Find the points on the curve____++The growth rate of a certain population?
374181
a) Equally inclined to both axes is parallel to y = x; hence you want dy/dx = 1.\n\n(y - 3)^2 = 4 - x^2\ny = (4 - x^2)^(1/2) + 3\n\ndy/dx = (1/2) * (4 - x^2)^(-1/2) * (-2x)\ndy/dx = x / (4 - x^2)^(1/2)\n\nSo:\n\nx^2 = (4 - x^2)\nx = +/- SQRT(2)\ny = 3 + SQRT(2)\n\n------------------------------------------------------\n\nb)  I might edit and do later if I've got time.
963964
P=(1,2,1) Q=(1,5,5)\n\nThe parameterisation X(t) with parameter t generates all points on the line (or line segment, for 0<=t<=1)\nwith X(0)=P, X(1)=Q :\n\nX(t) = t*P + (1-t)*Q\n       =t*(1,2,1) +(1-t)*(1,5,5)\n(multiplying in, which is generally not needed)\n       = (1, 5-3t, 5-4t)
729738
rate of change=dy/dx=slope
968183
I'm not really sure what you're asking.  The growth rate would depend on the formula itself, so without the formula, there's no way to determine the answer.  You'd need one to get the other.  I imagine that there could be an infinite number of growth rates and in infinite number of growth-rate-formulas.
838358
i can't give you the answer but i used my ti-83 and i can tell you the points of the graph...\n(1,0) (2,4) (3,8) (4,0)\n\nHope this helps
963751
find the first derivative for the curve, and then substitute the x coordinate for the slope at that point.\n\nfor example, for the graph y = x^2, the first derivative is 2x. to find the slope at where x = 1, sub x = 1 into 2x and the slope at that point is 2.\n\nalternatively, draw a tangent to the point where u want to find the point slope and the slope of the tangent is the point slope. note that this method's accuracy depends on how accurately you draw your tangent
812091
On 1 graph you could put age(x) and height(y).....and on another graph, you could probably put age(x) and weight(y)......I think :)
45317
The simplest model of population growth is based on the straightforward quantities of four properties of population change:\n\nB - Births or Natality \nD - Deaths or Mortality \nI - for Immigration and \nE - for Emigration\n\n\n\nNt+1 = Nt + B - D + I - E\n\n[Equation 3.1]\n\n\n\n\nThe net rate at which the population is increasing or decreasing is called the net rate of population growth - Ro. The net rate of population growth is simply the ratio of the size of the population at some time Nt+1 compared to the size of the population during some the previous moment in time Nt: \n\n\n\n\n[Equation 3.2]\n\n \n\nIf birth rates (b) and death rates (d) in a population are determined for a specific time interval we can calculate another measure of population growth call the intrinsic rate of increase - r from these two population parameters: \n\n\nr = b - d\n\n[Equation 3.3]\n\n\n\n\nThe rate of change of population size dN/dt is then given by the differential equation: \n\n\n\n\n[Equation 3.4]\n\n\n\n\nwhere N is the size of the population at a particular instant in time. To obtain the size of the population Nt at some time t, is integrated to give: \n\n\n\n\n[Equation 3.5]\n\n\n\n\nwhere No is the initial size of the population at some time designated zero, e is a constant = 2.718 (the base of natural logarithms), r is the intrinsic rate of increase and t is the time elapsed since zero.\n\nThe population growth described by equation [3.5] is geometric, which means the population size goes on doubling (or halving if d is greater than b) at a constant rate. This situation is clearly unrealistic, since the growth of all species populations must be limited by the availability of some finite resource.
759663
a) The curve you need is called a cycloid. Look that up on the web.\n\nb) The parametric form of the cycloid is\n\nx=12(a-sin a)      y=12(1-cos a) where a is then angular displacement. 24pi (circumference) will be the period and it passes through (0,0)
973905
To find a rate of change, you need to find out how much the average has changed from one point to the next, then divide it by how many units that change is spread across:\n\n1. ((78/10) - (48/4))/(10-4) = (7.8 - 12)/6 = -4.2/6 = -,7\nThe price decreased by $.70 per person\n\n2. (120/4 - 30/1)/(4-1) = (30 - 30)/3 = 0\nThe rate did not change\n\nSlopes:\nThe easiest way to do a point-slope problem is to use the slope-intercept form of the linear equation, y = mx + b, where m is the slope and b is the y-intercept (the place where x = 0, and the line crosses the y-axis).  Plug in your known values and solve:\n\n1. 5 = -2(3) + b\n5 = -6 + b\n11 = b\ny = -2x + 11\n\n2. -3 = (-4/3)(2) + b\n-3 = -8/3 + b\n-1/3 = b\ny = -4/3x - 1/3
189962
If you need the curve to pass through all and every point, then you need a polynomial whose grade is equal to the number of points less 1.\nThen then polynomial is contructed by setting \na0+a1 x1+a2 x1^2+ ...+ an x1^n = y1\na0+a2 x2+a2 x2^2+ ...+ an x2^n = y2\n...\na0+a2 xn+a2 xn^2+ ...+ an xn^n = yn\n\nYou have a set of lineal equations (since the unknwon are the coefficients a0...an) that you can easily solve.\n\n\nIf you have an already defined curve that has to pass "as near as possible" to the given points, then you have to go the way of regression.\nBasically, you calculate the "error" of each point as\nyn-f(xn); add up the square of the errors:\nsum((yn-f(xn))^2) and minimize the sum changing the function parameters
434195
That is the rate of change in an "instant" in time.  You calculate it by finding the slope of the curve at that point.  You do this by taking the first derivative at that point.  If it is a straight line, you can get it by doing the average rise over the average run:  Pick two points on the line. Call them x1,y1 and x2,y2.  You are probably looking at supply and demand so we'll do price and quantity, so call them Q1,P1 for the first point and Q2,P2 for the second.  The slope (instantaneous rate of change) is: (P2-P1)/(Q2-Q1).\nIf it's not a straight line, you can approximate it by picking one point to the left and one point to the right of the point you are trying to find.  Make these two points close to the point you are looking for, and equidistant from the point you are looking for.  Use these points in the above equation.
