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Need help with Rates involving two or more variables (1 Viewer)

kevinsta

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Here is the question:
If the volume of a cube is
increasing at the rate of 23mm^3/second
find the increase in its surface area
when its side is 140 mm.
 

aDimitri

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You need to get surface area in terms of volume so that you can find dA/dV
I'd rather not spoon feed, tell me if you still can't get it :)
 
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Let the sides of the cube be x, therefore volume = x^3

dv/dt = 23.

dv/dt = dv/dx x dx/dt

dv/dx = 3x^2

23/3x^2 = dx/dt (sub x = 140)

dx/dt = 23/58800

The question is asking to find ds/dt (S = surface area)
ds/dt = ds/dx x dx/dt

ds/dt = 12x (multiplied) by 23/58800

ds/dt = 12 x 140 x 23/58800
= 23/35
= 0.66mm^2/second (2 decimal place round off)
 

dunjaaa

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Let's call the side length of the cube 'x'
We are given: dV/dt = 23 & we seek to find: dA/dt
Generally what I do is express the Surface Area(A), Volume(V) in terms of 'x'
So A=6x^2 and V=x^3
To find dA/dt we need to somehow incorporate dV/dt via chain rule.
The way we can do that is by using the aforementioned expressions for surface area and volume.
dA/dt = (dA/dx) x (dx/dV) x (dV/dt) (Notice how I set my chains up so that the dx's and dV's will cancel. It takes practice, but it's very simple once you get your head around a few similar types of problems.)
dA/dt = (12x) x (1/3x^2) x 23 = 92/x
When x=140, dA/dt = 23/35 mm^2/second (Be careful of the units! Fortunately, everything is in mm so no conversion is needed)
 
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kevinsta

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Let the sides of the cube be x, therefore volume = x^3

dv/dt = 23.

dv/dt = dv/dx x dx/dt

dv/dx = 3x^2

23/3x^2 = dx/dt (sub x = 140)

dx/dt = 23/58800

The question is asking to find ds/dt (S = surface area)
ds/dt = ds/dx x dx/dt

ds/dt = 12x (multiplied) by 23/58800

ds/dt = 12 x 140 x 23/58800
= 23/35
= 0.66mm^2/second (2 decimal place round off)
Oh thanks i got up to finding dx/dt i didn't consider having to use another formula to find the answer.
 

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