## Friday, May 27, 2016

## Wednesday, May 11, 2016

Suppose one travels from point (2,3) to point (11,8) at a constant speed for one hour. How far did they travel in 35 minutes and where is their location in terms of x and y?

The key is to use the distance formula and slope. For the distance formula, recall it is the square root[(x2-x1)^2 + (y2-y1)^ You can apply that to get the total distance traveled at approximately 10.3, Now I use the slope to determine how much rise and run there is in 35 minutes. Take the slope of see that there is 5 rise and 9 run over the course of 60 minutes. Now take 5(35/60) and 9(35/60) to see how far the x and y coordinates move in the 35 minutes.

The key is to use the distance formula and slope. For the distance formula, recall it is the square root[(x2-x1)^2 + (y2-y1)^ You can apply that to get the total distance traveled at approximately 10.3, Now I use the slope to determine how much rise and run there is in 35 minutes. Take the slope of see that there is 5 rise and 9 run over the course of 60 minutes. Now take 5(35/60) and 9(35/60) to see how far the x and y coordinates move in the 35 minutes.

we have to know that to find concavity we need the second derivative and
since we are looking for concave up, it's where y'' > 0. Given that y
is an integral, y' is the integral evaluated at x, so it's y' =
6/(1+2x+ x^2), then for the second derivative I used the quotient rule. I
set the second derivative > 0 and see that x > -1 gives us
concave up

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