MATH1013 Calculus I. Derivatives V ( 4.7, 4.9) 1

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1 1 Based 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 MATH1013 Calculus I Derivatives V ( 4.7, 4.9) 1 Edmund Y. M. Chiang Department of Mathematics Hong Kong University of Science & Technology November 12, 2014

2 Optimization Anti-derivatives Initial value problems Motion problems

3 Optimization I (Briggs, et al, p. 293 a) Suppose an airline policy states that all baggage must be box-shaped with a sum of length, width and height not exceeding 64 in. What are the dimensions and volume of a square-based box with the greatest volume under these condition?

4 Optimization I (Briggs, et al, p. 293 b) We want to maximize the volume of a rectangular box under a constraint. Let V = w 2 h, 2w + h = 64, (0 w 32) where w is the width and h is the height of the box. That is, V = w 2 h = w 2 (64 2w) = 64w 2 2w 3. Assuming V has a maximum, then we have 0 = V (w) = 128 w 6 w 2 = 2w(64 3w), which holds only when w = 0, 64/ These are the critical points. V (w) = w so that V ( 64 ) ( 64 ) = < ( ) This implies that V , 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 , 709 is the abs. max.

5 Optimization I (Briggs, et al, p. 293 c) Figure: (Figure 4.54 (publisher))

6 Optimization Anti-derivatives Initial value problems Motion problems Optimization II (Briggs, et al, p. 294 a) Suppose one is standing on the shore of a circular pond with a radius of 1 mile and to get to a point on the shore directly opposite, first by swimming to a point P with speed 2 mile/hr and then walk along the shore with speed 3 mile/hr. Choose the point P to minimize the travel time.

7 Optimization II (Briggs, et al, p. 294 b) Figure: (Figure 4.56 (publisher))

8 Optimization II (Briggs, et al, p. 294 c) We see that the chord length is 2r sin(θ/2) and the arc length is r(π θ). Note that the radius is r = 1 mile. Thus the travel time is given by T (θ) = 2 sin(θ/2) 2 ( θ = sin 2) The critical point(s) is given by + π θ 3 + π θ, (0 θ π). 3 0 = dt dθ = 1 2 cos θ That is, when cos θ/2 = 2/3, or θ = arccos(2/3) 1.68 rad = 96. The end points give T (0) = π/ hr, and T (π) 1 hr. But T (1.68 rad) 1.23 hr.

9 Optimization II (Briggs, et al, p. 294 d) Figure: (Figure 4.57 (publisher))

10 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 and reaches the house? Assume that the vertical wall of the house and the horizontal ground have infinite extent.

11 Optimization III (Briggs, et al, p. 295 b) Figure: (Figure 4.58b (publisher))

12 Optimization III (Briggs, et al, p. 295 b) Let L be the length of the ladder, x be the distance of the from the foot of the ladder to the foot of the fence, and let b be the height of the house. It follows from the last slide that we apply Pythagoras theorem to obtain L 2 = (x + 3) 2 + b 2. But similar triangles consideration yield 8/x = b/(3 + x) so that L is a function of x only and its domain is x > 0: ( 8(x + 3) ) ( 2 L 2 = (x + 3) 2 + = (x + 3) ) x x 2 It is easy to check d dx L2 = 2(x + 3)(x 3 192) x 3 which equals zero if x 3 = 192 or x First order test implies that L(5.77) 15 ft is the minimum length.

13 Optimization IV (Stewart p. 330) Find the area of the largest rectangle that can be inscribed in a semicircle of radius r. Ans. We want to maximise the rectangular area A = 2xy under the constraint x 2 + y 2 = r 2 as indicated in the following figure.

14 Optimization IV (Stewart p. 330) We may rewrite the area function as It is routine to check that A = A(x) = 2xy r 2 x 2. A (x) = 2(r 2 2x 2 ) r 2 x 2, implying that a critical point occurs at 2x 2 = r 2 or x = r/ 2. First order or the concavity tests can confirm that this value gives the maximum. Or to use the extrema value theorem as explained in Stewart to get the same conclusion. That is, ( r ) A = r 2 2 is the value that we want.

15 Primitives Definition Let F (x) and f (x) be two given functions defined on an 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 of f (x). We use the notation F (x) = f (x) dx to denoted that F is a primitive of f, and we call the process of finding a primitive F for f indefinite integration (or simply integration).

16 Examples Remark We note that if F (x) is a primitive of f (x), then F (x) + C, where C is an arbitrary constant, is also a primitive of f (x) since (F (x) + C) = F (x) + 0 = f (x). We call C a constant of integration. Examples x dx = 1 2 x 2 + C, 2x dx = x 2 + C, x 2 dx = 1 3 x 3 + C, x 8 dx = 1 9 x 9 + C.

17 Non-uniqueness Recall from that Theorem 4.11 that if the derivatives of two functions F 1, F 2 agree on I : i.e., F 1 (x) = F 2 (x), then F 1, F 2 differ by a constant. That is, for some constant k. F 1 (x) = F 2 (x) + k, holds for all x in I Figure: (Publisher Figure 4.78)

18 Primitives of monomials x p Theorem Let p 1 be a real number. Then x p dx = 1 p + 1 x p+1 + C, for some arbitrary constant C. Examples 2 3 x dx = dx = x 1 x 2 dx = 1 x 4+1 dx = x C = x 3/2 dx = x 3/ C = 2 5 x 5/2 + C x 3/2 dx = x 3/2+1 3/ C = 2 x + C x 2 dx = x C = 1 x + C. x 4 dx = 1 3x 3 + C.

