so, i am working on some optimization problems and I fully understand the steps on how to find the dimensions that will minimize/maximize the amount of material needed, but i'm having trouble defining these concepts in normal English.
As an example, I will use a standard cylinder problem that gives you the volume and asks to determine how much material you will need to get the maximum/minimum surface area.
As I understanding it, the first derivative tells me that the surface area of the cylinder is changing at a specified rate when the radius is equal to a particular value. This rate of change varies, depending on the measurement of the radius.
It is the second derivative that gives me problems. As I understand it, the second derivative tells me that the rate of change of the surface area's rate of change varies depending on the value of the radius. Therefore, the minimum point of the second derivative tells me that the rate of change of the rate of change is at its least when the radius is a particular value (in graphical terms, the x value of the minimum point).
What I am struggling to understand is how this minimum point of the second derivative influences the amount of material needed for the surface area.
sorry if I overcomplicated this, but the books and classes I use do not do a good job of explaining this at all. they just give you the steps and computation for how to do it, but don't provide real evidence as to how it works or what the information tells you in 'real life' terms.
The first derivative tells you how the function value is changing. If df/dx is positive, f(x) is increasing as x increases. If df/dx is negative, f(x) is decreasing as x increases. At a point at which f(x) has a maximum (or minimum) value f is neither increasing nor decreasing and so df/dx must be 0 (or not exist).
The second derivative tells you how the first derivative is changing. If is positive, df/dx is increasing and if is negative, df/dx is decreasing.
Now, if x= a gives a maximum value for f (so df/dx(a)= 0) then f is increasing for x< a (so df/dx is positive) and decreasing for x> a (so df/dx is negative). That is, df/dx has gone from positive to negative and so df/dx is decreasing. That means must be negative.
If x= a gives a minimum value for f (so df/dx(a)= 0) then f is decreasing for x< a (so df/dx is negative) and increasing for x> a (so df/dx is positive). That is, df/dx has gone from negative to positive and so df/dx is increasing. That means must be positive.
That is why " at a maximum and at a minimum".
A good example of zero first derivative and non-zero second derivative is what happens when you toss a ball in the air. At the top of its arc it has reached maximum displacement, its velocity (first derivative) is zero, and its acceleration (2nd derivative) is constant at g = -9.8 m/s^2. So it has a non-zero 2nd deivative even as it's first derivate is zero. Remember that acceleration is rate of change of velocity, or the slope of the velocity curve.
Great, this clears up my second question, thanks a lot! so, in order for their to be a non-zero second derivative when their is a zero first derivative, you would need some sort of external force (e.g. gravity, mechanics of a car) to enable the first derivative to be positive? does this mean that when something starts out motionless, the second derivative must be positive before the first derivative can become something greater than zero?
The second derivative is acceleration, and from basic physics , so an external force (like gravity) causes acceleration. Velocity comes about from that acceleration occuring over a period of time: . Just keep in mind that the acceleration can be in the same direction as velocity (i.e. the object is accelerating, such as when you drop a rock off a cliff, or hit the gas when driving down the street) or in the opposite direction (the object decelerates, as when you throw a rock upward or when you hit the brakes on your car).