Answer:
The rate of decrease is: [tex]43.2mm^3/min[/tex]
Step-by-step explanation:
Given
[tex]l = 18mm[/tex]
[tex]\frac{dl}{dt} = -0.8mm/min[/tex] ---- We used minus because the rate is decreasing
Required
Rate of decrease when: [tex]l = 18mm[/tex]
The volume of the cube is:
[tex]V = l^3[/tex]
Differentiate
[tex]\frac{dV}{dl} = 3l^2[/tex]
Make dV the subject
[tex]dV = 3l^2 \cdot dl[/tex]
Divide both sides by dt
[tex]\frac{dV}{dt} = 3l^2 \cdot \frac{dl}{dt}[/tex]
Given that: [tex]l = 18mm[/tex] and [tex]\frac{dl}{dt} = -0.8mm/min[/tex]
[tex]\frac{dV}{dt} = 3 * (18mm)^2 * (-0.8mm/min)[/tex]
[tex]\frac{dV}{dt} = 3 * 18 *-0.8mm^3/min[/tex]
[tex]\frac{dV}{dt} = -43.2mm^3/min[/tex]
Hence, the rate of decrease is: 43.2mm^3/min
