math1013 calculus imachiang/1013/notes/1013... · math1013 calculus i derivatives v (x4.7, 4.9)1...

MATH1013 Calculus I

Derivatives V (§4.7, 4.9)1

Edmund Y. M. Chiang

Department of MathematicsHong Kong University of Science & Technology

November 12, 2014

1Based on Stewart, James, “Single Variable Calculus, Early Transcendentals”, 7th edition, Brooks/Coles, 2012

Briggs, Cochran and Gillett: Calculus for Scientists and Engineers: Early Transcendentals, Pearson 2013

Optimization Anti-derivatives Initial value problems Motion problems

Optimization

Anti-derivatives

Initial value problems

Motion problems


Optimization I (Briggs, et al, p. 293 a)Suppose an airline policy states that all baggage must bebox-shaped with a sum of length, width and height not exceeding64 in. What are the dimensions and volume of a square-based boxwith the greatest volume under these condition?

Figure: (Figure 4.53 (publisher))


Optimization I (Briggs, et al, p. 293 b)We want to maximize the volume of a rectangular box under aconstraint. Let

V = w2 h, 2w + h = 64, (0 ≤ w ≤ 32)

where w is the width and h is the height of the box. That is,

V = w2 h = w2 (64− 2w) = 64w2 − 2w3.

Assuming V has a maximum, then we have

0 = V ′(w) = 128 w − 6 w2 = 2w(64− 3w),

which holds only when w = 0, 64/3 ≈ 21.3. These are the criticalpoints. V ′′(w) = 128− 12w so that

V ′′(64

3

)= 128− 12

(64

3

)< 0.

This implies that V(643

)≈ 9, 709 is a local maximum. Since V is

a smooth function,so we need check the end points:

V (0) = 0, V (32) = 0. So V(643

)≈ 9, 709 is the abs. max.


Optimization I (Briggs, et al, p. 293 c)



Optimization II (Briggs, et al, p. 294 a)Suppose one is standing on the shore of a circular pond with aradius of 1 mile and to get to a point on the shore directlyopposite, first by swimming to a point P with speed 2 mile/hr andthen walk along the shore with speed 3 mile/hr. Choose the pointP to minimize the travel time.



Optimization II (Briggs, et al, p. 294 b)



Optimization II (Briggs, et al, p. 294 c)

We see that the chord length is 2r sin(θ/2) and the arc length isr(π − θ). Note that the radius is r = 1 mile. Thus the travel timeis given by

T (θ) =2 sin(θ/2)

2+π − θ

3

= sin(θ

2

)+π − θ

3, (0 ≤ θ ≤ π).

The critical point(s) is given by

0 =dT

dθ=

1

2cos

θ

2− 1

3.

That is, when cos θ/2 = 2/3, orθ = arccos(2/3) ≈ 1.68 rad = 96◦. The end points giveT (0) = π/3 ≈ 1.05 hr, and T (π) ≈ 1 hr. ButT (1.68 rad) ≈ 1.23 hr.


Optimization II (Briggs, et al, p. 294 d)



Optimization III (Briggs, et al, p. 295 a)An 8 ft height fence runs parallel to the side of a house 3 ft away.What is the lenght of the shortest ladder that clears the fence andreaches the house? Assume that the vertical wall of the house andthe horizontal ground have infinite extent.

Figure: (Figure 4.58a (publisher))


Optimization III (Briggs, et al, p. 295 b)

Figure: (Figure 4.58b (publisher))


Optimization III (Briggs, et al, p. 295 b)Let L be the length of the ladder, x be the distance of the fromthe foot of the ladder to the foot of the fence, and let b be theheight of the house. It follows from the last slide that we applyPythagoras theorem to obtain

L2 = (x + 3)2 + b2.

But similar triangles consideration yield 8/x = b/(3 + x) so that Lis a function of x only and its domain is x > 0:

L2 = (x + 3)2 +(8(x + 3)

x

)2= (x + 3)2

(1 +

64

x2

)It is easy to check

d

dxL2 =

2(x + 3)(x3 − 192)

x3

which equals zero if x3 = 192 or x ≈ 5.77. First order test impliesthat L(5.77) ≈ 15 ft is the minimum length.


Optimization IV (Stewart p. 330)Find the area of the largest rectangle that can be inscribed in asemicircle of radius r .

