Want this question answered?
The mean value theorem for differentiation guarantees the existing of a number c in an interval (a,b) where a function f is continuous such that the derivative at c (the instantiuous rate of change at c) equals the average rate of change over that interval. mean value theorem of integration guarantees the existing of a number c in an interval (a,b)where a function f is continuous such that the (value of the function at c) multiplied by the length of the interval (b-a) equals the value of a the definite integral from a to b. In other words, it guarantees the existing of a rectangle (whose base is the length of the interval b-a that has exactly the same area of the region under the graph of the function f (betweeen a and b).
This is done with a process of limits. Average rate of change is, for example, (change of y) / (change of x). If you make "change of x" smaller and smaller, in theory (with certain assumptions, a bit too technical to mention here), you get closer and closer to the instant rate of change. In the "limit", when "change of x" approaches zero, you get the true instantaneous rate of change.
During exercise an increase in heart rate corresponds to a shortening of the cardiac cycle (RR interval decreases). Most of this shortening occurs in the TP interval. The QT interval also shortens, but only slightly. then the interval shortens as the heart rate increases.
They are the same for a straight line but for any curve, the slope will change from point to point whereas the average rate of change (between two points) will remain the same.
Acceleration = rate of change of speed = (change of speed) / (time interval) = (25 - 5) / 4 = 20/4 = 5 m/s2
There have to be two (or more) ordered pairs for an average rate of change to make any sense. Your question does not.
Acceleration is the rate of change of velocity - in symbols, a = dv/dt. Or for average acceleration over a finite time: a(average) = delta v / delta twhere delta v is the change in velocity, and delta t is the time interval.
what exponential function is the average rate of change for the interval from x = 7 to x = 8.
The rate of changing the interval of 25 is 19.5. This is a math problem.
No. Average speed is the rate an object is moving measured over more than an instant, such as one second, one minute, or something like that. Instantaneous speed, however, is the limit of the average speed as the interval of time approaches zero, i.e. at a given instant.
Instantaneous speed:- It is the rate of change of position with respect to time,at that instant. Average speed:-Average speed is defined as the total path length travelled divided by the total time interval.
An annual percentage rate is the average percentage change over a period of a year. The percentage change is the change divided by the initial value, expressed as a percentage.
You cannot. Acceleration is the rate of change in velocity over time
yes, aka rise over run.
The mean value theorem for differentiation guarantees the existing of a number c in an interval (a,b) where a function f is continuous such that the derivative at c (the instantiuous rate of change at c) equals the average rate of change over that interval. mean value theorem of integration guarantees the existing of a number c in an interval (a,b)where a function f is continuous such that the (value of the function at c) multiplied by the length of the interval (b-a) equals the value of a the definite integral from a to b. In other words, it guarantees the existing of a rectangle (whose base is the length of the interval b-a that has exactly the same area of the region under the graph of the function f (betweeen a and b).
It is acceleration. The difference between final velocity and initial velocity, divided by the time is the AVERAGE acceleration. Remember, though that velocity is a vector. So if you are going round in a circle at a constant speed, your direction of motion is changing continuously and so you are always accelerating!
Depends. Slope of tangent = instantaneous rate of change. Slope of secant = average rate of change.