Math 231E. Homework 6. Solutions. Problem 1. You work for a pharmeceutical company and are tasked with designing the dosage regimen for your company’s new drug Xzaano’qaanz (pronounced “Zan-ah-cans”). The dosage regimen will be one special “initial” pill, then a sequence of identical “following” pills. The answer you are being asked for is the amount of the drug in the first pill, the amount of drug in all subsequent pills, and the time between dosages. Your answer should be reported in milligrams and hours. Your pharmacokinetic experiments show that the drug1 is metabolized in the bloodstream according to a linear decay model with decay rate constant k = −0.08664 hr−1 . You also know that the drug’s minimum therapeutic concentration in blood is 10 µg/ml and its toxicity level is 40 µg/ml. This means that you want the concentration in the patient’s blood to always be in the range 10 µg/ml — 40 µg/ml. You may assume that when the patient takes the pill, it instantaneously dissolves into their bloodstream, and that 7% of the average person’s mass is their blood. a. Design a regimen for a 150 lb patient that maximizes the time between dosages. b. Let us now say that the patient has a pathology that changes the rate at which the drug is metabolized to k = −0.5 day−1 . How does your regime change? c. What would you change about the pharmacology of your drug if you wanted a healthy patient to be able to take only one pill per day?

Solution: a. So, overall, we see that the minimum concentration is 1/4 the maximum concentration, which corresponds to two half-lifes for the drug. To compute the half-life, we can use the formula − ln 2 0.6931 = = 8 hr. k −0.08664 hr−1 So then two half-lifes is 16 hr, and this is the maximum time between dosages. τ1/2 =

Now, we need to figure out how much of the drug we need to get the patient to the maximum concentration of 40 µg/ml. The patient masses 1 kg = 68 kg. 2.204 lb If 7% of that is blood, that is 4.76 kg of blood. If blood were water, then this would be exactly 4.76 ` of blood. As it turns out, blood slightly denser than water with a density of 1.06, so in fact this corresponds to 4.5 ` of blood. This corresponds to 150 lb ×

1

For what it’s worth, these numbers are more or less the actual numbers for the antibiotic vancomycin.

1000 ml × 4.5 ` = 180, 000µg = 0.18 g. 1` So the initial pill should have about 0.18 g. Since subsequent pills need to bring the patient from 10 to 40 instead of 0 to 40, then it needs to have 3/4 of the drug, or about 0.135 g. 40 µg/ml ×

b. If the metabolizing rate is now k = −0.5 day, then the half-life is τ1/2 =

− ln 2 = 1.38 day = 33.3 hr, −0.5 day

so everything is the same, except now the patient takes the pill once every 66.6 hr. c. We can either slow down the metabolizing effect, or widen the range of acceptable concentrations. For example, if we do not change k, then we need the patient to be able to go three halflives while still being in the acceptable range, which means that the ratio between maximum and minimum concentrations should be 8 instead of 4. So if we could double the toxicity threshold, or halve the minimum therapeutic threshold, we would be good. Similarly, if we couldn’t change those thresholds, then we need 24 hours to be two half-lifes for the drug, which means that the half-life needs to be 12 hours. Then we need to change k to k=

− ln 2 = −0.5776 hr−1 . 12 hr

Problem 2. Set up and solve the following rate problems. For a) please include a diagram. a. A 5 m long ladder is propped up against a wall. The ladder begins to slip. At time t = 1 s the base of the ladder is 3 m from the wall and moving away from the wall at 1 m/s. How fast is the end of the ladder moving down the wall? Solution: If we draw a triangle, we see that the triangle, wall, and floor form a right triangle. Let us define x and the distance along the floor from the base of the ladder to the wall, and let y be the distance from the place the ladder touches the wall to the ground. We then have x2 + y 2 = 55 = 25. Differentiating this equation gives dx dy + 2y = 0, dt dt dx dy x +y = 0. dt dt

2x

The problem tells us that dx/dt = 1 m/s and x = 3 m. Using x2 + y 2 = 25, this means that y = 4 m, so we have dy (3 m)(1 m/s) + (4 m) = 0, dt so dy 3 = − m/s. dt 4

b. A rocket massing 30 kg initially is burning fuel at a rate of 1 kg/s. If the rocket is initially moving with a velocity of 10 m/s and is accelerating at 5 m/s2 what is the rate 1 of change of the rocket’s kinetic energy? Recall that kinetic energy is K = mv 2 . 2 Solution: Since we are given the formula for kinetic energy, let us differentiate this to obtain 1 dm 2 dv dK = v + mv , dt 2 dt  dt 1 kg  m 2 = −1 10 + (30 kg)(10 m/s)(5 m/s2 ) 2 s s kg · m2 kg · m2 = −50 + 1500 s3 s3 2 kg · m = 1450 s3 = 1450 J/s.