Ans. We want to maximise the rectangular area A = 2xy under theconstraint x2 + y2 = r2 as indicated in the following figure.

Figure: (Figure 4.7.9 (publisher))


Optimization IV (Stewart p. 330)

We may rewrite the area function as

A = A(x) = 2xy√

r2 − x2.

It is routine to check that

A′(x) =2(r2 − 2x2)√

r2 − x2,

implying that a critical point occurs at 2x2 = r2 or x = r/√

2.First order or the concavity tests can confirm that this value givesthe maximum. Or to use the extrema value theorem as explainedin Stewart to get the same conclusion. That is,

A( r√

2

)= r2

is the value that we want.


Primitives

Definition Let F (x) and f (x) be two given functions defined onan interval I . If

F ′(x) = f (x), holds for all x in I

then we say F (x) is called a primitive or an anti-derivative off (x). We use the notation

F (x) =

∫f (x) dx

to denoted that F is a primitive of f , and we call the process offinding a primitive F for f indefinite integration (or simplyintegration).


Examples

Remark We note that if F (x) is a primitive of f (x), thenF (x) + C , where C is an arbitrary constant, is also a primitive off (x) since

(F (x) + C )′ = F ′(x) + 0 = f (x).

We call C a constant of integration. Examples

•∫

x dx =1

2x2 + C ,

•∫

2x dx = x2 + C ,

•∫

x2dx =1

3x3 + C ,

•∫

x8dx =1

9x9 + C .


Non-uniquenessRecall from that Theorem 4.11 that if the derivatives of twofunctions F1, F2 agree on I : i.e., F ′1(x) = F ′2(x), then F1, F2 differby a constant. That is,

F1(x) = F2(x) + k , holds for all x in I

for some constant k .

Figure: (Publisher Figure 4.78)


Primitives of monomials xp

Theorem Let p 6= −1 be a real number. Then∫xp dx =

1

p + 1xp+1 + C ,

for some arbitrary constant C .Examples

•∫

2√

x3

dx =

∫x3/2dx =

x3/2+1

32 + 1

+ C =2

5x5/2 + C

•∫

12√

x3

dx =

∫x−3/2 dx =

x−3/2+1

−3/2 + 1+ C =

−2√x

+ C

•∫ 1

x2dx =

∫x−2 dx =

x−2+1

−2 + 1+ C =

−1

x+ C .

•∫ 1

x4dx =

x−4+1

−4 + 1+ C =

∫x−4 dx =

−1

3x3+ C .


Exercises

1.

∫x15 dx ,

2.

∫x−12 dx ,

3.

∫x−12 dx ,

4.

∫3x4 dx ,

5.

∫ √x dx ,

6.

∫4√

x5 dx ,

7.

∫1

7√

x8dx

8.

∫44√

xdx .

Remark Differentiate your answers to verify whether they arecorrect.


Linear combinationsSince

d(f + g)

dx=

df

dx+

dg

dxand

d(kf )

dx= k

df

dx,

where k is a constant. So we deduce∫f (x) + g(x) dx =

∫f (x) dx +

∫g(x) dx ,

and ∫k f (x) dx = k

∫f (x) dx ,

where k is a constant. We can easily generalize the aboveconsideration to linear combination of {f1, · · · , fn}∫

k1 f1(x) + k2 f2(x) + · · ·+ kn fn(x) dx

= k1

∫f1(x) dx + k2

∫f2(x) dx + · · ·+ kn

∫fn(x) dx

where {k1, · · · , kn}.


Finding primitive examples1.∫

x3/2 +1

x3/2+

2

x3dx =

∫x3/2dx +

∫x−3/2dx + 2

∫x−3dx

=(2

5x5/2 + c1

)− 2x−

12 + c2 + (−x−2 + c3)

=2

5x5/2 − 2x−

12 − x−2 + C .

2.∫y1/2(1 + y)2 dy =

∫y1/2(1 + 2y + y2) dy

=

∫y1/2dy +

∫2y3/2dy +

∫y5/2dy =

2

3y3/2 +

4

5y5/2 +

2

7y7/2 + C .

3.∫1

t1/2(1 + t)2 dt =

∫t−1/2(1 + 2t + t2) dt

=

∫t−1/2dt + 2

∫t1/2dt +

∫t3/2 dt = 2t1/2 +

4

3t3/2 +

2

5t5/2 + C .