19 1. x 15 dx, 2. x 12 dx, 3. x 12 dx, 4. 3x 4 dx, x 5. dx, 4 6. x 5 dx, dx x dx. x Exercises Remark Differentiate your answers to verify whether they are correct.

20 Linear combinations Since d(f + g) = df dx dx + dg d(kf ) and = k df dx dx 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 above consideration to linear combination of {f 1,, f n } k 1 f 1 (x) + k 2 f 2 (x) + + k n f n (x) dx = k 1 f 1 (x) dx + k 2 f 2 (x) dx + + k n f n (x) dx where {k 1,, k n }.

21 1. x 3/2 + 1 x 3/2 + 2 x 3 dx = Finding primitive examples = x 3/2 dx + x 3/2 dx + 2 x 3 dx ( 2 5 x 5/2 + c 1 ) 2x c2 + ( x 2 + c 3 ) = 2 5 x 5/2 2x 1 2 x 2 + C. 2. y 1/2 (1 + y) 2 dy = y 1/2 (1 + 2y + y 2 ) dy = y 1/2 dy + 2y 3/2 dy + y 5/2 dy = 2 3 y 3/ y 5/ y 7/2 + C t 1/2 (1 + t)2 dt = t 1/2 (1 + 2t + t 2 ) dt = t 1/2 dt + 2 t 1/2 dt + t 3/2 dt = 2t 1/ t3/ t5/2 + C.

22 Primitives of trigonometric functions Figure: (Briggs, et al Table 4.9)

23 Primitives of various special functions Figure: (Briggs, et al Table 4.10)

24 Examples (Briggs, et al) (p. 320) (p. 315) (p. 316) (p. 316) (p. 316) (sin 2y + cos 3y) dy (sec 2 3x + cos x 2 ) dx (e 10x + e x/10 ) dx 4 9 x 2 dx, 1 16t dt.

25 Initial value problems The simplest differential equation of first order is of the form f (x) = G(x), f (a) = b, where G(x) is a given function; where a, b are given initial condition. The f (x) = G(x) which is called a first order differential equation, together with the initial value condition is called an initial value problem (IVP).

26 An example of IVP Example (Briggs, et al, p. 317) Solve the IVP f (x) = x 2 2x, f (1) = 1 3. So a simple integration yields f (x) = (x 2 2x) dx = 1 3 x 3 x 2 + C. But 1 3 = f (1) = C implies that C = 1. So f (x) = 1 3 x 3 x

27 Sketch of the last IVP Figure: (Briggs, et al, Figure 4.87)

28 Example (Briggs, et al, p. 318) Race runner A begins at the point s(0) = 0 and runs with velocity v(t) = 2t. Runner B starts at the point S(0) = 8 and runs with velocity V (t) = 2. Find the positions of the runner for t 0 and determine who is ahead at t = 6. We have two IVP here. Namely, ds dt with solution s(t) = t 2 and = v(t) = 2t, s(0) = 0. ds dt = V (t) = 2, S(0) = 8 with solution S(t) = 2t + 8. Therefore, the two runners meet when s(t) = S(t), meaning that t 2 2t 8 = 0. That is, t = 4.

29 Example (Briggs, et al, p. 318) figure Figure: (Briggs, et al, Figure 4.88)

30 Example (Briggs, et al, p. 319) Neglecting air resistance, the motion of an object moving vertically near Earth s surface is determined by the acceleration due to gravity, which is approx. 9.8 m/s 2. Suppose a stone is thrown vertically upward at t = 0 with a velocity of 40 m/s from the edge of a cliff that is 100 m above a river. 1. Find the velocity v(t) of the object, for t 0, and in particular, when the object starts to fall back down. 2. Find the position s(t) of the object, for t Find the maximum height of the object above the river. 4. With what speed does the object strike the river?

31 Example (Briggs, et al, p. 319) We measure the height function s(t) from the sea level and adopt the upward direction to be our positive direction. Thus the initial height is s(0) = 100. (1) The acceleration dv due to gravity pointing to the centre of dt 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 The object starts to fall back down after it reached the maximum height, where v(t) = 0, that is, when v(t) = 9.8t + 40 = 0, giving t 4.1 s.

32 Example (Briggs, et al, p. 319) (2) The height s(t) satisfies the IVP ds(t) dt Solving the DE yields = v(t) = 9.8t + 40, s(0) = 100. s(t) = 4.9t t + C. The initial condition s(0) = 100 implies that 100 = s(0) = C. So s(t) = 4.9t t

33 Example (Briggs, et al, p. 319) (3) As a result the maximum height is reached when the parabolic s(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 quadratic Eqn s(t) = 0 gives us two roots, namely t 2 (which is to be discarded) and t So the velocity of the object when it strike the sea is given by v(10.2) 9.8(10.2) =

34 Example (Briggs, et al, p. 319) Figure: (Publisher Figure 4.89)

35 Example (Briggs, et al, p. 319) Figure: (Publisher Figure 4.90)

36 Example (Briggs, et al, p. 319) Figure: (Publisher Figure 4.91)