2nd derivatives
Last updated
Last updated
Important to know, especially for problems with a max or a min. How do you locate one? That's the 1st derivative's job. The 1st derivative is 0 if we have a min or max. When we set the derivative to 0 and solve, we find a max or min.
The 2nd derivative tells us the bending. It tells us if it's a min or a max. Now comes the derivative of the speed: that is the acceration. The rate at which we're speeding up or slowing down.
Distance = Height, speed = slope, acceleration = bending.
Other example:
We know y=sinx it's bending down thanks to the 2nd derivative which is below 0.
Let's continue these graphs over Pie/2 to see something:
the point where y=sinx = 0 is the point where the bending changes. Goes from bending down to bending up. That is because y''=0.
1 more example:
To find the min or max we take the derivative : y'=3x^2-2x and solve for when that equation = 0.
To know which one is max or min, we can look at the graph. Also we're talking about local max and mins.
At x=0 we expect to see the slope = 0. Yes that's true. Also that the second derivative is negative because it's bending down. It is! y'' at x=0 = y'' = 6*0 - 2 = -2. It is neg! So it's a max.
At x = 2/3 we're expecting a min, because Y'' is positive.
How do we find the inflection point? it's when y'' = 0. And on the graph, y'' is 0 when x=1/3.
Example: driving to work
Where a is the straight distance from home to the highway. b is that point to MIT. x is what we're trying to solve. We're looking for the shortest way to MIT. Road = 30mph and highway 60mph.
I minimize the equation by taking its derivative and setting x to 0. Derivative = what's under the line...
Setting to 0 means:
1 side is neg, the other positive... So this =
multiply by 60, and by the square root. Then we square everything to get rid of the square root. Now we have an equation. Final simplification, we get:
To finish the problem, we'd have to take the 2nd derivative and look at the sign of it. We would have found that it's a positive sign, so bending upwards so that this is a minimum time not maximum.
