ULTRA easy maximun/minumum question i dun get.. (1 Viewer)

[kooltrainer]

let say you're given the differential equation

dV/dt = -120 +2t

thats the rate at which the volume decreases .. and you're asked to find the fastest rate at which the volume decreases ..
so i differentiate it again ..

d2V/dt2 = 2 ..

so fastest rate should be at 2 second.. however, if u look at the dV/dt equation, its obvious that the fastest rate is when t = 0 and not t = 2 ... so why doesnt differentiating it again work?

 

[lyounamu]

let say you're given the differential equation

dV/dt = -120 +2t

thats the rate at which the volume decreases .. and you're asked to find the fastest rate at which the volume decreases ..
so i differentiate it again ..

d2V/dt2 = 2 ..

so fastest rate should be at 2 second.. however, if u look at the dV/dt equation, its obvious that the fastest rate is when t = 0 and not t = 2 ... so why doesnt differentiating it again work?

If you differentiate again, you are only finding the acceleration. You have to find the rate in relation to volume. So it should be dV/dt/dV not d^2V/dt^2.

 

[lolokay]

the fastest rate according to your second derivative is NOT 2 seconds. It implies that there is never a fastest rate.

This is mathematically true because if time were negative, then the rate of volume decrease would be greater than at t=0. In this case, you find the time of fastest decrease not be using the second derivative, but by using logic.

 

[Iruka]

I think the problem is that there is an implied domain for this function (t>=0), and calculus only works on the interior of an interval, not at the end points.

When you are using calculus to solve max/min problems, you always need to check the end points separately. (Sometimes you will have to infer the meaningful domain of your function from the constraints given in the question.)

You can have a maximum or minimum occurring at an end point but differentiation will not find it for you unless it also happens to be a stationary point of the curve.

 

[kooltrainer]

I think the problem is that there is an implied domain for this function (t>=0), and calculus only works on the interior of an interval, not at the end points.

When you are using calculus to solve max/min problems, you always need to check the end points separately. (Sometimes you will have to infer the meaningful domain of your function from the constraints given in the question.)

You can have a maximum or minimum occurring at an end point but differentiation will not find it for you unless it also happens to be a stationary point of the curve.

oh,so differention of linear funtions won't help you solve max/min problems?..
what if that differentiation equation was a very complicated quadratic .. would d2V/dt2 work to find the greatest change of rate?

 

[Iruka]

No, I am not saying that differentiation wont work.

I am saying that when you have a restricted domain for your function, you have to check the end points of the domain directly by substitution.

The difficulty in this case is that the restriction on the domain is only implied, not overtly stated in the question.

 

[lolokay]

oh,so differention of linear funtions won't help you solve max/min problems?..
what if that differentiation equation was a very complicated quadratic .. would d2V/dt2 work to find the greatest change of rate?

it won't help when differentiating a linear function because there is no max/min rate of change - so instead you find whether the least or greatest possible value as implied by the question has the greater gradient (t = initial or final)

if you're finding thg greatest rate of change in a non-linear function, then the second derivative will tell you then the max/min occurs - but you have to make sure that the time it occurs is a possible value, eg not a negative time or whatever else

 

[kooltrainer]

it won't help when differentiating a linear function because there is no max/min rate of change - so instead you find whether the least or greatest possible value as implied by the question has the greater gradient (t = initial or final)

if you're finding thg greatest rate of change in a non-linear function, then the second derivative will tell you then the max/min occurs - but you have to make sure that the time it occurs is a possible value, eg not a negative time or whatever else

let say dV/dt = 2t^2 + 77 t + 3
and u want to find the greatest change in volume..

can u do d2V/dt2 in this case to find the greatest value?
wouldnt that mean acceleration though?

 

[Iruka]

No, its not acceleration.

Acceleration is the second derivative of displacement with respect to time.

 

[lolokay]

let say dV/dt = 2t^2 + 77 t + 3
and u want to find the greatest change in volume..

can u do d2V/dt2 in this case to find the greatest value?
wouldnt that mean acceleration though?

yeah d2v/dt2 is the 'acceleration' (rate of change of rate of change in volume) - and this is what is used to find the greatest 'velocity' (change in volume)

to find the greatest (or least/stationary) value of something you find when it's change is zero - so you will be looking at acceleration for velocity etc