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All edges of a cube are expanding at a rate of 2 centimeters per second.
(a) How fast is the volume changing when each edge is 2 centimeter(s)?
(b) How fast is the volume changing when each edge is 14 centimeter(s)?​

All edges of a cube are expanding at a rate of 2 centimeters per seconda How fast is the volume changing when each edge is 2 centimetersb How fast is the volume class=

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absor201

Answer:

  • Part a → dv/dt = 24 cm³/per second
  • Part b →  dv/dt = 1176 cm³/per second

Step-by-step explanation:

As all edges of a cube are expanding at a rate of 2 centimeters per second.

i.e ds/dt = 2 cm/s

The formula of volume of cube: V = s³

Differentiating the volume:

dv/dt = 3s²

Write ds/dt following derivative of volume

dv/dt = 3s² ds/dt

plug in the value: ds/dt = 2 cm/s

dv/dt = (3s²) (2)

dv/dt = 6s²

Given the edge is 2cm.

so plug in the value: s=2

dv/dt = 6s²

          = 6(2)²=6(4)=24 cm³/per second

SIMILARLY,

When the edge is 14cm

dv/dt = 6s²

plug in the value: s=14

dv/dt = 6s²

         = 6(14)²=6(196)=1176 cm³/per second

Thus,

  • Part a → dv/dt = 24 cm³/per second
  • Part b →  dv/dt = 1176 cm³/per second
Cricetus

The volume will be:

(a) 24 cm³

(b) 1176 cm³

Given:

Expanding rate,

  • 2 cm/sec

Let,

  • The edge of a cube be "x cm".

(a)

When,

  • x = 2 cm

then,

→ [tex]\frac{dx}{dt} = 2 \ cm/sec[/tex]

Let,

  • The volume of cube be "V"

→ [tex]V = x^3[/tex]

→ [tex]\frac{dV}{dt} = 3x^2 \frac{dx}{dt}[/tex]

        [tex]= 3\times 2^2\times 2[/tex]

        [tex]= 3\times 4\times 2[/tex]

        [tex]= 24 \ cm^3[/tex]

(b)

When,

  • x = 14 cm

then,

→ [tex]\frac{dx}{dt} = 2 \ cm/sec[/tex]

Let,

  • The volume of cube be "V"

→ [tex]V = x^3[/tex]

→ [tex]\frac{dV}{dt} = 3x^2 \frac{dx}{dt}[/tex]

        [tex]=3\times 14^2\times 2[/tex]

        [tex]= 1176 \ cm^3[/tex]

Thus the above answer is correct.

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