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Work : It is the last application of integral which we'll be looking at under this course. In this section we'll be looking at the amount of work which is done through a force in moving an object.
Under a first course in Physics you classically look at the work as a constant force F, does while moving an object over a distance of d. In such cases the work,
W = Fd
Though, most forces are not constant and will depend upon in which exactly the force is acting. Therefore, let's assume that the force at any x is specified by F(x). Afterward the work complete by the force in moving an object from x = a to x = b is specified by,
Consider that if the force is constant we find the correct formula for a constant force.
Here b-a is only the distance moved or d.
Therefore, let's take a look at a couple of illustration of non-constant forces.
a) The distance d that can be seen from horizon to horizon from an airplane varies directly as the square root of the altitude h of the airplane. If d = 213 km for h = 3950
In the earlier section we introduced the Wronskian to assist us find out whether two solutions were a fundamental set of solutions. Under this section we will look at the other app
F(x)=2x+3
any example
Function of a Function Suppose y is a function of z, y = f(z) and z is a function of x, z = g(x)
commutative law
Each week Jaime saves $25. How long will it take her to save $350? Divide $350 by $25; 350 ÷ 25 = 14 weeks.
SQUARE 12 IN
Sally gets paid x dollars per hour for a 40-hour work week and y dollars for every hour she works over 40 hours. How much did Sally earn if she worked 48 hours? Since she worke
Vector Function The good way to get an idea of what a vector function is and what its graph act like is to look at an instance. Thus, consider the following vector function.
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