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Math Help - explaining what the second derivative test gives you in simple English

  1. #1
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    explaining what the second derivative test gives you in simple English

    Hi all,

    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.
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    Re: explaining what the second derivative test gives you in simple English

    Quote Originally Posted by hellothisismyname View Post
    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.
    Correct - and so if the rate of change of surface area is equal to zero you must be at a local minimum or amximum value.

    Quote Originally Posted by hellothisismyname View Post
    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.
    Yes, but it might be easier to think of the 2nd derivative as the rate of acceleration. If the second derivative is positive this means the surface area is accelerating with increasing value of radius. If negative, this means the area is decelerating with increasing radius. If zero this means the area is changing at a constant rate.

    Quote Originally Posted by hellothisismyname View Post
    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).
    Not quite. If the 2nd derivative is at a minimum and is negative that means it's rate if deceleration has bottomed out and is starting to decelerate less quickly. If it's at a minimum and positvie that means its acceleration is getting bigger.

    Quote Originally Posted by hellothisismyname View Post
    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.
    It doesn't really. In minima/maxima problems you typically look for the point where the 1st derivative is zero and then if the 2nd derivative is positive you know you have a local minimum and if the 2nd derivatie is negative it's a local maximum.
    Last edited by ebaines; May 30th 2013 at 10:20 AM.
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    Re: explaining what the second derivative test gives you in simple English

    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 d^2f/dx^2 is positive, df/dx is increasing and if d^2f/dx^2 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 d^2f/dx^2(a) 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 d^2f/dx^2(a) must be positive.

    That is why " d^2f/dx^2< 0 at a maximum and d^2f/dx^2> 0 at a minimum".
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    Re: explaining what the second derivative test gives you in simple English

    Quote Originally Posted by ebaines View Post

    Yes, but it might be easier to think of the 2nd derivative as the rate of acceleration. If the second derivative is positive this means the surface area is accelerating with increasing value of radius. If negative, this means the area is decelerating with increasing radius. If zero this means the area is changing at a constant rate.

    Not quite. If the 2nd derivative is at a minimum and is negative that means it's rate if deceleration has bottomed out and is starting to decelerate less quickly. If it's at a minimum and positvie that means its acceleration is getting bigger.

    It doesn't really. In minima/maxima problems you typically look for the point where the 1st derivative is zero and then if the 2nd derivative is positive you know you have a local minimum and if the 2nd derivatie is negative it's a local maximum.
    Thanks this is very helpful. What I still don't understand is how can their even be a second derivative if the first derivative is equal to 0. If the first derivative is zero, then their is no rate of change, so how can a rate of the rate of change exist?
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    Re: explaining what the second derivative test gives you in simple English

    Quote Originally Posted by hellothisismyname View Post
    Thanks this is very helpful. What I still don't understand is how can their even be a second derivative if the first derivative is equal to 0. If the first derivative is zero, then their is no rate of change, so how can a rate of the rate of change exist?
    You are confusing "equals 0" with "does not exist". A car, even with speed 0, can have non-zero acceleration else it would never be able to move!
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    Re: explaining what the second derivative test gives you in simple English

    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.
    Last edited by ebaines; May 30th 2013 at 11:38 AM.
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    Re: explaining what the second derivative test gives you in simple English

    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?
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    Re: explaining what the second derivative test gives you in simple English

    The second derivative is acceleration, and from basic physics  \vec F = m \vec a, so an external force (like gravity) causes acceleration. Velocity comes about from that acceleration occuring over a period of time: \normalsize \Delta \vec v(t) = \int \vec a dt. 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).
    Last edited by ebaines; May 30th 2013 at 12:03 PM.
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