PHYSICS HIGH SCHOOL

Volume of a Cube The volume V of a cube with sides of length x in. is changing with respect to time. At a certain instant of time, the sides of the cube are 7 in. long and increasing at the rate of 0.2 in./s. How fast is the volume of the cube changing (in cu in/s) at that instant of time?

Answers

Answer 1

Answer:

The volume of the cube is changing at the rate of 29.4 in³/s.

Time derivative:

The rate of change of the sides of the cube is 0.2 in/s, which can be mathematically represented as :

$(dx)/(dt)=0.2 \;in/s$

Now the volume of the cube is given by:

V = x³

If we take the time derivative of the above equation then it gives the rate of change of volume with time, so:

$(dV)/(dt)= (d)/(dt)x^3\n\n (dV)/(dt)= 3x^2(dx)/(dt)$

At the instant x = 7 in, the rate of change of volume will be:

$(dV)/(dt)= 3*7^2*0.2\;in^3/s\n\n(dV)/(dt)=29.4\;in^3/s$

So the volume is changing at 29.4 cubic inches per second.

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Answer 2

Answer:

Answer:

Therefore the volume of cube is change at the 29.4 cube in./s at that instant time.

Explanation:

Formula

$(dx^n)/(dx) =nx^(n-1)$

Cube :

The volume of a cube is = $side^3$

The side of length is x in.

Then volume of the cube is (V) = $x^3$

∴ V = $x^3$

Differentiate with respect to t

$(d)/(dt)(V)=(d)/(dt) (x^3)$

$\Rightarrow (dV)/(dt) =3x^2(dx)/(dt)$ ....(1)

Given that the side of the cube is increasing at the rate of 0.2 in/s.

i.e $(dx)/(dt) = 0.2$ in/s.

And the sides of the cube are 7 in i.e x= 7 in

Putting $(dx)/(dt) = 0.2$ and x= 7 in equation (1)

$\therefore (dV)/(dt) =3 * 7^2 * 0.2$ cube in./s

=29.4 cube in./s

Therefore the volume of cube is change at the 29.4 cube in./s at that instant time.

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Final answer:

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