Problem 3. The volume of a sphere of radius r is V (r) = 43 πr3 . a. Compute the derivative of V (r) with respect to r, b. The surface area of a sphere of radius r is 4πr2 . Compare this to your answer in a) and explain this geometrically.

Solution: a. If we differentiate we obtain V 0 (r) = 4πr2 .

b. We see that the derivative of volume with respect to radius is surface area. This makes sense by thinking about it in a couple of different ways. Let us imagine that we have a sphere of radius r and volume V = 43 πr3 . Let us increase the radius by a bit, from r to r + dr. Then the volume will increase from V to V + dV , and the theory of differentials tells us that dV = 4πr2 dr. Let’s think about what we have done: we increase the radius a little bit, which added a thin shell to the sphere. It is plausible that the volume of this shell should be equal to the surface area times the thickness, and this is exactly what we have shown. Looked at another (similar) way, let us imagine that the radius of the sphere is growing in some manner, so that r = r(t) for some (increasing) function. Then we have dV dr dr dV = = 4πr2 , dt dr dt dt which can also be interpreted as instantaneous change in volume = surface area × instantaneous change in radius.

Problem 4. Consider a circular disc with a sector of angle θ removed. Imagine rolling this shape into a conical cup, like one often finds at water coolers. θ and a. Show that if this is rolled into a cone the top of the cone will have radius 1 − 2π r θ 2 height 1 − (1 − ). 2π Hint: If we remove a sector of angle θ what is the circumference of the remaining piece? Solution: The circumference of the remaining piece is 2π − θ, and thus its radius is (2π − θ)/2π = 1 − θ/π. If we look at the cup from its side, we see that the height h forms one leg of a right triangle, and the radius is the other leg, and of course the hypotenuse has length 1. Therefore  2 θ 2 h + 1− = 1, 2π or

s h=

 2 θ 1− 1− . 2π

We will write y = 1 − θ/2π throughout, and this simplifies the formula for h to p h = 1 − y2.

b. How should one choose θ to maximize the volume of the cup? Recall that the vol1 ume of a right cone is bh. You might find it easier to work with the quantity 3 θ y =1− . 2π p Solution: We know h = 1 − y 2 . The base of the cone is a circle with radius r = 1 − θ/2π = y, so the area of the base is πy 2 and the volume of the cone is p 1 V (y) = πy 2 1 − y 2 . 3 Also, it is clear that the domain of our function is y ∈ [0, 1] — we cannot choose y < 0 and of course y cannot be larger than 1, because then h would be negative. Also notice that V (0) = V (1) = 0 and that V 0 (x) is differentiable in (0, 1) — thus Rolle’s Theorem tells us that there is an interior critical point. Since the function is zero on the boundary and positive in the middle, at least one of these interior must be the global maximum. Differentiating V (y) with respect to y gives 1 1 1 2 p (−2y) p V 0 (y) = πy 1 − y 2 + πy 2 3 3 2 1 − y2
π 2y(1 − y 2 ) − y 3 = p 3 1 − y2 πy (2 − 3y 2 ). = p 2 3 1−y

!

The critical points of this function are those y such that V 0 (y) = 0 or DNE, which are clearly the solutions to y = 0, 1 − y 2 = 0, 2 − 3y 2 = 0, or

r

2 . 3 Again, the boundary points y = 0, 1 give degenerate cones with apvolume of zero, and thus the maximum must occur at the only interior critical point y = 2/3. Thus our maximal volume is r p 1 2 1 2π V ( 2/3) = π = √ . 3 3 3 9 3 y = 0,

y = 1,

y=

Also, notice that y =

p 2/3 corresponds to θ = 1− 2π

r

2 , 3

or θ = 2π(1 −

p

2/3) ≈ 0.18(2π) = 0.18(360◦ ) = 66.1◦ .

c. Extra Credit:. Find one of these paper cups. Cut it open and flatten it to form a cone. Measure the angle of the missing sector. How close to the maximum volume is the cup?

