How do you find rate of change from 2 given coordinate points?
826422
Find the slope of the line that connects those points.
482492
So long as there is rate of change of motion
466006
Pick a few easy points like pi/2, pi, 3pi/2, 2pi.  You know the period.  Then just graph it.
904114
First, put your original equation as Y1\n\nFor Y2, type  nDeriv(Y1, x, x)\n\npress the buttons   (math)(8)   to get   nDeriv\nthe buttons (VARS)(1)(1)  to get Y1\n\n\nFor Y3, type  nDeriv(Y2, x, x)\n\nthen just display graph Y3\n\nIt will take quite a long while to graph, better if you manually differentiate the original equation once, then use the nDeriv to do second derivative.
766685
You need to differentiate the equation y=2x^2. \n\ndy/dx = 4x therfore the gradient at x=3 = 4*x = 12
150092
t=to\n_____\n√1-v2/c2
897839
The angular change ratio from the planar field
1132233
this question will require calcs using logarithm.\nhence will take time
839113
rate of change of demand or supply
1126609
Actually, no.\n\nIf dy/dx = 2, then I'd say yes, in moving from one coordinate pair to another, the y-value rises twice as fast as the x-value.  (That would be the case for a line with a slope of 2.)\n\nBut when dy/dx = 2x, as it does with a curve like y = x^2 + C, the ratio (of movement in the y-direction to movement in the x-direction) isn't constant -- it varies depending on where you are on the curve.  At the point (1, 2), for instance, that "slope," a.k.a. the slope of the tangent line at that point, is 2 -- the ratio of y-movement to x-movement is 2:1.  But at the point (3, 9), the slope of the tangent line is 6.\n\nHope that helps clear it up.  The important thing is that the slope of the tangent line is changing depending on where you are on the curve.
888670
First, draw a picture. Let C be Chris, B be Blaire, and A be the ball at some point. Thats a right angled triangle, where side BA is how far the ball has dropped; CB = 100 feet, and CA = the fishing line.\nNow, we need to know how fast the ball is dropping. Since its distance is 16t^2, its velocity is 32t (differentiating). So side BA is changing at 32t feet per second at any time.\nWe want to know the rate at which side CA is changing. But we know that CA^2 = CB^2 + AB^2, so CA = sqrt(CB^2 + AB^2) = sqrt(100^2 + AB^2).\nNow, we need to differentiate both sides with respect to time.\nThus, d/dt(CA) = d/dt(sqrt(100^2 + AB^2)).\nWe need the chain rule for the RHS: sqrt(something) differentiates to 0.5(something)^-0.5, then we multiply by the inside differentiated. AB^2 differentiated (with respect to t) is 2*AB *d/dt (AB) = 2*AB*32t.  So the overall equation becomes:\nd/dt(CA) = 0.5(100^2 + AB^2)^-0.5 * 2 * AB * 32t.\nNow, we know t = 3 and AB = 16*3^2, so substitute that in and you'll get the answer of 78.9 feet/second.\n\nI'm likely to have made a mistake, but hopefully that gives you the idea of what to do.
735022
There will be a term proportional to distance added to a constant delay term, so the form will be t=kD + c. Solving for k and c:\nt1=.152=100k + c\nt2=.229=170k + c\nt2-t1=.077=70k\nThen k=.0011\nt1=.152=.11 + c --> c=.042\nt2=.229=.187 + c --> c=.042\nResult is\nt=.0011*D + .042
449048
P1 x V1 / T1 = P2 x V2 / T2\n\nYou have all you need.  Calculate it.
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.
85418
dx/dt+ax=b :P\n\n\nmathematica had some online find-the-derivitive program, last I remember
887501
2 pi divided by the period is the angular rate.  2 pi divided by the new frequency after one year is the new angular rate.  Change in rate divided  by change in time is the angular acceleration.
977845
Put on a number plane with B initially at O (0. 0)\nTherefore A is initially at (-150, 0)\n\nNow A's position, t hours after midday, is (-150 + 35t, 0) and B's position is (0, 25t)\n\nSo t hours after midday \n\n|AB| = √[((-150 + 35t) - 0)² + (0 - 25t)²]\n= √[(22500 - 10500t + 1225t² + 625t²)\n= √(22500 - 10500t + 1850t²) \n\nVelocity of change = d(|AB|)/dt\n= 0.5 * (3700t - 10500) / √(22500 - 10500t + 1850t²)\n= 0.5 * (3700 * 4 - 10500) / √(22500 - 10500 * 4 + 1850 * 4²)\n= 0.5 * 4300 / √(10100)\n≈ 21.4 km/h