191015
Since the first 3 answerers didn't really comment on where they got their answers from, I thought a more detailed answer was called for.\n\nThere's a general formula for these types of growth, which also applies to things like radioactive decay rates and interest rates:\n\nP(t) = P(0)*e^(r*t)\n\nP(t) is the population at time t\nP(0) is the population you're starting out with (the population at time 0)\ne is the "natural base" number\nr is the rate of change\nt is the time\n\nIn this case, r is .005 (.5 percent), t is 3 years (878-875), P(0) is 360, and P(t) is our only unknown.  So plug them in:\n\nP(3) = 360*e^(.005*3) = 365.440703...\nWhich rounds to 365.\n\nThe first answerer's method becomes inaccurate after a while; for example, 50 years should be 462, not 450.  The second answerer is better.  The third is just plain wrong, with no explanation of his methods.
1144014
draw a picture of the graph over the specified interval.  divide the graph into vertical rectangles.  the area under the curve will be the sum of the areas of the rectangles.  saying that youve made "n" rectangles, the only way to get the exact area is to find the sum for the limit as n approaches infinity.  work from that concept.  if you already knew that, or need more help or something, email me and i can give you more steps and a much more detailed description.  that would take a really long time though so im gonna see if this helps first.
989515
A = 1000(.05)\nB= 1500(2)\nYou can graph this by solving for A and B using 1,2,3 in A as your y value and B as you x value.  Once you solve your equations you can graph your results to find when B will have 3 times as many cells as culture A.  Use the formula y2-y1/x2x1=m to find your slope.
466006
Pick a few easy points like pi/2, pi, 3pi/2, 2pi.  You know the period.  Then just graph it.
796769
The point P moves in such a way that its Cartesian coordinates with respect to an origin O at time t is given by x= e^(-t) , y= 2te^(-t). the distance OP is denoted by r and the angle between OP and the x-axis by θ.\n(a)Using an analytical method, find the value of t when r^2 is stationary. Determine the nature of the stationary point.\n\nr² = Δx² + Δy²\n= (e^(-t) - 0)² + (2te^(-t) - 0)²\n= e^(-2t) + 4t²e^(-2t)\n= e^(-2t)(1 + 4t²)\nd(r²)/dt = -2e^(-2t)(1 + 4t²) +8te^(-2t)\n= 2e^(-2t)(1 - 4t + 4t²)\n= 0 for stationary points\nSo 1 - 4t + 4t² = 0 as e^(-2t) > 0 for all t\nie (1 - 2t)² = 0, ie t = ½, ½\nWhen t <½, d(r²)/dt > 0 and when t >½, d(r²)/dt > 0\nThus when t = ½ there is a stationary inflexion\n\n(b) Find, in terms of t, the rate of change of θ with respect to\n\ntanθ = Δy/Δx\n= 2te^(-t)/e^(-t)\n=2t\n\nsec²θ * dθ/dt = 2\nie\n\ndθ/dt = 2/sec²θ\n= 2/(1 + tan²θ)\n= 2/(1 + 4t²)\n\nCan you show the non analytical method for part (a) too!\n\nx = e^(-t) → -t = lnx ie t = -lnx\n\ny = 2te^(-t) \n= -2xlnx\n\nx: .. 0 ...... 0.25 ... 0.5 .... 0.75 . 1.0 .. 1.25 .. 1.5\ny: undef . -0.15 . -0.36 . -0.22 ... 0 .... 0.28 . 0.61\n... (→0)\n\nLim x → 0 (x lnx) = 0 and OP increases with x \n(limiting min value of OP = 0)\n\nMin value of xlnx occurs when lnx + 1 = 0 ie x = -1/e and at this point OP² will show its inflexion as it cannot be a minimum and nor can it be a maximum so all it can do is slow its rate of growth.