Primitives of trigonometric functions

Figure: (Briggs, et al Table 4.9)


Primitives of various special functions

Figure: (Briggs, et al Table 4.10)


Examples (Briggs, et al)

• (p. 320)

∫(sin 2y + cos 3y) dy

• (p. 315)

∫(sec2 3x + cos

x

2) dx

• (p. 316)

∫(e−10x + ex/10) dx

• (p. 316)

∫4√

9− x2dx ,

• (p. 316)

∫1

16t2 + 1dt.


Initial value problems

The simplest differential equation of first order is of the form

f ′(x) = G (x), where G (x) is a given function;

f (a) = b, where a, b are given initial condition.

The f ′(x) = G (x) which is called a first order differentialequation, together with the initial value condition is called aninitial value problem (IVP).


An example of IVP

Example (Briggs, et al, p. 317) Solve the IVP

f ′(x) = x2 − 2x ,

f (1) =1

3.

So a simple integration yields

f (x) =

∫(x2 − 2x) dx =

1

3x3 − x2 + C .

But1

3= f (1) =

1

3· 13 − 12 + C implies that C = 1. So

f (x) =1

3x3 − x2 + 1.


Sketch of the last IVP

Figure: (Briggs, et al, Figure 4.87)


Example (Briggs, et al, p. 318)

Race runner A begins at the point s(0) = 0 and runs with velocityv(t) = 2t. Runner B starts at the point S(0) = 8 and runs withvelocity V (t) = 2. Find the positions of the runner for t ≥ 0 anddetermine who is ahead at t = 6.We have two IVP here. Namely,

ds

dt= v(t) = 2t, s(0) = 0.

with solution s(t) = t2 and

dS

dt= V (t) = 2, S(0) = 8

with solution S(t) = 2t + 8. Therefore, the two runners meetwhen s(t) = S(t), meaning that t2 − 2t − 8 = 0. That is, t = 4.


Example (Briggs, et al, p. 318) figure

Figure: (Briggs, et al, Figure 4.88)



Neglecting air resistance, the motion of an object moving verticallynear Earth’s surface is determined by the acceleration due togravity, which is approx. 9.8 m/s2. Suppose a stone is thrownvertically upward at t = 0 with a velocity of 40 m/s from the edgeof a cliff that is 100 m above a river.

1. Find the velocity v(t) of the object, for t ≥ 0, and inparticular, when the object starts to fall back down.

2. Find the position s(t) of the object, for t ≥ 0.

3. Find the maximum height of the object above the river.

4. With what speed does the object strike the river?



We measure the height function s(t) from the sea level and adoptthe upward direction to be our positive direction. Thus the initialheight is s(0) = 100.

(1) The accelerationdv

dtdue to gravity pointing to the centre of

Earth, which is therefore negative. In fact we have the IVP:

dv(t)

dt= v ′(t) = −9.8, v(0) = 40

Solving the DE gives v(t) = −9.8t + C . Thus

40 = v(0) = −9.8(0) + C ,

giving C = 40. Hence v(t) = −9.8t + 40.The object starts to fall back down after it reached the maximumheight, where v(t) = 0, that is, when v(t) = −9.8t + 40 = 0,giving t ≈ 4.1 s.



(2) The height s(t) satisfies the IVP

ds(t)

dt= v(t) = −9.8t + 40, s(0) = 100.

Solving the DE yields

s(t) = −4.9t2 + 40t + C .

The initial condition s(0) = 100 implies that 100 = s(0) = C . So

s(t) = −4.9t2 + 40t + 100.



(3) As a result the maximum height is reached when the parabolics(t) is at its critical point:

0 =ds

dt= v(t) = −9.8t + 40,

That is, when t ≈ 4.1 s. Thus the maximum height is

s(4.1) ≈ 182 m.

(4) The object hits the sea when s(t) = 0. Solving the quadraticEqn s(t) = 0 gives us two roots, namely t ≈ −2 (which is to bediscarded) and t ≈ 10.2. So the velocity of the object when itstrike the sea is given by

v(10.2) ≈ −9.8(10.2) + 400 = −59.96 ≈ −60.

math1013 calculus imachiang/1013/notes/1013... · math1013 calculus i derivatives v (x4.7, 4.9)1...

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