Related rates of change (1 Viewer)
- Thread starter xibu34
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let s be the side length of the cube and V be the volume of the cube
V = s^3
dV/ds = 3s^2
ds/dt = 0.12
dV/dt = dV/ds x ds/dt = 3s^2 x 0.12 = 0.36s^2
dV/dt when s=150 --> 0.36(150)^2 = 8100mm^3/s
5uckerberg
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@xibu34
In this question, we are told that the side length is increasing at 0.12mm/s. and that the side is 150mm.
To start off we can see that
noting that s represents the side and t represents the time.
Now, what is the volume of a cube?
, and 
and we are trying to find 
so now using the chain rule
. Here we can say 
giving us 
In this question, we are told that the side length is increasing at 0.12mm/s. and that the side is 150mm.
To start off we can see that
Now, what is the volume of a cube?
chilli 412
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here's how i remember how to do these ones
1. what are we trying to find? (here we want to find the increase (rate of change) in a cube's volume when it's side length is 150mm - this is the same as dV/dt)
1. what differentials are we given? (here we are given the derivative of the side length with respect to time - ds/dt = 0.12
2. what functions (equations) can we use to find another differential? (we're looking at a cube, and the volume of a cube can be given by Volume = (the side length)^3, or just V=s^3
with this equation V = s^3, you can find the derivative of V with respect to s: dV/ds = 3s^2
so know we know dV/ds, and we know ds/dt, and we are trying to find dV/dt
we can multiply our two known differentials together, and the 'ds' will cancel from both and we will be left with dV/ds
so therefore dV/ds is equal to (dV/ds)*(ds/dt)
so now we just replace our two differentials, because we have expressions for them
dV/ds = (3s^2)*(0.12)
= 0.36s^2
know we want to know this rate of increase when s = 150mm, so we just plug in 150mm
rate of increase of the volume = 0.36(150)^2
= 8100mm^3/s , or you could write 8100mm^3s^-1, always remember the units of measurement
1. what are we trying to find? (here we want to find the increase (rate of change) in a cube's volume when it's side length is 150mm - this is the same as dV/dt)
1. what differentials are we given? (here we are given the derivative of the side length with respect to time - ds/dt = 0.12
2. what functions (equations) can we use to find another differential? (we're looking at a cube, and the volume of a cube can be given by Volume = (the side length)^3, or just V=s^3
with this equation V = s^3, you can find the derivative of V with respect to s: dV/ds = 3s^2
so know we know dV/ds, and we know ds/dt, and we are trying to find dV/dt
we can multiply our two known differentials together, and the 'ds' will cancel from both and we will be left with dV/ds
so therefore dV/ds is equal to (dV/ds)*(ds/dt)
so now we just replace our two differentials, because we have expressions for them
dV/ds = (3s^2)*(0.12)
= 0.36s^2
know we want to know this rate of increase when s = 150mm, so we just plug in 150mm
rate of increase of the volume = 0.36(150)^2
= 8100mm^3/s , or you could write 8100mm^3s^-1, always remember the units of measurement