course Mth 173 ??·? ??????w··??assignment #010
......!!!!!!!!...................................
10:29:47 query problem 1.6.7 (was 1.10.16)cubic polynomial representing graph. What cubic polynomial did you use to represent the graph?
......!!!!!!!!...................................
RESPONSE --> Since the x-intercepts are -3, 1, and 4, the basic formula is: y = k(x+3)(x-1)(x-4) where k is an amplitude modifier constant.
.................................................
......!!!!!!!!...................................
10:30:38 *&*& The given function has zeros at x = -2, 1, 5, so y = k (x+2)(x-1)(x-5). At x = 0 the function has value 2, so 2 = k (0+2)(0-1)(0-5), or 2 = 10 k. Thus k = .2 and the function is y = .2 ( x+2)(x-1)(x-5). **
......!!!!!!!!...................................
RESPONSE --> Slightly different question in 4th edition. No value at x=0 was given, so I can't solve for k.
.................................................
......!!!!!!!!...................................
10:38:18 Query problem 1.6.15 (formerly 1.4.19) s = .01 w^.25 h^.75what is the surface area of a 65 kg person 160 cm tall?
......!!!!!!!!...................................
RESPONSE --> Surface area is 1.277m^2.
.................................................
......!!!!!!!!...................................
10:38:23 ** Substituting we get s = .01 *65^.25 *160^.75 = 1.277meters^2 **
......!!!!!!!!...................................
RESPONSE --> Ok.
.................................................
......!!!!!!!!...................................
10:38:36 What is the weight of a person 180 cm tall whose surface area is 1.5 m^2?
......!!!!!!!!...................................
RESPONSE --> 86.8kg
.................................................
......!!!!!!!!...................................
10:38:46 ** Substituting the values we get 1.5 = .01 w^.25*180^.75 . Dividing both sides by 180: 1.5/180^.75 .01w^.25. Dividing both sides by .01: 3.05237 = w^.25 Taking the fourth power of both sides: w = 3.052^4 = 86.806 **
......!!!!!!!!...................................
RESPONSE --> Ok.
.................................................
......!!!!!!!!...................................
10:39:22 For 70 kg persons what is h as a function of s?
......!!!!!!!!...................................
RESPONSE --> H = ( s / (.01*70^.25))^(4/3)
.................................................
......!!!!!!!!...................................
10:42:11 ** Substituting 70 for the weight we get s = .01 *70^.25 h^.75 s = .02893 h^.75 s/.02893 = h^.75. Taking the 1/.75 = 1.333... power of both sides: (s/.02893)^1.333 = h h = 110.7s^1.333... = 110.7 s^(4/3). **
......!!!!!!!!...................................
RESPONSE --> Okay, I think I see how you got that. I should have reduced the denominator of the fraction and then raised 1/0.0289265 to the 4/3 power.
.................................................
......!!!!!!!!...................................
10:43:31 query problem 1.6.20 flow at 5 cm/s thru rect cross section 3 cm by x cm. What is the expression for the volume emerging from the pipe in 1 second and how did you obtain it?
......!!!!!!!!...................................
RESPONSE --> 4th edition 1.6.19. The expression I found was: R = kr^4 This is because the Rate is directly proportional (k) to the fourth power of the radius (r).
.................................................
......!!!!!!!!...................................
10:45:17 ** The pipe has rectangular cross-section with dimensions 3 cm by x cm, so its cross-sectional area is 3 x. Every second the fluid filling a 5 cm section of the pipe exits the pipe. The volume of this section is the product of its length and its cross-section, or 5 cm * (3 x cm) = 15 x cm^2. If you express x in cm, then 15 x cm^2 will be in cm^3. **
......!!!!!!!!...................................
RESPONSE --> Oh. This is a completely different question from the one I answered. I see how you found it, I just thought it was another question.
.................................................
......!!!!!!!!...................................
10:54:55 query problem 1.6.36 Box with square ends, length + girth < 108. Vol of max box with square of side s. What is your expression for the volume in terms of s?
......!!!!!!!!...................................
RESPONSE --> Not in 4th edition. Initial formula: V = Length * s^2 Length < 108 - 4s because Girth = 4s Therefore: Volume = (108 - 4s) * s^2
.................................................
......!!!!!!!!...................................
11:02:36 ** If s is the side of the square base then the volume is length * s^2. Since length + girth < 108 and the girth is the perimeter of the square we have length + 4 s < 108 so that the max length is Lmax = 108 - 4s. Thus the max volume with side s is V = L * cross-sectional area = (108 - 4 s) * s^2 = 108 s^2 - 4 s^3. The graph of this V vs. s passes through (0, 0) and (27, 0), first increasing at a decreasing rate until it reaches a peak then decreasing at an increasing rate. Note for future reference that we can find the max possible volume: This max possible volume changes as s changes. To find the max possible volume note that the peak of the graph occurs where the graph levels off before beginning to decrease. The slope of the graph is given by the derivative of V with respect to s. find the derivative of V with respect to s, which we will denote V '. Since the derivative of x^3 is 3 s^2 and the derivative of s^2 is 2s we get V ' = 216 s - 12 s^2. This derivative also represents the rate at which volume changes with respect to s. The max of V will occur when V ' = 0 so we get 216 s - 12 s^2 = 0, which has solution s = 18. The peak of the graph occurs at s = 18, so the max volume occurs when s = 18. If s = 18 then L = 108 - 4 s = 108 - 4 * 18 = 108 - 72 = 36. Thus the volume is V = s^2 * L = 18^2 * 36 = 11664.**
......!!!!!!!!...................................
RESPONSE --> Ok. Interesting. I never thought of using derivatives like that. Useful process.
.................................................
"