Monday, October 20, 2014

Moving left and right, speeding up or slowing down?

From Khan Academy

Let's deal with moving left and right to start.
The particle is moving to the right when its velocity is positively valued; that is, when v(t)>0.
The particle is moving to the left when its velocity is negatively valued; that is, when v(t)<0.
Now we consider when the particle is speeding up and slowing down.
The particle is speeding up for those times when the product of its velocity and its acceleration is greater than zero. Therefore we want
v(t)a(t)>0.
For this condition to be satisfied, we want those times when v(t) and a(t) are both positively valued or both negatively valued.
In contrast, he particle is slowing down for those times when the product of the velocity and the acceleration is less than zero. Therefore we want
v(t)a(t)<0.

Friday, October 10, 2014

The fundamental theorem of calculus

The Fundamental Theorem of Calculus
Suppose  f  is continuous on  [a,b].
1   If  y=xaf(t)dt, then  dydx=f(x).
2 baf(t)dt=F(b)F(a), where  F  is an antiderivative of  f.

The theorem can be applied in many instances. One of which is as shown, for finding a function that increases or is concaved upwards.
The function  F  is increasing on the interval(s) where its derivative  f  is positive in value
The function  F  is concave up on the interval(s) where its derivative  f  is increasing


    Fundamental theorem of calculus



    Swapping the bounds for definite integral