contestada

A car is being compacted into a rectangular solid. The volume is decreasing at a rate of 2 m3/sec. The length and width of the compactor are square, but the height is not the same length as the length and width. If the length and width walls move toward each other at a rate of 0.25 m/ sec, find the rate at which the height is changing when the length and width are 2 m and the height is 1.5 m.

Respuesta :

Using implicit differentiation, it is found that the height is changing at a rate of -0.875 m/sec.

The volume of a rectangular solid of length l, width w and height h is given by:

[tex]V = lwh[/tex]

Applying implicit differentiation, the rate of change of the volume is given by:

[tex]\frac{dV}{dt} = wh\frac{dl}{dt} + lh\frac{dw}{dt} + lw\frac{dh}{dt}[/tex]

In this problem:

  • Volume is decreasing at a rate of 2 m3/sec, hence [tex]\frac{dV}{dt} = -2[/tex]
  • The length and width walls move toward each other at a rate of 0.25 m/ sec, hence [tex]\frac{dl}{dt} = \frac{dw}{dt} = 0.25[/tex]
  • Length and width are 2 m and the height is 1.5 m, hence [tex]l = w = 2, h = 1.5[/tex]

Then:

[tex]\frac{dV}{dt} = wh\frac{dl}{dt} + lh\frac{dw}{dt} + lw\frac{dh}{dt}[/tex]

[tex]-2  = 2(1.5)(0.25) + 2(1.5)(0.25) + 2(2)\frac{dh}{dt}[/tex]

[tex]4\frac{dh}{dt} = -3.5[/tex]

[tex]\frac{dh}{dt} = -\frac{3.5}{4}[/tex]

[tex]\frac{dh}{dt} = -0.875[/tex]

The height is changing at a rate of -0.875 m/sec.

A similar problem is given at https://brainly.com/question/9543179