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Question 987649: Suppose the side length of a cube is x = 11 cm. If x changes by an amount δx = 0.01 cm, what is the corresponding approximate change in the volume of the cube δV?
THANK YOU :-)
Found 2 solutions by jim_thompson5910, solver91311:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! V = x^3 ... start with the volume of a cube formula
dV/dx = 3x^2 ... apply the derivative with respect to x
dV = 3x^2*dx
dV = 3*11^2*0.01 ... plug in x = 11 and dx = 0.01
dV = 3.63
If the change in x is 0.01 cm, then the approximate change in volume is 3.63 cubic cm
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Alternative non-calculus based way to do it
Original Volume:
V = x^3
V = 11^3
V = 1,331
Say the side length is x=11+0.01 = 11.01 now. That makes the volume become
V = 11.01^3
V = 1,334.633301
The difference in the two volumes is
1,334.633301 - 1,331 = 3.63330099999984 which is approximately 3.63 that we got before
The same applies if you did x = 11-0.01 = 10.99 (the only real difference is that the result of the subtraction would be -3.63, but the absolute value of that leads to the same result)
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
"...changes by an amount..." is non-specific. Did the measure of the edge of the cube get larger or smaller? Either one is a change, and the amount specified is a positive number does not alter the ambiguity. "approximate" without a specification of required precision is non-specific. "A little bit bigger (or smaller)" is an unassailably correct answer to this question as posed.
Be that as it may, the volume of a cube with edges that measure  units has a volume of  cubic units. If the measure of the edge is increased by some value  , then the new volume is ^3\ =\ x^3\ +\ 3x^2h\ +\ 3xh^2\ +\ h^3) , and the difference between the original volume and the new volume is  . Similarly, if the edge is decreased some value , then the new volume is ^3\ =\ x^3\ -\ 3x^2h\ +\ 3xh^2\ -\ h^3) , and the difference between the original volume and the new volume is  .
Once you decide whether you are getting larger or smaller, you can plug your values of  and  into the appropriate expression for  , do the arithmetic, and round to whatever precision you need or desire.
John
My calculator said it, I believe it, that settles it
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