θ 1

Figure 1: A blueprint for a cone? Or Pac–Man?!? We’ll let our viewers decide, after these messages.

Problem 5. We discussed in class (and in the notes) what the Taylor series of sin(sin(· · · (sin(x)))) looks like, and here we study this function in more depth. In all parts, we are considering Taylor series at a = 0. We will use the notation fk (x) to be the kth iterate of sin(·), so that f1 (x) = sin(x), f2 (x) = sin(sin(x)), etc. In general, fk (x) = sin(fk−1 (x)) = fk−1 (sin(x)). a. First write down the first nonzero two terms in the Taylor series for sin(x). (Note here and below that this means writing down the series and including the thirdorder term.) b. Now we write down the first two terms in the Taylor series for f2 (x) = sin(sin(x)). First obtain it by plugging the Taylor series for sin(x) into itself. Then compute it directly, i.e. compute the first three derivatives of f2 (x) at 0 and obtain the Taylor series that way. Do your answers agree?

c. Now compute the first two terms in the Taylor series for f3 (x). Hint: Do not try to do this directly, but do this by plugging in! d. Now compute the first two terms in the Taylor series for f4 (x). Do you see a pattern? e. Can you conjecture what the first two terms in the Taylor series for fk (x) are? Can you prove this conjecture? f. Using this conjecture, compute the third derivative of fk (x) at x = 0. Solution: a. We have x3 sin(x) = x − + O(x5 ). 6 b. If we plug in, we get  3  1 x3 x3 5 5 + O(x ) − + O(x ) x− x− 6 6 6 x3 1 3 =x− − x + O(x5 ) 6 6 x3 =x− + O(x5 ). 3 We can also compute directly. We compute derivatives: 

f20 (x) = cos(sin(x)) cos(x) f200 (x) = − sin(sin(x)) cos2 (x) − cos(sin(x)) sin(x) f2000 (x) = − cos(sin(x)) cos3 (x) − 2 sin(sin(x)) cos(x)(− sin(x)) + sin(sin(x)) cos(x) sin(x) − cos(sin(x)) cos(x) = − cos(sin(x)) cos3 (x) + 3 sin(sin(x)) sin(x) cos(x) − cos(sin(x)) cos(x). Plugging in x = 0 we obtain f2 (0) = 0,

f20 (0) = 1,

f200 (0) = 0,

f2000 (0) = −2,

which gives x3 + O(x4 ), 3 which agrees with the other term. It took a lot more work though, and we’d hate to do this directly for f3 (·)! sin(sin(x)) = x −

c. Since f3 (x) = f2 (f (x)), let us plug the Taylor series for sin(·) into the formula from (b): 

  3 x3 1 x3 5 5 x− + O(x ) − x− + O(x ) 6 3 6 x3 1 3 =x− − x + O(x5 ) 6 3 3 x =x− + O(x5 ). 2

d. Since f4 (x) = f3 (f (x)), let us plug the Taylor series for sin(·) into the formula from (c): 

  3 x3 1 x3 5 5 x− + O(x ) − x− + O(x ) 6 2 6 x3 1 3 =x− − x + O(x5 ) 6 2 2x3 =x− + O(x5 ). 3

e. We see that in each case, the linear term has coefficient one, but the cubic coefficient of fk (·) has always been k/6. So let us make the conjecture that k fk (x) = x − x3 + O(x5 ). 6 We can prove this by induction! Assume that this formula for fk is true, and let us compute the formula for fk+1 by writing fk+1 (x) = fk (f (x)):  3   k x3 x3 5 5 + O(x ) − + O(x ) x− x− 6 6 6 x3 k 3 =x− − x + O(x5 ) 6 6 k+1 3 x + O(x5 ). =x− 6 Since the formula works for k = 1, then it follows. f. If the first two terms of the Taylor series of fk (x) are x − (k/6)x3 + O(x5 ), then this means that fk (0) = 0,

fk0 (0) = 1,

Good luck trying to get this directly!

fk00 (0) = 0,

fk000 (0) = 6! ·

−k = −5! · k. 6