞
⎠

Differentiating the above with respect to a yields

⌠
⌡

∞

0

ln x x^a

1 + x³

dx =

−π²

cos

⎛
⎝

1 + a

⎞
⎠

sin²

⎛
⎝

1 + a

⎞
⎠

Differentiating again to get

⌠
⌡

∞

0

(ln x)² x^a

1 + x³

dx =

π³

cos²

⎛
⎝

1 + a

⎞
⎠

+ 1

sin³

⎛
⎝

1 + a

⎞
⎠

By substituting a = 0 in the above, one gets

⌠
⌡

∞

0

(ln x)²

1 + x³

dx =

10 π³

81 √3

6. Problem Show that

(a)

y(x) =

⌠
⌡

x

a

(x − t)³ f(t) d t

satisfies the initial-value problem y""(x) = f(x); y(a) = y′(a) = y"(a) = y"′(a) = 0.

Solution (a) Using the Leibniz rule,

y′(x)

⌠
⌡

x

a

3 (x − t)² f(t) dt

y"(x)

⌠
⌡

x

a

6 (x − t) f(t) dt,

y"′(x)

⌠
⌡

x

a

f(t) dt,

and y""(x) = f(x) . If you substitute x for a in the above, the initial conditions are all satisfied.

7. Problem Evaluate ∫₀¹ x^0.7 ln x dx by differentiating the known formula

⌠
⌡

1

0

x^a dx =

a + 1

(a > −1).

Solution Differentiating ∫₀¹ x^a dx = [1/(1 + a)] with respect to a yields

⌠
⌡

1

0

x^a ln x dx = −

(1 + a)²

Therefore,

⌠
⌡

1

0

x^0.7 ln x dx = −

1.7²

8. Problem Show that

⌠
⌡

1

0

x³ − 1

ln x

dx = ln 4.

HINT: Considering

I(a) =

⌠
⌡

1

0

x^a−1

ln x

dx, (a ≥ 0)

show that I′(a)=1/(a + 1) with the "initial condition" I(0) = 0. Solve for I(a) and set a = 3.

Solution

I(a) ≡

⌠
⌡

1

0

x^a − 1

ln x

dx.

I′(a)

⌠
⌡

1

0

x^a ln x

ln x

⌠
⌡

1

0

x^a dx

1 + a

By solving this differential equation, one gets I(a) = ln (1+a) + C where C is an integral constant. Since I(0) = 0, C = 0. Therefore,

⌠
⌡

1

0

x^a−1

ln x

dx = ln (1 + a)

and

⌠
⌡

1

0

x³ − 1

ln x

dx = ln 4.

10. Problem Consider a one-dimensional fluid flow, say flow in a channel of cross-sectional area A, with velocity v(x, t) m/sec and mass density σ(x,t) grams/m³; x is measured positive downstream and t is the time. Consider a "control-volume" a(t) ≤ x ≤ b(t) that drifts with the fluid; ie., da/dt = v(a(t), t) and da/dt = v(b(t), t). Then show that the principle of conservation of mass (i.e., that the amount of mass within the control volume remains constant) can be expressed as

⌠
⌡

b(t)

a(t)

σ(x, t) A dx = A

⌠
⌡

b(t)

a(t)

⎡
⎣

∂σ

∂t

∂

∂x

(v σ)

⎤
⎦

dx = 0.

Solution This problem will be discussed in class (i.e. equation of continuity).

⌠
⌡

b(t)

a(t)

σ(x, t) A dx

⎛
⎝

⌠
⌡

b(t)

a(t)

∂σ

∂t

dx+(b′(t) σ_b − a′(t) σ_a)

⎞
⎠

⎛
⎝

⌠
⌡

b(t)

a(t)

∂σ

∂t

dx+ [ v σ]_a^b

⎞
⎠

⎛
⎝

⌠
⌡

b(t)

a(t)

∂σ

∂t

dx+

⌠
⌡

b(t)

a(t)

∂

∂x

(v σ) dx

⎞
⎠

(10)

File translated from T_EX by T_TH, version 4.03.
On 28 Aug 2022, 18:18.