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Second partial derivatives - what does it actually mean?

I understand how you take partial derivatives but I'm really confused about what the second partial derivative (one derivative for each variable) actually tells you. I don't think that's very clear so I'll give an example (volume of a cylinder).

V(r,h)=πr2hV(r,h)=\pi r^2h
∂V∂r)h=2πrh\frac{\partial V}{\partial r})_h = 2\pi rh
∂2V∂r∂h=2πr\frac{\partial^2 V}{\partial r \partial h} = 2\pi r

∂V∂r)h\frac{\partial V}{\partial r})_h tells us the rate of change of volume with respect to radius when we keep h constant but what does ∂2V∂r∂h\frac{\partial^2 V}{\partial r \partial h} mean in this context? Thanks!
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Avatar for TeeEm
TeeEm
19
Original post by Plagioclase
I understand how you take partial derivatives but I'm really confused about what the second partial derivative (one derivative for each variable) actually tells you. I don't think that's very clear so I'll give an example (volume of a cylinder).

V(r,h)=πr2hV(r,h)=\pi r^2h
∂V∂r)h=2πrh\frac{\partial V}{\partial r})_h = 2\pi rh
∂2V∂r∂h=2πr\frac{\partial^2 V}{\partial r \partial h} = 2\pi r

∂V∂r)h\frac{\partial V}{\partial r})_h tells us the rate of change of volume with respect to radius when we keep h constant but what does ∂2V∂r∂h\frac{\partial^2 V}{\partial r \partial h} mean in this context? Thanks!


the rate of change with respect to the radius, of the rate of change of the volume with respect to the height.


or for nice functions as these commute

the rate of change with respect to height, of the rate of change of the volume with respect to the radius.


how quickly the gradient of V in the r direction changes as we move in the h direction if we treat this as a "surface"
(edited 8 years ago)
Original post by TeeEm
the rate of change with respect to the radius, of the rate of change of the volume with respect to the height.


or for nice functions as these commute

the rate of change with respect to height, of the rate of change of the volume with respect to the radius.


how quickly the gradient of V in the r direction changes as we move in the h direction in treat this as a "surface"


Okay so I've visualised the surface here (r=x, h=y). The first partial derivative (with respect to x) gives us the gradient function for constant y. So the second partial derivative tells us how quickly the gradient of the surface for constant y changes as we change y? That looks like it makes sense to me, the second partial derivative would be negative for x=-1 and indeed the gradient for x=-1 decreases as you increase y and the opposite for x=1.
Reply 3
Avatar for TeeEm
TeeEm
19
Original post by Plagioclase
Okay so I've visualised the surface here (r=x, h=y). The first partial derivative (with respect to x) gives us the gradient function for constant y. So the second partial derivative tells us how quickly the gradient of the surface for constant y changes as we change y? That looks like it makes sense to me, the second partial derivative would be negative for x=-1 and indeed the gradient for x=-1 decreases as you increase y and the opposite for x=1.


You have good attitude trying to look for an explanation for things, something which I lacked when I was at University.
(or perhaps the "I need to understand list" grew too long and I simply gave up.)

Sometimes I wonder whether most people at this level truly understand things or simply got used to them.

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